g: \(x-\frac{7}{12}x+\frac38x=\frac{5}{24}\)

=>\(x\left(1-\frac{7}{12}+\frac38\right)=\frac{5}{24}\)

=>\(x\left(\frac{24}{24}-\frac{14}{24}+\frac{9}{24}\right)=\frac{5}{24}\)

=>\(x\cdot\frac{19}{24}=\frac{5}{24}\)

=>\(x=\frac{5}{24}:\frac{19}{24}=\frac{5}{19}\)

h: \(\left(x-\frac13\right)^2-\frac12=1\frac34\)

=>\(\left(x-\frac13\right)^2-\frac12=\frac74\)

=>\(\left(x-\frac13\right)^2=\frac12+\frac74=\frac94\)

=>\(\left[\begin{array}{l}x-\frac13=\frac32\\ x-\frac13=-\frac32\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac32+\frac13=\frac{9+2}{6}=\frac{11}{6}\\ x=\frac{-3}{2}+\frac13=\frac{-9+2}{6}=-\frac76\end{array}\right.\)

ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

\(A=\left(\dfrac{3\sqrt{x}}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+13}{x+6\sqrt{x}+9}\)

\(=\left(\dfrac{3}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+13}{\left(\sqrt{x}+3\right)^2}\)

\(=\dfrac{3\sqrt{x}+9-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}\)

\(=\dfrac{\sqrt{x}+13}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}\)

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

Vậy...

Lâu ko ôn lại cũng hơi miss tích phân r :v

\(\int\limits^{\dfrac{-\pi}{4}}_{\dfrac{\pi}{4}}\tan x.dx\)

\(\int\tan x.dx=\int\dfrac{\sin x}{\cos x}.dx=-\int\dfrac{1}{\cos x}.d\left(\cos x\right)=-ln\left|\cos x\right|\)

\(\Rightarrow\int\limits^{\dfrac{-\pi}{4}}_{\dfrac{\pi}{4}}\tan x.dx=-ln\left|\cos\dfrac{-\pi}{4}\right|+ln\left|\cos\dfrac{\pi}{4}\right|\)

a)

= 8/22 + 5/22

= 11/22

= 1/2

b)

= 24/8 - 3/8

= 21/8

c)

= 45/25

= 9/5

d)

= 3/7 x 1/4

= 3/28

:)

dùng latex đi

B

B

Ta có: \(\left(x^2+3\right)^2-\left(x+3\right)\left(x-3\right)\left(x^2+9\right)\)

\(=x^4+6x^2+9-\left(x^2-9\right)\left(x^2+9\right)\)

\(=x^4+6x^2+9-x^4+81\)

\(=6x^2+90\)

a: \(3x^2\left(x^2-2x+5\right)\)

\(=3x^2\cdot x^2-3x^2\cdot2x+3x^2\cdot5\)

\(=3x^4-6x^3+15x^2\)

b: (x+1)(2x-3)

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

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

c: \(\left(2x^3-3x^2+x+15\right):\left(2x+3\right)\)

\(=\left(2x^3+3x^2-6x^2-9x+10x+15\right):\left(2x+3\right)\)

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

1/2* x+2/3=9/2

1/2 * x = 9/2 - 2/3 

1/2 * x= 23/6

x= 23/6 : 1/2

x= 23/6 x 2= 23/3

___

1/2*x-1/3=2/3

1/2*x = 2/3 + 1/3

1/2 * x= 1

x= 1: 1/2 

x= 2

____

1/4+3/4:x=3

3/4 : x = 3 - 1/4

3/4 : x= 11/4

x= 11/4 : 3/4

x= 11/3

\(\dfrac{1}{2}\)\(\times\)\(x\) + \(\dfrac{2}{3}\) = \(\dfrac{9}{2}\)

\(\dfrac{1}{2}\)\(\times\)\(x\)        = \(\dfrac{9}{2}\) - \(\dfrac{2}{3}\)

\(\dfrac{1}{2}\)\(\times\)\(x\)       = \(\dfrac{23}{6}\)

      \(x\)       = \(\dfrac{23}{6}\):\(\dfrac{1}{2}\)

      \(x\)      = \(\dfrac{23}{3}\) 

\(\dfrac{1}{2}\)\(\times\)\(x\) - \(\dfrac{1}{3}\) = \(\dfrac{2}{3}\) 

\(\dfrac{1}{2}\)\(\times\)\(x\)       = \(\dfrac{2}{3}\) + \(\dfrac{1}{3}\)

\(\dfrac{1}{2}\times\)\(x\)      =  1

     \(x\)       = 1 : \(\dfrac{1}{2}\)

   \(x\)         = 2

\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\): \(x\) = 3

          \(\dfrac{3}{4}\): \(x\) = 3 - \(\dfrac{1}{4}\) 

          \(\dfrac{3}{4}\):\(x\) = \(\dfrac{11}{4}\)

              \(x\) = \(\dfrac{3}{4}\): \(\dfrac{11}{4}\)

             \(x\) = \(\dfrac{3}{11}\)

Lời giải:
$x-\frac{x}{3}\times \frac{3}{2}=2-\frac{1}{2}$

$x-x\times \frac{1}{2}=\frac{3}{2}$

$x\times (1-\frac{1}{2})=\frac{3}{2}$

$x\times \frac{1}{2}=\frac{3}{2}$

$x=\frac{3}{2}: \frac{1}{2}=3$