a.

\(\left\{{}\begin{matrix}x^4+y^4=34\\y=2-x\end{matrix}\right.\)

\(\Rightarrow x^4+\left(x-2\right)^4=34\)

Đặt \(x-1=t\)

\(\Rightarrow\left(t+1\right)^4+\left(t-1\right)^4=34\)

\(\Leftrightarrow t^4+6t^2-16=0\Rightarrow\left[{}\begin{matrix}t^2=2\\t^2=-8\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}t=\sqrt{2}\Rightarrow x=\sqrt{2}+1\Rightarrow y=1-\sqrt{2}\\t=-\sqrt{2}\Rightarrow x=1-\sqrt{2}\Rightarrow y=1+\sqrt{2}\end{matrix}\right.\)

b.

\(\left\{{}\begin{matrix}xy^2-x^2y+6x-y^2-y-6=0\\x^2y-xy^2+6y-x^2-x-6=0\end{matrix}\right.\) (1)

Lần lượt cộng 2 vế và trừ 2 vế ta được:

\(\left\{{}\begin{matrix}-x^2-y^2+5x+5y-12=0\\2xy\left(y-x\right)+7\left(x-y\right)+\left(x-y\right)\left(x+y\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-5\left(x+y\right)+12=0\\\left(y-x\right)\left(2xy-x-y-7\right)=0\end{matrix}\right.\)

Th1: \(\left\{{}\begin{matrix}x=y\\x^2+y^2-5\left(x+y\right)+12=0\end{matrix}\right.\)

\(\Rightarrow2x^2-10x+12=0\Rightarrow...\)

TH2: \(\left\{{}\begin{matrix}2xy-\left(x+y\right)-7=0\\x^2+y^2-5\left(x+y\right)+12=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2xy-\left(x+y\right)-7=0\\\left(x+y\right)^2-2xy-5\left(x+y\right)+12=0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2v-u-7=0\\u^2-2v-5u+12=0\end{matrix}\right.\)

\(\Rightarrow u^2-6u+5=0\)

\(\Leftrightarrow...\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-1\right)\left(x+y-2\right)=6\\\left(x-1\right)^2+\left(y-1\right)^2=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-1\right)\left(x+y-2\right)=6\\\left(x+y-2\right)^2-2\left(x-1\right)\left(y-1\right)=5\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\left(x-1\right)\left(y-1\right)=v\\x+y-2=u\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}uv=6\\u^2-2v=5\end{matrix}\right.\) \(\Rightarrow u^2-\dfrac{12}{u}=5\)

\(\Rightarrow u^3-5u-12=0\)

\(\Leftrightarrow\left(u-3\right)\left(u^2+3u+4\right)=0\)

\(\Leftrightarrow u=3\Rightarrow v=2\)

\(\Rightarrow\left\{{}\begin{matrix}x+y-2=3\\\left(x-1\right)\left(y-1\right)=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=5-x\\\left(x-1\right)\left(y-1\right)=2\end{matrix}\right.\)

\(\Rightarrow\left(x-1\right)\left(5-x-1\right)=2\)

\(\Leftrightarrow...\) em tự hoàn thành bài toán

1.

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2-2m\left(x+\dfrac{1}{x}\right)-1+2m=0\)

Đặt \(x+\dfrac{1}{x}=t\Rightarrow\left|t\right|\ge2\)

\(\Rightarrow t^2-1-2mt+2m=0\)

\(\Leftrightarrow\left(t-1\right)\left(t+1\right)-2m\left(t-1\right)=0\)

\(\Leftrightarrow\left(t-1\right)\left(t+1-2m\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=1\left(loại\right)\\t=2m-1\end{matrix}\right.\)

Pt có nghiệm \(\Leftrightarrow\left[{}\begin{matrix}2m-1\ge2\\2m-1\le-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}m\ge\dfrac{3}{2}\\m\le-\dfrac{1}{2}\end{matrix}\right.\)

2.

Cộng vế với vế: \(3\left|x\right|=3\Rightarrow\left|x\right|=1\)

\(\Rightarrow\left|y\right|=-1< 0\) (không thỏa mãn)

Vậy hệ pt vô nghiệm

5) \(\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)\)

\(=\left(x-y\right)^2-2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)

\(=\left[\left(x-y\right)-\left(x+y\right)\right]^2\)

\(=\left(x-y-x-y\right)^2\)

\(=\left(-2y^2\right)\)

\(=4y^2\)

6) \(\left(5-x\right)^2+\left(x+5\right)^2-\left(2x+10\right)\left(x-5\right)\)

\(=\left(x-5\right)^2-2\left(x-5\right)\left(x+5\right)+\left(x+5\right)^2\)

\(=\left[\left(x-5\right)-\left(x+5\right)\right]^2\)

\(=\left(x-5-x-5\right)^2\)

\(=\left(-10\right)^2=100\)

7) \(\left(x-2\right)^2+\left(x+1\right)^2+2\left(x-2\right)\left(-1-x\right)\)

\(=\left(x-2\right)^2-2\left(x-2\right)\left(x+1\right)+\left(x+1\right)^2\)

\(=\left[\left(x-2\right)-\left(x+1\right)\right]^2\)

\(=\left(-3\right)^2=9\)

8) \(-\left(2x+3y\right)^2+\left(2x-3y\right)^2-2\left(4x^2-9y^2\right)\)

\(=\left(2x-3y\right)^2+2\left(2x+3y\right)\left(2x-3y\right)+\left(2x+3y\right)^2\)

\(=\left[\left(2x+3y\right)+\left(2x-3y\right)\right]^2\)

\(=\left(4x\right)^2=16x^2\)