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4 tháng 7 2021

a)Pt \(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\dfrac{1}{3}+\dfrac{1}{2}\)

\(\Leftrightarrow\left|2x-1\right|=\dfrac{5}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{5}{6}\\2x-1=-\dfrac{5}{6}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{12}\\x=\dfrac{1}{12}\end{matrix}\right.\)

Vậy...

b)Đk:\(x\ge3\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\x-4=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=4\left(tm\right)\\x=2\left(ktm\right)\end{matrix}\right.\)

Vậy...

c)Đk:\(x\ge1\)

\(x+\sqrt{x-1}=13\)

\(\Leftrightarrow\sqrt{x-1}=13-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}13-x\ge0\\x-1=x^2-26x+169\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\x^2-27x+170=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\x^2-17x-10x+170=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\\left(x-17\right)\left(x-10\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}13\ge x\\\left[{}\begin{matrix}x=17\\x=10\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow x=10\) (tm)

Vậy...

30 tháng 6 2021

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a) Ta có: \(A=\left(\dfrac{x-\sqrt{x}+2}{x-\sqrt{x}-2}-\dfrac{x}{x-2\sqrt{x}}\right):\dfrac{1-\sqrt{x}}{2-\sqrt{x}}\)

\(=\left(\dfrac{x-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)

\(=\dfrac{x-\sqrt{x}+2-x-\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)

\(=\dfrac{-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)

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

b) Ta có: \(B=\left(\dfrac{1}{2\sqrt{x}+1}+\dfrac{1}{2\sqrt{x}-1}\right):\dfrac{1}{1-4x}\)

\(=\dfrac{2\sqrt{x}-1+2\sqrt{x}+1}{\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}\cdot\dfrac{-\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{1}\)

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

18 tháng 10 2021

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 \(x=\dfrac{9}{4}\)

18 tháng 10 2021

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

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

\(\Leftrightarrow\left|x+1\right|=3\)

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

b) Ta có: \(3\sqrt{4x+4}-\sqrt{9x-9}-8\sqrt{\dfrac{x+1}{16}}=5\)

\(\Leftrightarrow6\sqrt{x+1}-3\sqrt{x-3}-2\sqrt{x+1}=5\)

\(\Leftrightarrow4\sqrt{x+1}=5+3\sqrt{x-3}\)

\(\Leftrightarrow16\left(x+1\right)=25+30\sqrt{x-3}+9\left(x-3\right)\)

\(\Leftrightarrow16x+16=25+9x-27+30\sqrt{x-3}\)

\(\Leftrightarrow30\sqrt{x-3}=16x+16+2-9x\)

\(\Leftrightarrow30\sqrt{x-3}=7x+18\)

\(\Leftrightarrow x-3=\left(\dfrac{7x+18}{30}\right)^2\)

\(\Leftrightarrow x-3=\dfrac{49x^2}{900}+\dfrac{7}{25}x+\dfrac{9}{25}\)

\(\Leftrightarrow\dfrac{49}{900}x^2-\dfrac{18}{25}x+\dfrac{84}{25}=0\)

\(\Delta=\left(-\dfrac{18}{25}\right)^2-4\cdot\dfrac{49}{900}\cdot\dfrac{84}{25}=-\dfrac{16}{75}< 0\)

Vậy: Phương trình vô nghiệm

6 tháng 7 2021

a)Pt\(\Leftrightarrow\left|x+1\right|=3\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

b)Đk:\(x\ge-1\)

Sửa đề: \(3\sqrt{4x+4}-\sqrt{9x+9}-8\sqrt{\dfrac{x+1}{16}}=5\)

Pt \(\Leftrightarrow6\sqrt{x+1}-3\sqrt{x+1}-2\sqrt{x+1}=5\)

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

\(\Leftrightarrow x=24\left(tm\right)\)

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

6 tháng 2 2022

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

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

4 tháng 7 2021

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

4 tháng 7 2021

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...

Ta có: \(\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)

\(=\dfrac{8+x\left(1+\sqrt{x}-1\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x\sqrt{x}+8}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}+2}{x-4}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)

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

\(=\dfrac{2\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{2\sqrt{x}+4+x-2\sqrt{x}}{2\left(x-4\right)}\)

\(=\dfrac{x+4}{2x-8}\)