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CAPITULO XXVIII

Problemas sobre ecuaciones simultaneas
Ejercicio 208
1. $\begin{array}{cc}{x}^{2}–2x+1& ={\left(\left[{x}^{2}–2x\right]+1\right)}^{2}\\ & ={\left({x}^{2}–2x\right)}^{2}+2\left({x}^{2}–2x\right)+1\\ & ={x}^{4}–4{x}^{3}+4{x}^{2}+2{x}^{2}–4x+1\\ & ={x}^{4}–4{x}^{3}+6{x}^{2}–4x+1\end{array}$
2. $\begin{array}{cc}2{x}^{2}+x+1& ={\left(\left[2{x}^{2}+x\right]+1\right)}^{2}\\ & ={\left(2{x}^{2}+x\right)}^{2}+2\left(2{x}^{2}+x\right)+1\\ & =4{x}^{4}+4{x}^{3}+{x}^{2}+4{x}^{2}+2x+1\\ & =4{x}^{4}+4{x}^{3}+5{x}^{2}+2x+1\end{array}$
3. $\begin{array}{cc}{x}^{2}–5x+2& ={\left(\left[{x}^{2}–5x\right]+2\right)}^{2}\\ & ={\left({x}^{2}–5x\right)}^{2}+4\left({x}^{2}–5x\right)+{2}^{2}\\ & ={x}^{4}–10{x}^{3}+25{x}^{2}+4{x}^{2}–20x+4\\ & ={x}^{4}–10{x}^{3}+29{x}^{2}–20x+4\end{array}$
4. $\begin{array}{cc}{x}^{3}–5{x}^{2}+6& ={\left(\left[{x}^{3}–5{x}^{2}\right]+6\right)}^{2}\\ & ={\left({x}^{3}–5{x}^{2}\right)}^{2}+12\left({x}^{3}–5{x}^{2}\right)+{6}^{2}\\ & ={x}^{6}–10{x}^{5}+25{x}^{4}+12{x}^{3}–60{x}^{2}+36\end{array}$
5. $\begin{array}{cc}4{a}^{4}–3{a}^{2}+5& ={\left(\left[4{a}^{4}–3{a}^{2}\right]+5\right)}^{2}\\ & ={\left(4{a}^{4}–3{a}^{2}\right)}^{2}+10\left(4{a}^{4}–3{a}^{2}\right)+{5}^{2}\\ & =16{a}^{8}–24{a}^{6}+9{a}^{4}+40{a}^{4}–30{a}^{2}+25\\ & =16{a}^{8}–24{a}^{6}+49{a}^{4}–30{a}^{2}+25\end{array}$
6. $\begin{array}{cc}x+2y–z& ={\left(\left[x+2y\right]–z\right)}^{2}\\ & ={\left(x+2y\right)}^{2}–2z\left(x+2y\right)+{z}^{2}\\ & ={x}^{2}+4xy+4{y}^{2}–2xz–4yz+{z}^{2}\end{array}$
7. $\begin{array}{cc}3–{x}^{3}–{x}^{6}& ={\left(\left[3–{x}^{3}\right]–{x}^{6}\right)}^{2}\\ & ={\left(3–{x}^{3}\right)}^{2}–2{x}^{6}\left(3–{x}^{3}\right)+{x}^{12}\\ & =9–6{x}^{3}+{x}^{6}–6{x}^{6}+2{x}^{9}+{x}^{12}\\ & =9–6{x}^{3}–5{x}^{6}+2{x}^{9}+{x}^{12}\end{array}$
8. $\begin{array}{cc}5{x}^{4}–7{x}^{2}+3x& ={\left(\left[5{x}^{4}–7{x}^{2}\right]+3x\right)}^{2}\\ & ={\left(5{x}^{4}–7{x}^{2}\right)}^{2}+6x\left(5{x}^{4}–7{x}^{2}\right)+{\left(3x\right)}^{2}\\ & =25{x}^{8}–70{x}^{6}+49{x}^{4}+30{x}^{5}–42{x}^{3}+9{x}^{2}\\ & =25{x}^{8}–70{x}^{6}+30{x}^{5}+49{x}^{4}–42{x}^{3}+9{x}^{2}\end{array}$
9. $\begin{array}{cc}2{a}^{2}+2ab–3{b}^{2}& ={\left(\left[2{a}^{2}+2ab\right]–3{b}^{2}\right)}^{2}\\ & ={\left(2{a}^{2}+2ab\right)}^{2}–6{b}^{2}\left(2{a}^{2}+2ab\right)+{\left(–3{b}^{2}\right)}^{2}\\ & =4{a}^{4}+8{a}^{3}b+4{a}^{2}{b}^{2}–12{a}^{2}{b}^{2}–12a{b}^{3}+9{b}^{4}\\ & =4{a}^{4}+8{a}^{3}b–8{a}^{2}{b}^{2}–12a{b}^{3}+9{b}^{4}\end{array}$
10. $\begin{array}{cc}{m}^{3}–2{m}^{2}n+2{n}^{4}& ={\left(\left[{m}^{3}–2{m}^{2}n\right]+2{n}^{4}\right)}^{2}\\ & ={\left({m}^{3}–2{m}^{2}n\right)}^{2}+4{n}^{4}\left({m}^{3}–2{m}^{2}n\right)+{\left(2{n}^{4}\right)}^{2}\\ & ={m}^{6}–4{m}^{5}n+4{m}^{4}{n}^{2}+4{m}^{3}{n}^{4}–8{m}^{2}{n}^{5}+4{n}^{8}\end{array}$
11. $\begin{array}{cc}\frac{a}{2}–b+\frac{c}{4}& ={\left(\left[\frac{a}{2}–b\right]+\frac{c}{4}\right)}^{2}\\ & ={\left(\frac{a}{2}–b\right)}^{2}+\frac{c}{2}\left(\frac{a}{2}–b\right)+{\left(\frac{c}{4}\right)}^{2}\\ & =\frac{{a}^{2}}{4}–ab+{b}^{2}+\frac{ac}{4}–\frac{bc}{2}+\frac{{c}^{2}}{16}\\ & =\frac{{a}^{2}}{4}+{b}^{2}+\frac{{c}^{2}}{16}–ab+\frac{ac}{4}–\frac{bc}{2}\end{array}$
12. $\begin{array}{cc}\frac{x}{5}–5y+\frac{5}{3}& ={\left(\left[\frac{x}{5}–5y\right]+\frac{5}{3}\right)}^{2}\\ & ={\left(\frac{x}{5}–5y\right)}^{2}+\frac{10}{3}\left(\frac{x}{5}–5y\right)+{\left(\frac{5}{3}\right)}^{2}\\ & =\frac{{x}^{2}}{25}–2xy+25{y}^{2}+\frac{2}{3}x–\frac{50}{3}y+\frac{25}{9}\end{array}$
13. $\begin{array}{cc}\frac{1}{2}{x}^{2}–x+\frac{2}{3}& ={\left(\left[\frac{1}{2}{x}^{2}–x\right]+\frac{2}{3}\right)}^{2}\\ & ={\left(\frac{1}{2}{x}^{2}–x\right)}^{2}+\frac{4}{3}\left(\frac{1}{2}{x}^{2}–x\right)+{\left(\frac{2}{3}\right)}^{2}\\ & =\frac{{x}^{4}}{4}–{x}^{3}+{x}^{2}+\frac{2}{3}{x}^{2}–\frac{4}{3}x+\frac{4}{9}\\ & =\frac{{x}^{4}}{4}–{x}^{3}+\frac{5}{3}{x}^{2}–\frac{4}{3}x+\frac{4}{9}\end{array}$
14. $\begin{array}{cc}\frac{a}{x}–\frac{1}{3}+\frac{x}{a}& ={\left(\left[\frac{a}{x}–\frac{1}{3}\right]+\frac{x}{a}\right)}^{2}\\ & ={\left(\frac{a}{x}–\frac{1}{3}\right)}^{2}+\frac{2x}{a}\left(\frac{a}{x}–\frac{1}{3}\right)+{\left(\frac{x}{a}\right)}^{2}\\ & =\frac{{a}^{2}}{{x}^{2}}–\frac{2a}{3x}+\frac{1}{9}+2–\frac{2x}{3a}+\frac{{x}^{2}}{{a}^{2}}\\ & =\frac{{a}^{2}}{{x}^{2}}–\frac{2a}{3x}–\frac{2x}{3a}+\frac{{x}^{2}}{{a}^{2}}+\frac{19}{9}\end{array}$
15. $\begin{array}{cc}\frac{3}{4}{a}^{2}–\frac{1}{2}a+\frac{4}{5}& ={\left(\left[\frac{3}{4}{a}^{2}–\frac{1}{2}a\right]+\frac{4}{5}\right)}^{2}\\ & ={\left(\frac{3}{4}{a}^{2}–\frac{1}{2}a\right)}^{2}+\frac{8}{3}\left(\frac{3}{4}{a}^{2}–\frac{1}{2}a\right)+{\left(\frac{4}{5}\right)}^{2}\\ & =\frac{9}{16}{a}^{4}–\frac{3}{4}{a}^{3}+\frac{1}{4}{a}^{2}+2{a}^{2}–\frac{4}{3}a+\frac{16}{25}\\ & =\frac{9}{16}{a}^{4}–\frac{3}{4}{a}^{3}+\frac{29}{20}{a}^{2}–\frac{4}{3}a+\frac{16}{25}\end{array}$
16. $\begin{array}{cc}\frac{{a}^{2}}{4}–\frac{3}{5}+\frac{{b}^{2}}{9}& ={\left(\left[\frac{{a}^{2}}{4}–\frac{3}{5}\right]+\frac{{b}^{2}}{9}\right)}^{2}\\ & ={\left(\frac{{a}^{2}}{4}–\frac{3}{5}\right)}^{2}+\frac{2{b}^{2}}{9}\left(\frac{{a}^{2}}{4}–\frac{3}{5}\right)+{\left(\frac{{b}^{2}}{9}\right)}^{2}\\ & =\frac{{a}^{4}}{16}–\frac{3}{10}{a}^{2}+\frac{9}{25}+\frac{{a}^{2}{b}^{2}}{18}–\frac{2{b}^{2}}{15}+\frac{{b}^{4}}{81}\\ & =\frac{{a}^{4}}{16}–\frac{3}{10}{a}^{2}+\frac{{a}^{2}{b}^{2}}{18}–\frac{2{b}^{2}}{15}+\frac{{b}^{4}}{81}+\frac{9}{25}\end{array}$
17. $\begin{array}{cc}{x}^{3}–{x}^{2}+x+1& ={\left(\left[{x}^{3}–{x}^{2}\right]+\left[x+1\right]\right)}^{2}\\ & ={\left({x}^{3}–{x}^{2}\right)}^{2}+2\left({x}^{3}–{x}^{2}\right)\left(x+1\right)+{\left(x+1\right)}^{2}\\ & ={x}^{6}–2{x}^{5}+{x}^{4}+2\left({x}^{4}+\overline{){x}^{3}}–\overline{){x}^{3}}–{x}^{2}\right)+{x}^{2}+2x+1\\ & ={x}^{6}–2{x}^{5}+{x}^{4}+2{x}^{4}–2{x}^{2}+{x}^{2}+2x+1\\ & ={x}^{6}–2{x}^{5}+3{x}^{4}–{x}^{2}+2x+1\end{array}$
18. $\begin{array}{cc}{x}^{3}–3{x}^{2}–2x+2& ={\left(\left[{x}^{3}–3{x}^{2}\right]–\left[2x–2\right]\right)}^{2}\\ & ={\left({x}^{3}–3{x}^{2}\right)}^{2}–2\left({x}^{3}–3{x}^{2}\right)\left(2x–2\right)+{\left(2x–2\right)}^{2}\\ & ={x}^{6}–6{x}^{5}+9{x}^{4}–2\left(2{x}^{4}–2{x}^{3}–6{x}^{3}+6{x}^{2}\right)+4{x}^{2}–8x+4\\ & ={x}^{6}–6{x}^{5}+9{x}^{4}–4{x}^{4}+16{x}^{3}–12{x}^{2}+4{x}^{2}–8x+4\\ & ={x}^{6}–6{x}^{5}+5{x}^{4}+16{x}^{3}–8{x}^{2}–8x+4\end{array}$
19. $\begin{array}{cc}{x}^{4}+3{x}^{2}–4x+5& ={\left(\left[{x}^{4}+3{x}^{2}\right]–\left[4x–5\right]\right)}^{2}\\ & ={\left({x}^{4}+3{x}^{2}\right)}^{2}–2\left({x}^{4}+3{x}^{2}\right)\left(4x–5\right)+{\left(4x–5\right)}^{2}\\ & ={x}^{8}+6{x}^{6}+9{x}^{4}–2\left(4{x}^{5}–5{x}^{4}+12{x}^{3}–15{x}^{2}\right)+16{x}^{2}–40x+25\\ & ={x}^{8}+6{x}^{6}+9{x}^{4}–8{x}^{5}+10{x}^{4}–24{x}^{3}+30{x}^{2}+16{x}^{2}–40x+25\\ & ={x}^{8}+6{x}^{6}–8{x}^{5}+19{x}^{4}–24{x}^{3}+46{x}^{2}–40x+25\end{array}$
20. $\begin{array}{cc}{x}^{4}–4{x}^{3}+2x–3& ={\left(\left[{x}^{4}–4{x}^{3}\right]+\left[2x–3\right]\right)}^{2}\\ & ={\left({x}^{4}–4{x}^{3}\right)}^{2}+2\left({x}^{4}–4{x}^{3}\right)\left(2x–3\right)+{\left(2x–3\right)}^{2}\\ & ={x}^{8}–8{x}^{7}+16{x}^{6}+2\left(2{x}^{5}–3{x}^{4}–8{x}^{4}+12{x}^{3}\right)+4{x}^{2}–12x+9\\ & ={x}^{8}–8{x}^{7}+16{x}^{6}+4{x}^{5}–22{x}^{4}+24{x}^{3}+4{x}^{2}–12x+9\end{array}$
21. $\begin{array}{cc}3–6a+{a}^{2}–{a}^{3}& ={\left(\left[3–6a\right]+\left[{a}^{2}–{a}^{3}\right]\right)}^{2}\\ & ={\left(3–6a\right)}^{2}+2\left(3–6a\right)\left({a}^{2}–{a}^{3}\right)+{\left({a}^{2}–{a}^{3}\right)}^{2}\\ & =9–36a+36{a}^{2}+2\left(3{a}^{2}–3{a}^{3}–6{a}^{3}+6{a}^{4}\right)+{a}^{4}–2{a}^{5}+{a}^{6}\\ & =9–36a+36{a}^{2}+6{a}^{2}–18{a}^{3}+12{a}^{4}+{a}^{4}–2{a}^{5}+{a}^{6}\\ & =9–36a+42{a}^{2}–18{a}^{3}+13{a}^{4}–2{a}^{5}+{a}^{6}\end{array}$
22. $\begin{array}{cc}\frac{1}{2}{x}^{3}–{x}^{2}+\frac{2}{3}x+2& ={\left(\left[\frac{1}{2}{x}^{3}–{x}^{2}\right]+\left[\frac{2}{3}x+2\right]\right)}^{2}\\ & ={\left(\frac{1}{2}{x}^{3}–{x}^{2}\right)}^{2}+2\left(\frac{1}{2}{x}^{3}–{x}^{2}\right)\left(\frac{2}{3}x+2\right)+{\left(\frac{2}{3}x+2\right)}^{2}\\ & =\frac{{x}^{6}}{4}–{x}^{5}+{x}^{4}+2\left(\frac{1}{3}{x}^{4}+{x}^{3}–\frac{2}{3}{x}^{3}–2{x}^{2}\right)+\frac{4}{9}{x}^{2}+\frac{8}{3}x+4\\ & =\frac{{x}^{6}}{4}–{x}^{5}+{x}^{4}+\frac{2}{3}{x}^{4}+\frac{2}{3}{x}^{3}–4{x}^{2}+\frac{4}{9}{x}^{2}+\frac{8}{3}x+4\\ & =\frac{{x}^{6}}{4}–{x}^{5}+\frac{5}{3}{x}^{4}+\frac{2}{3}{x}^{3}–\frac{32}{9}{x}^{2}+\frac{8}{3}x+4\end{array}$
23. $\begin{array}{cc}\frac{1}{2}{a}^{3}–\frac{2}{3}{a}^{2}+\frac{3}{4}a–\frac{1}{2}& ={\left(\left[\frac{1}{2}{a}^{3}–\frac{2}{3}{a}^{2}\right]+\left[\frac{3}{4}a–\frac{1}{2}\right]\right)}^{2}\\ & ={\left(\frac{1}{2}{a}^{3}–\frac{2}{3}{a}^{2}\right)}^{2}+2\left(\frac{1}{2}{a}^{3}–\frac{2}{3}{a}^{2}\right)\left(\frac{3}{4}a–\frac{1}{2}\right)+{\left(\frac{3}{4}a–\frac{1}{2}\right)}^{2}\\ & =\frac{{a}^{6}}{4}–\frac{2{a}^{5}}{3}+\frac{4{a}^{4}}{9}+2\left(\frac{3{a}^{4}}{8}–\frac{{a}^{3}}{4}–\frac{{a}^{3}}{2}+\frac{{a}^{2}}{3}\right)+\frac{9{a}^{2}}{16}–\frac{3a}{4}+\frac{1}{4}\\ & =\frac{{a}^{6}}{4}–\frac{2{a}^{5}}{3}+\frac{4{a}^{4}}{9}+\frac{3{a}^{4}}{4}–\frac{3{a}^{3}}{2}+\frac{2{a}^{2}}{3}+\frac{9{a}^{2}}{16}–\frac{3a}{4}+\frac{1}{4}\\ & =\frac{{a}^{6}}{4}–\frac{2{a}^{5}}{3}+\frac{43{a}^{4}}{36}–\frac{3{a}^{3}}{2}+\frac{59{a}^{2}}{48}–\frac{3a}{4}+\frac{1}{4}\end{array}$
24. $\begin{array}{cc}{x}^{5}–{x}^{4}+{x}^{3}–{x}^{2}+x–2& ={\left(\left[{x}^{5}–{x}^{4}+{x}^{3}\right]–\left[{x}^{2}–x+2\right]\right)}^{2}\\ & ={\left({x}^{5}–{x}^{4}+{x}^{3}\right)}^{2}–2\left({x}^{5}–{x}^{4}+{x}^{3}\right)\left({x}^{2}–x+2\right)+{\left({x}^{2}–x+2\right)}^{2}\\ & ={\left(\left[{x}^{5}–{x}^{4}\right]+{x}^{3}\right)}^{2}–2\left({x}^{7}–{x}^{6}+2{x}^{5}–{x}^{6}+{x}^{5}–2{x}^{4}+{x}^{5}–{x}^{4}+2{x}^{3}\right)+{\left(\left[{x}^{2}–x\right]+2\right)}^{2}\\ & ={\left({x}^{5}–{x}^{4}\right)}^{2}+2{x}^{3}\left({x}^{5}–{x}^{4}\right)+{x}^{6}–2\left({x}^{7}–2{x}^{6}+4{x}^{5}–3{x}^{4}+2{x}^{3}\right)+{\left({x}^{2}–x\right)}^{2}+4\left({x}^{2}–x\right)+{2}^{2}\\ & ={x}^{10}–2{x}^{9}+{x}^{8}+2{x}^{8}–2{x}^{7}+{x}^{6}–2{x}^{7}+4{x}^{6}–8{x}^{5}+6{x}^{4}–4{x}^{3}+{x}^{4}–2{x}^{3}+{x}^{2}+4{x}^{2}–4x+4\\ & ={x}^{10}–2{x}^{9}+3{x}^{8}–4{x}^{7}+5{x}^{6}–8{x}^{5}+7{x}^{4}–6{x}^{3}+5{x}^{2}–4x+4\end{array}$