câu 2 có nghiệm x=2 , liên hợp đi 

câu 2 đề sai

ok tớ sẽ giải nhunh ! sửa câu 2 đi rồi tớ sẽ làm cho bn !

câu 1 ) thì đúng

câu 2 sai đề

d)\(2x^2+4x=\sqrt{\frac{x+3}{2}}\)

ĐK:\(x\ge-3\)

\(\Leftrightarrow4x^4+16x^3+16x^2=\frac{x+3}{2}\)

\(\Leftrightarrow\frac{8x^4+32x^3+32x^2-x-3}{2}=0\)

\(\Leftrightarrow8x^4+32x^3+32x^2-x-3=0\)

\(\Leftrightarrow\left(2x^2+3x-1\right)\left(4x^2+10x+3\right)=0\)

d)\(2x^2+4x=\sqrt{\frac{x+3}{2}}\)

ĐK:\(x\ge-3\)

\(\Leftrightarrow4x^4+16x^3+16x^2=\frac{x+3}{2}\)

\(\Leftrightarrow\frac{8x^4+32x^3+32x^2-x-3}{2}=0\)

\(\Leftrightarrow8x^4+32x^3+32x^2-x-3=0\)

\(\Leftrightarrow\left(2x^2+3x-1\right)\left(4x^2+10x+3\right)=0\)

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

\(\Leftrightarrow4\left(x-1\right)\left(x+1\right)-3x=\sqrt[3]{x^2\left(x-1\right)\left(x+1\right)}\)

dat \(\left(x-1\right)\left(x+1\right)=y\)

\(4y-3x=\sqrt[3]{x^2y}\)

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

\(\Leftrightarrow64y^3-144y^2x+108yx^2-27x^3=x^2y\)

\(\Leftrightarrow64y^3-144y^2x+107yx^2-27x^3=0\)

\(\Leftrightarrow64y^3-64y^2x-80y^2x+80x^2y+27x^2y-27x^3=0\)

\(\Leftrightarrow\left(y-x\right)\left(64y^2-80xy+27x^2\right)=0\)

de thay \(64y^2-80xy+27x^2=\left(8y\right)^2-2.8y.5x+25x^2+2x^2=\left(8y-5x\right)^2+2x^2>0\)

\(\Rightarrow y=x\)hay \(\left(x-1\right)\left(x+1\right)=x\Rightarrow x^2-x-1=0\) 

\(\left(x-\frac{1}{2}\right)^2-\frac{5}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{5}+1}{2}\\x=\frac{-\sqrt{5}+1}{2}\end{cases}}\)

câu b tương tự nhé bạn

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

giúp mk bài này với

câu 2 có thể là am-gm 2016 số 

\(\Leftrightarrow\sqrt{x^2-\frac{1}{4}+\sqrt{\left(x+\frac{1}{2}\right)^2}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)\(\Leftrightarrow\sqrt{x^2+x+\frac{1}{4}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\)\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\Leftrightarrow2x+1=2x^3+x^2+2x+1\)\(\Leftrightarrow2x^3+x^2=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)

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

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

\(x^2+1\ge1\forall x\Rightarrow2x+1\ge0!2x+1!=2x+1\)

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

\(\left(1\right)\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\)

\(\left(1\right)\Leftrightarrow2x+1=\left(2x+1\right)\left(x^2+1\right)\Leftrightarrow\left(2x+1\right).\left(1-\left(x^2+1\right)\right)=0\)

\(\hept{\begin{cases}2x+1=0\\-x^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\x=0\end{cases}}}\)

Chúc bạn học tốt !!!