b: ĐKXĐ: \(x\in R\)

c: ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\x\le0\end{matrix}\right.\)

ĐKXĐ: \(x\ge\frac{1}{3}\)

\(x^2+5x=x\sqrt{3x-1}+\left(x+1\right)\sqrt{5x}\)

\(\Leftrightarrow2x^2+10x-2x\sqrt{3x-1}-2\left(x+1\right)\sqrt{5x}=0\)

\(\Leftrightarrow\left(x^2-2x\sqrt{3x-1}+3x-1\right)+\left[\left(x+1\right)^2-2\left(x+1\right)\sqrt{5x}+5x\right]=0\)\(\Leftrightarrow\left(x-\sqrt{3x-1}\right)^2+\left(x+1-\sqrt{5x}\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-\sqrt{3x-1}=0\\x+1-\sqrt{5x}=0\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-3x+1=0\\x^2-3x+1=0\end{matrix}\right.\Leftrightarrow x=\frac{3\pm\sqrt{5}}{2}\left(tm\right)\)

a: ĐKXĐ: \(\left[{}\begin{matrix}x\ge2\\x\le-5\end{matrix}\right.\)

b: ĐKXĐ: \(x=2\)

c: ĐKXĐ: \(x\ge4\)

\(\left|B\right|\)+3<2x-1

\(\Leftrightarrow\)\(\left[{}\begin{matrix}B+3< 2x-1\\-B+3< 2X-1\end{matrix}\right.\)

+ B+3<2x-1

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

\(\Leftrightarrow\dfrac{4-3x}{4}+\dfrac{12}{4}< 2x-1\)

\(\Leftrightarrow4-3x+12< 2x-1\)

\(\Leftrightarrow16-3x< 2x-1\)

\(\Leftrightarrow-5x< -17\)

\(\Leftrightarrow x>\dfrac{17}{5}\)

+ -B+3<2x-1

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

\(\Leftrightarrow\dfrac{-4+3x}{-4}+\dfrac{-12}{-4}< 2x-1\)

\(\Leftrightarrow-4+3x-12< 2x-1\)

\(\Leftrightarrow x< 15\)

Vậy:ĐK của x để \(\left|B\right|+3< 2x-1\)

                             B=\(\dfrac{4-3x}{4}\)                

                            \(\left(x|x>\dfrac{17}{5};x< 15\right)\)

Lời giải:

Ta có:

\(P=\int \frac{x^2-1}{x\sqrt{x^3+x}}dx=\int \frac{\frac{x^2-1}{x^2}}{\frac{\sqrt{x^3+x}}{x}}dx\)

\(=\int \frac{(1-\frac{1}{x^2})dx}{\frac{\sqrt{x^3+x}}{x}}=\int \frac{d\left(x+\frac{1}{x}\right)}{\frac{\sqrt{x^3+x}}{x}}\)

Đặt \(\frac{\sqrt{x^3+x}}{x}=t\Rightarrow t^2=\frac{x^3+x}{x^2}=x+\frac{1}{x}\)

Khi đó: \(P=\int \frac{d(t^2)}{t}=\int \frac{2tdt}{t}=\int 2dt=2t+c=\frac{2\sqrt{x^3+x}}{x}+c\)

Làm xong k lun

\(\dfrac{3}{2}\sqrt{3x}-3x-5=-\dfrac{1}{2}\sqrt{3x}\left(đk:x\ge0\right)\)

\(\Leftrightarrow3x-2\sqrt{3x}+5=0\)

\(\Leftrightarrow\left(\sqrt{3x}-1\right)^2+4=0\)(vô lý do \(\left(\sqrt{3x}-1\right)^2+4\ge4>0\))

Vậy \(S=\varnothing\)

a,3/4 . (-5/12)+3/4.(-7/12)

` 3/4 . [ - ( 5/12 + 7/12  ) ] `

`3/4 . (-1) = -3/4 `

`2/3 . x - 0,5 = 3/4 `

`        x - 0,5 = 3/4 - 2/3 `

`        x-0,5 = 1/12 `

`        x =       1/12 + 0,5 `

`        x= 7/12 `

a, 3/4 . (-5/12)+3/4.(-7/12) 

= 3/4 . [(-5/12) + (-7/12)]

= 3/4 . (-1)

= -3/4

----------------------------------------------------------------------------

a, 2/3 .x-0,5=3/4

2/3 . x = 3/2 + 0,5

2/3 . x = 2 

x = 2 : 2/3

x = 3 

vậy x = 3

Bài 2:

a: =>2/3x=3/4+1/2=3/4+2/4=5/4

=>x=5/4:2/3=5/4*3/2=15/8

b:=>-2x+4=3x-12

=>-5x=-16

=>x=16/5