1) \(ĐK:x\in R\)

2) \(ĐK:x< 0\)

3) \(ĐK:x\in\varnothing\)

4) \(=\sqrt{\left(x+1\right)^2+2}\) 

\(ĐK:x\in R\)

5) \(=\sqrt{-\left(a-4\right)^2}\)

\(ĐK:x\in\varnothing\)

\(a,ĐK:\left\{{}\begin{matrix}x\ge5\\x\le3\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Vậy pt vô nghiệm

\(b,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow0x=2\Leftrightarrow x\in\varnothing\)

\(c,ĐK:x\ge-\dfrac{3}{2}\\ PT\Leftrightarrow x^2+4x+5-2\sqrt{2x+3}=0\\ \Leftrightarrow\left(2x+3-2\sqrt{2x+3}+1\right)+\left(x^2+2x+1\right)=0\\ \Leftrightarrow\left(\sqrt{2x+3}-1\right)^2+\left(x+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3=1\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\left(tm\right)\\ d,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

a) \(\sqrt{x-5}=\sqrt{3-x}\)

⇔\(\left(\sqrt{x-5}\right)^2=\left(\sqrt{3-x}\right)^2\)

⇔\(x-5=3-x\)

⇔\(x=4\)

b) \(\sqrt{4-5x}=\sqrt{2-5x}\)

⇔\(\left(\sqrt{4-5x}\right)^2=\left(\sqrt{2-5x}\right)^2\)

⇔\(4-5x=2-5x\)

⇔\(2=0\) (Vô lí)

\(a,\)

          \(\sqrt{x-5}=3\)

\(\Leftrightarrow\)\(x-5=3^2\)

\(\Leftrightarrow\)\(x=14\)

\(b,\)

            \(\sqrt{x-10}=-2\)

\(x\)không có giá trị  ( vì \(\sqrt{x-10}\ge0\forall\))

\(c,\)

          \(\sqrt{2x-1}=\sqrt{5}\)

\(\Leftrightarrow2x-1=\sqrt{5^2}\)

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

\(d,\)

               \(\sqrt{4-5x}=12\)

\(\Leftrightarrow4-5x=12^2\)

\(\Leftrightarrow5x=4-144\)

\(\Leftrightarrow5x=-140\)

\(\Leftrightarrow x=-\frac{140}{5}=-28\)

a, \(\sqrt{x-5}=3;ĐK:x-5\ge0\Leftrightarrow x\ge5\)

Ta có: \(\sqrt{x-5}=3\Leftrightarrow x-5=9\Leftrightarrow x=14\)

b, \(\sqrt{x-10}=-2;ĐK:x-10\ge0\Leftrightarrow x\ge10\) 

Vì: \(\sqrt{x-10}\ge0\) nên không có giá trị nào của x để \(\sqrt{x-10}=-2\)

c, \(\sqrt{2x-1}=\sqrt{5};ĐK:2x-1\ge0\Leftrightarrow x\ge0,5\)

Ta có: \(\sqrt{2x-1}=\sqrt{5}\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x=6\Leftrightarrow x=3\)

d, \(\sqrt{4-5x}=12;ĐK:4-5x\ge0\Leftrightarrow x\le\frac{4}{5}\)

Ta có: \(\sqrt{4-5x}=12\Leftrightarrow4-5x=144\)

\(\Leftrightarrow-5x=140\Leftrightarrow x=-28\)

b: \(\Leftrightarrow\left(x^2+5x+4\right)=5\sqrt{x^2+5x+28}\)

Đặt \(x^2+5x+4=a\) 

Theo đề, ta có \(5\sqrt{a+24}=a\)

=>25a+600=a²

=>a=40 hoặc a=-15

=>x²+5x-36=0

=>(x+9)(x-4)=0

=>x=4 hoặc x=-9

c: \(\Leftrightarrow x^2+5x=2\sqrt[3]{x^2+5x-2}-2\)

Đặt \(x^2+5x=a\)

Theo đề, ta có: \(a=2\sqrt[3]{a}-2\)

\(\Leftrightarrow\sqrt[3]{8a}=a+2\)

=>(a+2)³=8a

=>\(a^3+6a^2+12a+8-8a=0\)

\(\Leftrightarrow a^3+6a^2+4a+8=0\)

Đến đây thì bạn chỉ cần bấm máy là xong

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. ĐKXĐ: \(4-5x\ge0\) \(\Leftrightarrow-5x\ge-4\Leftrightarrow5x\le4\Leftrightarrow x\le\dfrac{4}{5}\)

\(\sqrt{4-5x}=12\)

\(\Leftrightarrow4-5x=2\sqrt{3}\)

\(\Leftrightarrow-5x=-4-2\sqrt{3}\)

\(\Leftrightarrow x=\dfrac{-4-2\sqrt{3}}{-5}\)

\(\Leftrightarrow x=\dfrac{4+2\sqrt{3}}{5}\left(KTMĐKXĐ\right)\)

Vậy x không tồn tại

b. \(10-2\sqrt{2x+1}=4\) (1)

\(ĐKXĐ:2x+1\ge0\Leftrightarrow2x\ge-1\Leftrightarrow x\ge-\dfrac{1}{2}\)

(1) => \(-2\sqrt{2x+1}=-6\)

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

\(\Leftrightarrow2x+1=\sqrt{3}\)

\(\Leftrightarrow2x=\sqrt{3}-1\)

\(\Leftrightarrow x=\dfrac{\sqrt{3}-1}{2}\left(TMĐKXĐ\right)\)

c. \(5-\sqrt{x-1}=7\) (1)

ĐKXĐ: \(x-1\ge0\Leftrightarrow x\ge1\)

(1) <=> \(-\sqrt{x-1}=2\) (vô lí)

Vậy không tồn tại x

bài kia làm sai rùi:

a. \(\sqrt{4-5x}=12\) (1)

ĐKXĐ: \(4-5x\ge0\Leftrightarrow x\le\dfrac{4}{5}\)

\(\Leftrightarrow4-5x=144\)

\(\Leftrightarrow5x=-140\)

\(\Leftrightarrow x=-28\left(TMĐKXĐ\right)\)

Vậy phương trình có nghiệm là \(S=\left\{-28\right\}\)

b. \(10-2\sqrt{2x+1}=4\) (1)

ĐKXĐ: \(2x+1\ge0\Leftrightarrow x\ge-\dfrac{1}{2}\)

\(\left(1\right)\Leftrightarrow2\sqrt{2x+1}=6\)

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

\(\Leftrightarrow2x+1=9\)

\(\Leftrightarrow2x=8\)

\(\Leftrightarrow x=4\left(TMĐKXĐ\right)\)

Vậy phương trình có nghiệm là: \(S=\left\{4\right\}\)

c. Ở dưới làm đúng rồi

d. \(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\) (1)

ĐKXĐ: \(3x\ge0\Leftrightarrow x\ge0\)

(1) \(\Leftrightarrow10+\sqrt{3x}=\left(2+\sqrt{6}\right)^2\)

\(\Leftrightarrow10+\sqrt{3x}=10+4\sqrt{6}\)

\(\Leftrightarrow\sqrt{3x}=-10+10+4\sqrt{6}\)

\(\Leftrightarrow\sqrt{3x}=4\sqrt{6}\)

\(\Leftrightarrow3x=96\)

\(\Leftrightarrow x=32\left(TMĐKXĐ\right)\)

Vậy phương trình có nghiệm là: \(S=\left\{32\right\}\)

e. \(\sqrt{x+1}+10=2\sqrt{x+1}-2\) (1)

ĐKXĐ: \(x+1\ge0\Leftrightarrow x\ge-1\)

\(\left(1\right)\Leftrightarrow\sqrt{x+1}-2\sqrt{x+1}=-10-2\)

\(\Leftrightarrow-\sqrt{x+1}=-12\)

\(\Leftrightarrow\sqrt{x+1}=12\)

\(\Leftrightarrow x+1=144\)

\(\Leftrightarrow x=143\left(TMĐKXĐ\right)\)

Vậy phương trình có nghiệm là \(S=\left\{143\right\}\)

f. \(\sqrt{16x+32}-5\sqrt{x+2}=-2\) (1)

ĐKXĐ: \(\left[{}\begin{matrix}\sqrt{16x+32\ge0}\\\sqrt{x+2\ge0}\end{matrix}\right.\left[{}\begin{matrix}x\ge-2\\x\ge-2\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\sqrt{16\left(x+2\right)}-5\sqrt{x+2}=-2\)

\(\Leftrightarrow4\sqrt{x+2}-5\sqrt{x+2}=-2\)

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

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

\(\Leftrightarrow x+2=4\)

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

Vậy phương trình có nghiệm là \(S=\left\{2\right\}\)

a) Ta có: \(\sqrt{4-5x}=12\)

\(\Leftrightarrow4-5x=144\)

\(\Leftrightarrow5x=-140\)

hay x=-28

b) Ta có: \(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\)

\(\Leftrightarrow\sqrt{3x}+10=10+4\sqrt{6}\)

\(\Leftrightarrow\sqrt{3x}=4\sqrt{6}\)

\(\Leftrightarrow3x=96\)

hay x=32

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

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1