Câu 2,3,4 nx thôi ạ. Câu 1 có bạn giúp r ạ 

1)\(\sqrt{4x^2+12x+9}=2-x\)

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

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

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

\(\)

\(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3\left(x^2-x-1\right)}-\sqrt{x^2-3x+4}\)

\(\Leftrightarrow\left(\sqrt{3x^2-5x+1}-\sqrt{3}\right)-\left(\sqrt{x^2-2}-\sqrt{2}\right)=\left(\sqrt{3\left(x^2-x-1\right)}-\sqrt{3}\right)-\left(\sqrt{x^2-3x+4}-\sqrt{2}\right)\)

\(\Leftrightarrow\frac{3x^2-5x+1-3}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x^2-2-2}{\sqrt{x^2-2}+\sqrt{2}}=\frac{3\left(x^2-x-1\right)-3}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}-\frac{x^2-3x+4-2}{\sqrt{x^2-3x+4}+\sqrt{2}}\)

\(\Leftrightarrow\frac{3x^2-5x-2}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x^2-4}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3x^2-3x-6}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{x^2-3x+2}{\sqrt{x^2-3x+4}+\sqrt{2}}=0\)

\(\Leftrightarrow\frac{\left(x-2\right)\left(3x+1\right)}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{\left(x-2\right)\left(x+2\right)}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3\left(x-2\right)\left(x+1\right)}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{\left(x-1\right)\left(x-2\right)}{\sqrt{x^2-3x+4}+\sqrt{2}}=0\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{3x+1}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x+2}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3\left(x+1\right)}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{x-1}{\sqrt{x^2-3x+4}+\sqrt{2}}\right)=0\)

Dễ thấy: \(\frac{3x+1}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x+2}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3\left(x+1\right)}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{x-1}{\sqrt{x^2-3x+4}+\sqrt{2}}=0\) vô nghiệm

\(\Rightarrow x-2=0\Rightarrow x=2\)

sao cái từ "dễ thấy" nó khó thấy quá v 

c:

ĐKXĐ: 6-5x>=0

=>5x<=6

=>x<=1,2

\(2\sqrt[3]{3x-2}-3\cdot\sqrt{6-5x}+16=0\)

=>\(2\cdot\sqrt[3]{3x-2}+4+12-3\cdot\sqrt{6-5x}=0\)

=>\(2\cdot\left(\sqrt[3]{3x-2}+2\right)+3\left(4-\sqrt{6-5x}\right)=0\)

=>\(2\cdot\frac{3x-2+8}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{16-6+5x}{4+\sqrt{6-5x}}=0\)

=>\(2\cdot\frac{3x+6}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{5x+10}{4+\sqrt{6-5x}}=0\)

=>\(\left(2\cdot\frac{3}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{5}{4+\sqrt{6-5x}}\right)\left(x+2\right)=0\)

=>x+2=0

=>x=-2(nhận)

d: ĐKXĐ: x>=1

\(\sqrt[3]{x+6}-2\cdot\sqrt{x-1}=4-x^2\)

=>\(\sqrt[3]{x+6}-2-2\cdot\sqrt{x-1}+2=4-x^2\)

=>\(\frac{x+6-8}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+2\left(1-\sqrt{x-1}\right)=\left(2-x\right)\left(2+x\right)\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+2\cdot\frac{1-x+1}{1+\sqrt{x-1}}=\left(2-x\right)\left(2+x\right)\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-2\cdot\frac{x-2}{1+\sqrt{x-1}}-\left(2-x\right)\left(2+x\right)=0\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-2\cdot\frac{x-2}{1+\sqrt{x-1}}+\left(x-2\right)\left(2+x\right)=0\)

=>\(\left(x-2\right)\left(\frac{1}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-\frac{2}{1+\sqrt{x-1}}+\left(2+x\right)\right)=0\)

=>x-2=0

=>x=2(nhận)

cái đoạn y = 1/(x^2+can x) là không có đâu nhá 

Xem lại đề nha bạn.

\(\sqrt{-x^2-x+6}\) mới đúng chứ b

a)\(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)

ĐK:tự xác định 

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

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

Suy ra x=-1 là nghiệm và pt \(\sqrt{2\left(x+3\right)}+\sqrt{x-1}=2\sqrt{x+1}\)

\(\Leftrightarrow2\left(x+3\right)+x-1+2\sqrt{2\left(x+3\right)\left(x-1\right)}=4\left(x+1\right)\)

\(\Leftrightarrow2\sqrt{2\left(x+3\right)\left(x-1\right)}=x-1\)

\(\Leftrightarrow8\left(x+3\right)\left(x-1\right)-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(8x+24-x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x+25\right)=0\Rightarrow x=1\) (thỏa và 7x+25=0 loại do điều kiện....)

b nghiệm xấu quá để mình xem lại :v

\(\Leftrightarrow\sqrt{2x+6}+\sqrt{x-1}=2\sqrt{x+1}\)

\(\Leftrightarrow\sqrt{2x+6}-2\sqrt{2}+\sqrt{x-1}=2\sqrt{x+1}-2\sqrt{2}\)

\(\Leftrightarrow\frac{2\left(x-1\right)}{\sqrt{2x+6}+2\sqrt{2}}+\sqrt{x-1}=\frac{2\sqrt{x-1}}{\sqrt{x+1}+2\sqrt{2}}\)

\(\Leftrightarrow\frac{2\sqrt{x-1}}{\sqrt{2x+6}+2\sqrt{2}}+1=\frac{2\sqrt{x-1}}{\sqrt{x+1}+1\sqrt{2}}\)

đến đây thì chịu 

tìm đc 1 nghiệm là -1;1,nên bình phương lên