ta có pt

<=> \(2\left(2x+1\right)\sqrt{x+8}=4x^2+4x+1+x+8-x^2+2x-1\)

\(\Leftrightarrow2\left(2x+1\right)\sqrt{x+8}=\left(2x+1\right)^2+x+8-\left(x-1\right)^2\)

\(\Leftrightarrow\left(2x+1\right)^2-2\left(2x+1\right)\sqrt{x+8}+x+8-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(2x+1-\sqrt{x+8}\right)^2-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(2x+1-\sqrt{x+8}+x-1\right)\left(2x+1-\sqrt{x+8}-x+1\right)=0\)

\(\Leftrightarrow\left(3x-\sqrt{x+8}\right)\left(x+2-\sqrt{x+8}\right)=0\)

đến đây thì dễ rồi nhé

(3x-1) (x² +2) = (3x-1)(7x-10)

=> (3x-1) (x²+2)-(3x-1)(7x-10)=0

=>(3x-1)(x²+2-7x+10)=0

=>(3x-1)(x²-7x+12)=0

=>(3x-1)(x-3)(x-4)=0

=>3x-1=0 => x= 1/3

    x-3=0 =>  x=3

    x-4=0 =>  x=4

vậy pt có tập nghiệm S={ 1/3; 3; 4}

mk lam xong roi ban moi giai

vg còn bài nào ko ak

nhiều thế giải ko đổi đâu bạn

vậy trả lời câu a thôi

đkxđ : \(\frac{1}{2}\le x\le7\)

\(x^2-5x+3\sqrt{2x-1}=2\sqrt{14-2x}+5\)

\(\Leftrightarrow\left(x^2-5x\right)+3\left(\sqrt{2x-1}-3\right)=2\left(\sqrt{14-2x}-2\right)\)

\(\Leftrightarrow x\left(x-5\right)+\frac{3.\left(2x-10\right)}{\sqrt{2x-1}+3}+\frac{2.\left(2x-10\right)}{\sqrt{14-2x}+2}=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+\frac{6}{\sqrt{2x-1}+3}+\frac{4}{\sqrt{14-2x}+2}\right)=0\)

\(\Leftrightarrow x=5\)

còn bài a,c lười đánh lắm

ĐK: \(x\ge1\)

Đặt \(\sqrt{3x-2}+2\sqrt{x-1}=t\left(t\ge1\right)\)

\(pt\Leftrightarrow3t=t^2-4\)

\(\Leftrightarrow t^2-3t-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=4\\t=-1\left(l\right)\end{matrix}\right.\)

\(t=4\Leftrightarrow\sqrt{3x-2}+2\sqrt{x-1}=4\)

\(\Leftrightarrow7x-6+4\sqrt{\left(3x-2\right)\left(x-1\right)}=16\)

\(\Leftrightarrow4\sqrt{3x^2-5x+2}=22-7x\)

\(\Leftrightarrow\left\{{}\begin{matrix}48x^2-80x+32=484+49x^2-308x\\22-7x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}452+x^2-228x=0\\x\le\dfrac{22}{7}\end{matrix}\right.\)

\(\Leftrightarrow x=2\left(tm\right)\)

ĐKXĐ:...

\(\sqrt{3x^2-5x-1}-\sqrt{3x^2-7x+9}+\sqrt{x^2-2}-\sqrt{x^2-3x+13}=0\)

\(\Leftrightarrow\frac{2\left(x-5\right)}{\sqrt{3x^2-5x-1}+\sqrt{3x^2-7x+9}}+\frac{3\left(x-5\right)}{\sqrt{x^2-2}+\sqrt{x^2-3x+13}}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\frac{2}{\sqrt{3x^2-5x-1}+\sqrt{3x^2-7x+9}}+\frac{3}{\sqrt{x^2-2}+\sqrt{x^2-3x+13}}\right)=0\)

\(\Leftrightarrow x-5=0\) (ngoặc to phía sau luôn dương)

\(\Rightarrow x=5\)

Akai Haruma @Nguyễn Việt Lâm

vô nghiện

theo mik thì vô no

chắc sai đề

\(\Leftrightarrow\sqrt{12-7x}-\sqrt{x^2-x}=\sqrt{3x^2-5x-1}-\sqrt{x^2-3x+4}\)

\(\Rightarrow-\sqrt{3x^2-5x-1}-\sqrt{x^2-x}+\sqrt{x^2-3x+4}+\sqrt{12-7x}=0\)

=>\(x\approx-3,4579061804411\)

ra số rất lẻ

bấm máy nữa chứ gì =.=