<=>x^2-y^2+x^2-xy=8

<=>(x-y)(2x+y)=8

2x+y>x-y

tự xét tiếp

lớp 9 kém thế

\(4-2xy+\left(2-x-y\right)^2=0\)

\(\Leftrightarrow y^2+x^2-4x-4y+8=0\)

\(\Leftrightarrow\left(y-2\right)^2+\left(x-2\right)^2=0\)

A=(\(\frac{x-2}{x+2\sqrt{x}}\)+\(\frac{1}{\sqrt{x}+2}\)).\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

A=(\(\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\)+\(\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\)).\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

A=\(\frac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\).\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

A=\(\frac{x-\sqrt{x}+2\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\).\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

A=\(\frac{\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\).\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

A=\(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\).\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

A=\(\frac{\sqrt{x}-1}{\sqrt{x}}\).\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

A=\(\frac{\sqrt{x}+1}{\sqrt{x}}\)

Vậy............................

điều kiện 1-x²>=0 <=>-1<=x<=1

Đặt a=\(\sqrt{1-x^2}\) \(\left(0\le a\le1\right)\)  =>a²+x²=1 =>(a+x)²=1+2ax ,=>ax=\(\frac{\left(a+x\right)^2-1}{2}\) (1)

Thay vào phương trình ta được x³+a³=ax\(\sqrt{2}\) <=>(a+x)(a²+x²-ax)=ax\(\sqrt{2}\) <=>(a+x)(1-ax)=ax\(\sqrt{2}\)

Thay (1) vào ta được (a+x)(1-\(\frac{\left(a+x\right)^2-1}{2}\))=\(\frac{\left(a+x\right)^2-1}{2}.\sqrt{2}\) <=>(a+x)³+\(\sqrt{2}\)(a+x)²-3(a+x)-\(\sqrt{2}\)=0

Đặt a+x=m; m²\(\le2\left(a^2+x^2\right)\)=2 <=>\(-\sqrt{2}\le m\le\sqrt{2}\)   (2)

thay vào phương trình trên ta được m³+m²\(\sqrt{2}\)-3m-\(\sqrt{2}\)=0 <=>(m-\(\sqrt{2}\))(m\(-\sqrt{2}-1\))(\(m-\sqrt{2}+1\))=0

<=>\(m=\sqrt{2};m=1-\sqrt{2};m=1+\sqrt{2}\)( loại vì không thỏa mãn (2))

*Xét m=a+x=\(\sqrt{2}\) ta có\(\hept{\begin{cases}a+x=\sqrt{2}\\a^2+x^2=1\end{cases}}\)<=>\(\hept{\begin{cases}a+x=\sqrt{2}\\\left(\sqrt{2}-x\right)^2+x^2=1\end{cases}}\)<=>\(x=\frac{1}{\sqrt{2}}\)(thỏa mãn x\(\in\left[-1;1\right]\)

*Xét m=a+x=1-\(\sqrt{2}\) ta có \(\hept{\begin{cases}a+x=1-\sqrt{2}\\a^2+x^2=1\end{cases}}\)<=>...  <=>  x=\(\frac{\sqrt{2\sqrt{2}-1}-1-\sqrt{2}}{2}\)

vậy pt có 2 nghiệm x1=..;x2=...

\(DK:x\in\left[-2;2\right]\)

\(\Leftrightarrow\left(2-x\right)+\sqrt{4-x^2}\left(3x-1\right)=0\)

\(\Leftrightarrow\left(2-x\right)+\sqrt{\left(2+x\right)\left(2-x\right)}\left(3x-1\right)=0\)

\(\Leftrightarrow\sqrt{2-x}\left[\sqrt{2-x}+\sqrt{2+x}\left(3x-1\right)\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(1\right)\\\sqrt{2-x}+\sqrt{2+x}\left(3x-1\right)=0\left(2\right)\end{cases}}\)

Xet PT(2)

\(\sqrt{2-x}+\sqrt{2+x}\left(3x-1\right)=0\)

\(\Leftrightarrow\left(2+x\right)\left(3x-1\right)^2=2-x\)

\(\Leftrightarrow\left(2+x\right)\left(9x^2-6x+1\right)=2-x\)

\(\Leftrightarrow9x^3+12x^2-10x=0\)

\(\Leftrightarrow x\left(9x^2+12x-10\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(n\right)\\9x^2+12x-10=0\end{cases}}\)

Xet PT 

\(9x^2+12x-10=0\)

Ta co:

\(\Delta^`=6^2-9.\left(-10\right)=126>0\)

\(\Rightarrow\hept{\begin{cases}x_1=\frac{-4+\sqrt{14}}{6}\left(n\right)\\x_2=\frac{-4-\sqrt{14}}{6}\left(n\right)\end{cases}}\)

Vay tap nghiem cua PT la \(S=\left\{\frac{-4-\sqrt{14}}{6};\frac{-4+\sqrt{14}}{6};0;2\right\}\)

\(\sqrt{x^2+7}\)có nghĩa

\(\Leftrightarrow x^2+7\ge0\)

Mà \(x^2\ge0\Rightarrow x^2+7\ge7>0\)

Vậy bt có nghĩa với mọi x