a: \(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)

\(\Leftrightarrow\sqrt{x-2}=4\)

=>x-2=16

hay x=18

b: \(\Leftrightarrow\left|3x+2\right|=4x\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x\left(x>=-\dfrac{2}{3}\right)\\3x+2=-4x\left(x< -\dfrac{2}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{2}{7}\left(nhận\right)\end{matrix}\right.\)

c: \(\Leftrightarrow3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)

\(\Leftrightarrow4\sqrt{x-2}=40\)

=>x-2=100

hay x=102

d: =>5x-6=9

hay x=3

\(a,\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\left(dk:x\ge2\right)\)

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

\(\Leftrightarrow\sqrt{x-2}=4\)

\(\Leftrightarrow x-2=16\)

\(\Leftrightarrow x=18\left(tmdk\right)\)

b,\(\sqrt{9x^2-12x+4=3x\left(dk:x\ge0\right)}\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}3x-2=3x\\3x-2=-3x\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\in\varnothing\\x=\dfrac{1}{3}\left(tmdk\right)\end{matrix}\right.\)

Các câu còn lại làm tương tự nhé 

\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: x≥2)

\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+6\sqrt{\dfrac{1}{81}\left(x-2\right)}=-4\)

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

\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{4}{3}\sqrt{x-2}=-4\)

\(-\sqrt{x-2}=-4\)

\(\sqrt{x-2}=4\)

\(\left|x-2\right|=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=16\\x-2=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\left(TM\right)\\x=-14\left(L\right)\end{matrix}\right.\)

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

\(\sqrt{4x^2+5x+1}-2\sqrt{x^2-x+1}=9x-3\)

\(\Leftrightarrow\sqrt{4x^2+5x+1}-\dfrac{2\sqrt{7}}{3}-\left(2\sqrt{x^2-x+1}-\dfrac{2\sqrt{7}}{3}\right)=9x-3\)

\(\Leftrightarrow\dfrac{4x^2+5x+1-\dfrac{28}{9}}{\sqrt{4x^2+5x+1}+\dfrac{2\sqrt{7}}{3}}-\dfrac{4\left(x^2-x+1\right)-\dfrac{28}{9}}{2\sqrt{x^2-x+1}+\dfrac{2\sqrt{7}}{3}}=9x-3\)

\(\Leftrightarrow\dfrac{\dfrac{36x^2+45x-19}{9}}{\sqrt{4x^2+5x+1}+\dfrac{2\sqrt{7}}{3}}-\dfrac{\dfrac{36x^2-36x+8}{9}}{2\sqrt{x^2-x+1}+\dfrac{2\sqrt{7}}{3}}=3\left(3x-1\right)\)

\(\Leftrightarrow\dfrac{\dfrac{\left(3x-1\right)\left(12x+19\right)}{9}}{\sqrt{4x^2+5x+1}+\dfrac{2\sqrt{7}}{3}}-\dfrac{\dfrac{4\left(3x-2\right)\left(3x-1\right)}{9}}{2\sqrt{x^2-x+1}+\dfrac{2\sqrt{7}}{3}}-3\left(3x-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(\dfrac{12x+19}{9\left(\sqrt{4x^2+5x+1}+\dfrac{2\sqrt{7}}{3}\right)}-\dfrac{4\left(3x-2\right)}{9\left(2\sqrt{x^2-x+1}+\dfrac{2\sqrt{7}}{3}\right)}-3\right)=0\)

Dễ thấy: \(\dfrac{12x+19}{9\left(\sqrt{4x^2+5x+1}+\dfrac{2\sqrt{7}}{3}\right)}-\dfrac{4\left(3x-2\right)}{9\left(2\sqrt{x^2-x+1}+\dfrac{2\sqrt{7}}{3}\right)}-3< 0\)

\(\Rightarrow3x-1=0\Rightarrow3x=1\Rightarrow x=\dfrac{1}{3}\)

Ace Legona cái dễ thấy của bạn mình nghĩ lại là mấu chốt của bài này

Bạn tham khảo:

Câu hỏi của Nguyễn Thị Bình Yên - Toán lớp 9 | Học trực tuyến

Bạn lưu ý:

\(a=\sqrt{4x^2+5x+1}\ge0\)

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

Do đó \(a+b>1\) hay \(a+b-1>0\)

ĐKXĐ: \(\left[{}\begin{matrix}x\le-1\\x\ge-\frac{1}{4}\end{matrix}\right.\)

\(\sqrt{4x^2+5x+1}-\sqrt{4x^2-4x+4}+9x-3=0\)

Đặt \(\left\{{}\begin{matrix}a=\sqrt{4x^2+5x+1}\ge0\\b=\sqrt{4x^2-4x+4}>0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=9x-3\)

Phương trình trở thành:

\(a-b+a^2-b^2=0\)

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

\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)

\(\Leftrightarrow a-b=0\) (do \(a;b>0\Rightarrow a+b+1>0\))

\(\Leftrightarrow a=b\Rightarrow\sqrt{4x^2+5x+1}=\sqrt{4x^2-4x+4}\)

\(\Leftrightarrow4x^2+5x+1=4x^2-4x+4\)

\(\Leftrightarrow9x=3\Rightarrow x=\frac{1}{3}\)

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ĐKXĐ: \(\left[{}\begin{matrix}x\le-1\\x\ge-\frac{1}{4}\end{matrix}\right.\)

\(\sqrt{4x^2+5x+1}-\sqrt{4x^2-4x+4}+9x-3=0\)

Đặt \(\left\{{}\begin{matrix}a=\sqrt{4x^2+5x+1}\ge0\\b=\sqrt{4x^2-4x+4}>0\end{matrix}\right.\) \(\Rightarrow a^2-b^2=9x-3\)

Phương trình trở thành:

\(a-b+a^2-b^2=0\)

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

\(\Leftrightarrow\left(a-b\right)\left(1+a+b\right)=0\)

\(\Leftrightarrow a=b\) (do \(\left\{{}\begin{matrix}a\ge0\\b>0\end{matrix}\right.\Rightarrow1+a+b>0\))

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

\(\Leftrightarrow9x=3\)

\(\Rightarrow x=\frac{1}{3}\)