Bài 2 

b, `\sqrt{3x^2}=x+2`          ĐKXĐ : `x>=0`

`=>(\sqrt{3x^2})^2=(x+2)^2`

`=>3x^2=x^2+4x+4`

`=>3x^2-x^2-4x-4=0`

`=>2x^2-4x-4=0`

`=>x^2-2x-2=0`

`=>(x^2-2x+1)-3=0`

`=>(x-1)^2=3`

`=>(x-1)^2=(\pm \sqrt{3})^2`

`=>` $\left[\begin{matrix} x-1=\sqrt{3}\\ x-1=-\sqrt{3}\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=1+\sqrt{3}\\ x=1-\sqrt{3}\end{matrix}\right.$

Vậy `S={1+\sqrt{3};1-\sqrt{3}}`

mình nghĩ ĐKXĐ là như này : 

x+2≥0

➩ x≥-2

có phải k

ĐKXĐ: \(-2\le x\le3\)

\(\dfrac{\sqrt{-x^2+x+6}}{2x+5}-\dfrac{\sqrt{-x^2+x+6}}{x-4}\ge0\)

\(\Leftrightarrow\sqrt{-x^2+x+6}\left(\dfrac{1}{2x+5}-\dfrac{1}{x-4}\right)\ge0\)

\(\Leftrightarrow\dfrac{\left(-x-9\right)\sqrt{x^2+x+6}}{\left(2x+5\right)\left(x-4\right)}\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x^2+x+6=0\\\dfrac{-x-9}{\left(2x+5\right)\left(x-4\right)}\ge0\end{matrix}\right.\) \(\Leftrightarrow-2\le x\le3\)

Hoặc có thể biện luận như sau:

Ta có: \(\left\{{}\begin{matrix}2x+5>0;\forall x\in\left[-2;3\right]\\x-4< 0;\forall x\in\left[-2;3\right]\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{-x^2+x+6}}{2x+5}\ge0\\\dfrac{\sqrt{-x^2+x+6}}{x-4}\le0\end{matrix}\right.\) ; \(\forall x\in\left[-2;3\right]\)

Do đó nghiệm của BPT là \(-2\le x\le3\)

???

\(B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)+\dfrac{2-2\sqrt{x}}{\sqrt{x}}(x \geq 0,x \neq 1\)
`=((2x+1-x+\sqrtx)/(x\sqrtx-1))(((\sqrtx+1)(x-\sqrtx+1))/(\sqrtx+1)-\sqrtx)+(2-2sqrtx)/sqrtx`
`=((x-\sqrtx+1)/((\sqrtx-1))(x+sqrtx+1)))(x-2\sqrtx+1)-(2\sqrtx-2)/sqrtx`
`=(1/(\sqrtx-1))(\sqrtx-1)^2-(2(\sqrtx-1))/sqrtx`
`=\sqrtx-1-(2(\sqrtx-1))/sqrtx`
`=(x-\sqrtx-2\sqrtx+2)/sqrtx`
`=(x-3sqrtx+2)/sqrtx`

a: ĐKXĐ: -21<=x<=21 và x<>0

Ta có: \(\frac{\sqrt{21+x} + \sqrt{21-x}}{\sqrt{21+x} - \sqrt{21-x}} = \frac{21}{x}\)

=>\(\frac{(\sqrt{21+x} + \sqrt{21-x})^2}{(21+x) - (21-x)} = \frac{21}{x}\)

=>\(\frac{(21+x) + (21-x) + 2\sqrt{(21+x)(21-x)}}{2x} = \frac{21}{x}\)

=>\(\frac{42 + 2\sqrt{441 - x^2}}{2x} = \frac{21}{x}\)

=>\(\frac{21 + \sqrt{441 - x^2}}{x} = \frac{21}{x}\)

=>\(21+\sqrt{441-x^2}=21\)

=>\(\sqrt{441-x^2}=0\)

=>\(441-x^2=0\)

=>\(x^2=441\)

=>x=21(nhận) hoặc x=-21(nhận)

b: ĐKXĐ: x∈R

\(\left(\sqrt[3]{x+1}+\sqrt[3]{3x+1}\right)^3\)

\(=x+1+3x+1+3\cdot\sqrt[3]{\left(x+1\right)\left(3x+1\right)}\cdot\left(\sqrt[3]{x+1}+\sqrt[3]{3x+1}\right)\)

=4x+2+\(3\cdot\sqrt[3]{\left(x+1\right)\left(3x+1\right)}\cdot\left(\sqrt[3]{x+1}+\sqrt[3]{3x+1}\right)\)

Phương trình ban đầu sẽ trở thành:

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

=>\(3\sqrt[3]{(x^2-1)(3x+1)}=-3x-3\)

=>\(\sqrt[3]{(x^2-1)(3x+1)}=-(x+1)\)

=>\((x^2-1)(3x+1) = -(x+1)^3\)

=>\((x-1)(x+1)(3x+1) + (x+1)^3 = 0\)

=>(x+1)(3x^2+x-3x-1+x^2+2x+1)=0

=>(x+1)(4x^2)=0

=>x=0 hoặc x=-1

Khi x=0 thì \(\sqrt[3]{1}+\sqrt[3]{1}=2<>\sqrt[3]{-1}=-1\) (loại)

Khi x=-1 thì \(\sqrt[3]{0}+\sqrt[3]{-2}=\sqrt[3]{-2}\) (nhận)

b: ĐKXĐ: 8/3<=x<=7

Ta có: \(\sqrt{x+3} - \sqrt{7-x} > \sqrt{3x-8}\)

=>\(\sqrt{x+3}>\sqrt{3x-8}+\sqrt{7-x}\)

=>\(x + 3 > (3x - 8) + (7 - x) + 2\sqrt{(3x-8)(7-x)}\)

=>\(x + 3 > 2x - 1 + 2\sqrt{(3x-8)(7-x)}\)

=>\(4 - x > 2\sqrt{-3x^2 + 29x - 56}\) (1)

Để (1) có nghiệm thì 4-x>0

=>x<4

=>8/3<=x<4

(1) sẽ tương đương: \((4 - x)^2 > 4(-3x^2 + 29x - 56)\)

=>\(16 - 8x + x^2 > -12x^2 + 116x - 224\)

=>\(13x^2 - 124x + 240 > 0\)

=>\(x^2-\frac{124}{13}x+\frac{240}{13}>0\)

=>\(x^2-2\cdot x\cdot\frac{62}{13}+\frac{3844}{169}-\frac{3844}{169}+\frac{240}{13}>0\)

=>\(\left(x-\frac{62}{13}\right)^2>\frac{3844}{169}-\frac{240}{13}=\frac{724}{169}\)

=>\(\left[\begin{array}{l}x-\frac{62}{13}>\frac{\sqrt{724}}{13}\\ x-\frac{62}{13}<-\frac{\sqrt{724}}{13}\end{array}\right.\Rightarrow\left[\begin{array}{l}x>\frac{62+\sqrt{724}}{13}\\ x<\frac{62-\sqrt{724}}{13}\end{array}\right.\)

mà 8/3<=x<4

nên \(\frac{8}{3} \le x < \frac{62 - 2\sqrt{181}}{13}\)

Câu 1: \(x\sqrt{\dfrac{y^2}{x}}=\sqrt{\dfrac{y^2}{x^3}}\)

cho m hỏi là sao ra đc cái đó vậy ạ, bn có thể lm kĩ hơn 1 xíu đc kh

\(c,=2+2\sqrt{3}-\left(2+\sqrt{2}\right)=2\sqrt{3}-\sqrt{2}\\ d,=\sqrt{\left(2x-3\right)^2}-2x+1=\left|2x-3\right|-2x+1\\ =2x-3-2x+1=-2\left(x\ge\dfrac{3}{2}\Leftrightarrow2x-3\ge0\right)\)

a: ĐKXĐ: x>=2

Ta có: \(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\frac{x-2}{4}}=26\)

=>\(4\sqrt{x-2}+3\sqrt{x-2}-\frac12\cdot\sqrt{x-2}=26\)

=>\(6,5\cdot\sqrt{x-2}=26\)

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

=>x-2=16

=>x=18(nhận)

b: ĐKXĐ: x∈R

\(3x+\sqrt{4x^2-8x+4}=1\)

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

=>3x+|2x-2|=1

=>3x-1+|2x-2|=0(1)

TH1: x>=1

(1) sẽ trở thành: 3x-1+2x-2=0

=>5x-3=0

=>5x=3

=>x=3/5(loại)

TH2: x<1

(1) sẽ trở thành: 3x-1-2x+2=0

=>x+1=0

=>x=-1(nhận)

c: ĐKXĐ: x>=0

Ta có: \(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\)

=>\(2x-4\sqrt{x}+\sqrt{x}-2-7=0\)

=>\(2x-3\sqrt{x}-9=0\)

=>\(2x-6\sqrt{x}+3\sqrt{x}-9=0\)

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

=>\(\sqrt{x}-3=0\)

=>\(\sqrt{x}=3\)

=>x=9(nhận)