\(\sqrt{6+\sqrt{32}}-\sqrt{11-\sqrt{72}}\)

\(=\sqrt{6+2\sqrt{8}}-\sqrt{11-2.3\sqrt{2}}\)

\(=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)

\(=2+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}-1\)

\(\hept{\begin{cases}x^2+y^2=2x\left(1\right)\\\left(x-1\right)^3+y^3=1\left(2\right)\end{cases}}\)

Pt (1): \(x^2+y^2=2x\)

\(\Leftrightarrow x^2-2x+1=1-y^2\)

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

\(\Leftrightarrow\left(x-1\right)^6=\left(1-y^2\right)^3\)(*)

PT(2): \(\left(x-1\right)^3+y^3=1\)

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

\(\Leftrightarrow\left(x-1\right)^6=\left(1-y^3\right)^2\)(**)

Từ (*) và (**) \(\Rightarrow\left(1-y^2\right)^3=\left(1-y^3\right)^2\)

\(\Leftrightarrow\left(1-y\right)^3\left(1+y\right)^3-\left(1-y\right)^2\left(1+y+y^2\right)^2=0\)

\(\Leftrightarrow\left(1-y\right)^2\left[\left(1-y\right)\left(1+y\right)^3-\left(1+y+y^2\right)^2\right]=0\)

\(\Leftrightarrow\left(1-y\right)^2\left(-2y^4-4y^3-3y^2\right)=0\)

\(\Leftrightarrow-y^2\left(1-y\right)^2\left(2y^2+4y+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y^2=0\\\left(1-y\right)^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}y=0\\y=1\end{cases}}\)

Thay y = 0 vào pt (2) ta đc

\(\left(x-1\right)^3=1\)

\(\Leftrightarrow x=2\)

Thay y = 1 vào pt(2) ta đc

\(\left(x-1\right)^3=0\)

\(\Leftrightarrow x=1\)

Vậy hpt có nghiệm (x,y) là (2,0);(1,1)

ĐK: \(x\ge-\frac{2}{3}\).

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

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

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

\(ĐK:x\ge\frac{-2}{3}\)

\(\hept{\begin{cases}x\ge\frac{1}{2}\\\left(2x-1\right)^2\left(3x+2\right)=\left(2x-1\right)^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge\frac{1}{2}\\\left(2x-1\right)^2\left(3x+1\right)=0\end{cases}}\)

\(\Rightarrow x=\frac{1}{2}\left(tm\right)\)

\(8.a,\sqrt{-5x-10}\)

\(-5x-10\ge0\)

\(x\le-2\)

\(b,\sqrt{x^2-2x+1}\)

\(\sqrt{\left(x-1\right)^2}\)

\(\left|x-1\right|\ge0\left(\forall x\right)\)pt vô số nghiệm

\(c,\sqrt{2x^2+4x+5}\)

\(\sqrt{\left(\sqrt{2}x+\sqrt{2}\right)^2+3}\)

\(\sqrt{\left(\sqrt{2}x+\sqrt{2}\right)^2+3}\ge\sqrt{3}>0\)

\(\sqrt{\left(\sqrt{2}x+\sqrt{2}\right)^2+3}>0\left(\forall x\right)\)pt vô số nghiệm

\(d,\sqrt{-x^2+4x-4}\)

\(-x^2+4x-4\ge0\)

\(-\left(x-2\right)^2\ge0\)

\(\hept{\begin{cases}-\left(x-2\right)^2\ge0\\-\left(x-2\right)^2\le0\end{cases}< =>-\left(x-2\right)=0}\)

\(x=2\)

Các bạn ơi vào đây giải toán có thưởng nè!

https://tailieugiaoduc.edu.vn/DienDan/Topic/27

Các bạn ơi vào đây giải toán có thưởng nè!

https://tailieugiaoduc.edu.vn/DienDan/Topic/27