Sửa lại câu c) đặt \(\sqrt{x}+1=\)t \(\Rightarrow\left[2\left(t+\dfrac{1}{2}\right)\right]\left(t-3\right)\)=7⇒\(\left\{{}\begin{matrix}t=3\\t=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=4\\x=\dfrac{9}{4}\end{matrix}\right.\)

a) \(\left(\sqrt{4-3x}\right)^2=8^2\)\(\Leftrightarrow4-3x=64\Rightarrow x=-20\)

b) \(\sqrt{4x-8}+1=12\sqrt{\dfrac{x-2}{9}}\Leftrightarrow2\sqrt{x-2}+1\)\(=\left(12\sqrt{\left(x-2\right).\dfrac{1}{9}}\right)\)

\(\Leftrightarrow2t+1=12.\dfrac{1}{3}t\) (Đặt t = \(\sqrt{x-2}\))

\(\Rightarrow t=\dfrac{1}{2}\) \(\Rightarrow\sqrt{x-2}=\dfrac{1}{2}\)\(\Rightarrow x=\dfrac{9}{4}\)

c) pt\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x}+1=7\\\sqrt{x}-2=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\x=4\end{matrix}\right.\)

a) Ta có: \(\sqrt{4-3x}=8\)

\(\Leftrightarrow4-3x=64\)

\(\Leftrightarrow3x=4-64=-60\)

hay x=-20

b) Ta có: \(\sqrt{4x-8}-12\cdot\sqrt{\dfrac{x-2}{9}}=-1\)

\(\Leftrightarrow2\cdot\sqrt{x-2}-12\cdot\dfrac{\sqrt{x-2}}{3}=-1\)

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

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

\(\Leftrightarrow x-2=\dfrac{1}{4}\)

hay \(x=\dfrac{9}{4}\)

2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)

\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)

Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)

\(\Rightarrow x=3\)

c,\(pt\Leftrightarrow3\left(x-1\right)+\frac{x-1}{4x}+\left(2-\sqrt{3x+1}\right)=0\)

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

\(\Rightarrow x=1\)

\(3+\frac{1}{4x}+\frac{1}{2+\sqrt{3x+1}}=0\)

bạn làm nốt pần này nhá

a/

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x^2+x+1}=b>0\end{matrix}\right.\)

\(\Leftrightarrow6a^2+\sqrt{3}ab-3b^2=0\)

\(\Leftrightarrow\left(3a-\sqrt{3}b\right)\left(2a+\sqrt{3}b\right)=0\)

\(\Leftrightarrow3a-\sqrt{3}b=0\Rightarrow b=\sqrt{3}a\)

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

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

b/ ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}x+3=a\\\sqrt{\left(4-x\right)\left(12+x\right)}=b\end{matrix}\right.\)

\(\Rightarrow a^2+b^2=x^2+6x+9+48-8x-x^2=57-2x=2\left(28-x\right)+1\)

\(\Rightarrow28-x=\frac{a^2+b^2-1}{2}\)

Phương trình trở thành:

\(ab=\frac{a^2+b^2-1}{2}\Leftrightarrow\left(a-b\right)^2=1\Leftrightarrow\left[{}\begin{matrix}a+1=b\\a-1=b\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=\sqrt{\left(4-x\right)\left(12+x\right)}\\x+2=\sqrt{\left(4-x\right)\left(12+x\right)}\end{matrix}\right.\) \(\Leftrightarrow...\)

c/ ĐKXĐ: ...

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{x^2-1}=b\ge0\end{matrix}\right.\)

\(ab=2a^2-b^2\Leftrightarrow2a^2-ab-b^2=0\)

\(\Leftrightarrow\left(a-b\right)\left(2a+b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\2a+b=0\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow a=b\Leftrightarrow\sqrt{x}=\sqrt{x^2-1}\)

\(\Leftrightarrow x^2-x-1=0\)

d/ Là \(2x^2+5\) hay \(2x+5\) bạn?