a, \(\Leftrightarrow\left|2x-1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\)

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

Vậy ...

b, ĐKXĐ : \(x\ge-1\)

\(\Leftrightarrow2\sqrt{x+1}-3\sqrt{x+1}-2\sqrt{x+1}=5\)

\(\Leftrightarrow\sqrt{x+1}=-\dfrac{5}{3}\)

Vậy phương trình vô nghiệm

a)Pt \(\Leftrightarrow\left|2x-1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Vậy...

b)Đk:\(x\ge-1\)

Pt\(\Leftrightarrow2\sqrt{x+1}-3\sqrt{x+1}-2\sqrt{x+1}=5\)

\(\Leftrightarrow-3\sqrt{x+1}=5\) (vô nghiệm)

Vậy...

a\(\sqrt{\left(2x-1\right)^2}=4\)

\(\Leftrightarrow\left|2x-1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\)

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

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

Vậy S={\(\dfrac{5}{2};-\dfrac{3}{2}\)}

Bài 1 :

a, ĐKXĐ : \(3-2x\ge0\)

\(\Rightarrow x\le\dfrac{3}{2}\)

Vậy ...

b, ĐKXĐ : \(\left\{{}\begin{matrix}-\dfrac{5}{2x+1}\ge0\\2x+1\ne0\end{matrix}\right.\)

\(\Rightarrow2x+1< 0\)

\(\Rightarrow x< -\dfrac{1}{2}\)

Vậy ...

a,ĐKXĐ \(3-2\text{x}>0\Leftrightarrow-2x>-3\Leftrightarrow\text{x}< \dfrac{3}{2}\)

b,\(\dfrac{-5}{2x+1}>0\Leftrightarrow2x+1< 0\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\)

( bây giờ mình bận nên làm trước 2 bài =))

a, \(x\le\dfrac{3}{2}\)

b, \(x< -\dfrac{1}{2}\)

*a, \(\sqrt{\left(2x-3\right)^2}=5=>|2x-3|=5=>\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)

b, \(\sqrt{9x+9}+\sqrt{4x+4}-\sqrt{16x+16}=3\)

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

\(< =>\sqrt{x+1}=3=>x+1=9=>x=8\left(tm\right)\)

a: Ta có: \(\sqrt{4-3x}=8\)

\(\Leftrightarrow4-3x=64\)

\(\Leftrightarrow3x=-60\)

hay x=-20

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

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

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

hay \(x=\dfrac{9}{4}\)

\(\left\{{}\begin{matrix}8>0\left(luondung\right)\\4-3x=64\end{matrix}\right.\) \(\Leftrightarrow x=-20\left(ktm\right)\)

f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)

\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)

\(\Leftrightarrow\left|x+1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)

\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

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

a) \(\sqrt{\left(x-3\right)^2}=2\Rightarrow\left|x-3\right|=2\Rightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Rightarrow\sqrt{9\left(x+2\right)}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25\left(x+2\right)}=6\)

\(\Rightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Rightarrow2\sqrt{x+2}=6\Rightarrow\sqrt{x+2}=3\Rightarrow x+2=9\Rightarrow x=7\)

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

Ta có: \(x-2\sqrt{x}+3=x-2\sqrt{x}+1+2=\left(\sqrt{x}-1\right)^2+2\ge2\)

\(\Rightarrow\dfrac{1}{x-2\sqrt{x}+3}\le2\Rightarrow Q_{max}=2\) khi \(x=1\)

a) Ta có: \(\sqrt{\left(x-3\right)^2}=2\)

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

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

b) ĐKXĐ: \(x\ge-2\)

Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+\dfrac{4}{5}\cdot5\sqrt{x+2}=6\)

\(\Leftrightarrow2\sqrt{x+2}=6\)

\(\Leftrightarrow x+2=9\)

hay x=7(thỏa ĐK)

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

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

Vậy:.....

b) ĐKXĐ: x ≥ -2

 \(\Leftrightarrow\sqrt{9}.\sqrt{x+2}-5.\sqrt{x+2}+\dfrac{4}{5}.\sqrt{25}.\sqrt{x+2}=6\)

<=> \(\sqrt{x+2}.\left(3-5+\dfrac{4}{5}.5\right)=6\)

\(\Leftrightarrow2.\sqrt{x+2}=6\)

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

<=> x + 2 = 9

<=> x = 7

Tham khảo ạ:

a) Ta có: 

⇔[x−3=2x−3=−2⇔[x=5x=1⇔[x−3=2x−3=−2⇔[x=5x=1

b) ĐKXĐ: 

Ta có: ⇔3√x+2−5√x+2+45⋅5√x+2=6⇔3x+2−5x+2+45⋅5x+2=6

hay x=7(thỏa ĐK)