Toán 7

N

niemkieuloveahbu

[TEX]\frac{x^2+y^2}{10}=\frac{x^2-2y^2}{7}(1) \ and\ x^4y^4=81(2)\\ (1) \Leftrightarrow x^2=9y^2\\ The\ vao\ (2):\\ 81y^4.y^4=81 \Leftrightarrow y^8=1 \Leftrightarrow y=\pm 1 \Rightarrow x^2=9 \Leftrightarrow x=\pm 3[/TEX]
 
B

braga

Đặt [TEX]x^2=a(a \geq 0) \ ; \ y^2=b(b \geq 0)[/TEX]

Ta có: [TEX]\frac{a+b}{10}=\frac{a-2b}{7}[/TEX] và [TEX]a^2b^2=81[/TEX]

[TEX]\frac{a+b}{10}=\frac{a-2b}{7}=\frac{(a+b)-(a-2b)}{10-7}=\frac{3b}{3}=b \ \ \ \ \ (1)[/TEX]

[TEX]\frac{a+b}{10}=\frac{a-2b}{7}=\frac{2a+2b}{20}=\frac{(2a+2b)+(a-2b)}{20+7}=\frac{3a}{27}=\frac{a}{9} \ \ (2)[/TEX]

Từ (1) và (2) [TEX]\Rightarrow \frac{a}{9}=b \Rightarrow a=9b[/TEX]

Do [TEX]a^2b^2=81 \Rightarrow (9b)^2b^2=81 \Rightarrow 81b^4=81 \Rightarrow b^4=1 \Rightarrow b=1 \Rightarrow a=1.9=9[/TEX]

Ta có: [TEX]x^2=9 \ ; \ y^2=1 \Rightarrow \fbox{x=\pm 3;y=\pm 1}[/TEX]
 
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