\(231-2.\left(x-6\right)=1339:13\)

\(231-2x+12=103\)

\(2x+12=128\)

\(2x=116\)

\(x=58\)

Vậy \(x=58\)

     231 - 2(x - 6) = 1339 : 13

=> 231 - 2(x - 6) = 103

=> 2(x - 6)          = 231 - 103

=> 2(x - 6)          = 128

=> x - 6              = 128 : 2

=> x - 6              = 64

=> x                   = 64 + 6

=> x                   = 70

231 - 2 ( x - 6 ) = 1339 : 13

231 - 2x + 12 = 103

2x + 12 = 231 - 103

2x + 12 = 128

2x = 128 - 12

2x = 116

x = 116 : 2 

x = 58

Vậy x = 58

Ta có: \(\int\dfrac{xdx}{x^2+3}\)

Đặt \(u=x^2+3\left(u>0\right)\) 

Có \(du=2xdx\)

\(\Rightarrow\int\dfrac{xdx}{x^2+3}=\)\(\int\dfrac{du}{2u}=\dfrac{1}{2}ln\left(u\right)=\dfrac{1}{2}ln\left(x^2+3\right)\)

\(\Leftrightarrow4x^2-20x-4x^2+3x+12x-3=5\)

\(\Leftrightarrow-5x=8\)

hay \(x=-\dfrac{8}{5}\)

Áp dụng dãy tỉ số bằng nhau:

b.

\(\dfrac{x}{2}=\dfrac{y}{-5}=\dfrac{x-y}{2-\left(-5\right)}=\dfrac{-7}{7}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-1\right)=-2\\y=-5.\left(-1\right)=5\end{matrix}\right.\)

d.

\(\dfrac{4}{x}=\dfrac{7}{y}\Rightarrow\dfrac{y}{7}=\dfrac{x}{4}=\dfrac{y-x}{7-4}=\dfrac{-12}{3}=-4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.\left(-4\right)=-16\\y=7.\left(-4\right)=-28\end{matrix}\right.\)

65 - 5 . ( x + 2 ) = 15 

5 . ( x + 2 ) = 65 - 15 

5 . ( x + 2 ) = 50 

( x + 2 ) = 50 : 5 

x + 2 = 10 

x = 8 

Vậy x = 8 

\(65-5.\left(x+2\right)=15\)

\(5\left(x+2\right)=50\)

\(x+2=10\)

\(x=8\)

65 - 5(x + 2) = 15

=> 5(x + 2) = 50

=> 5x + 10 = 50

=> 5x = 40

=> x = 8

vậy_

=> 72 - 20x - 36x - 84 = 30x - 240 - 6x + 84

=> (72 - 84 )  - (20x + 36x ) = (30x - 6x ) - 240 + 84

=> -12 - 56x = 24x - 156

=> -12 + 156 = 24x + 56x 

=> 144 = 80x

=> x = 144  : 80

=> x = 9/5

=> 72 - 20x - 36x - 84 = 30x - 240 - 6x + 84

=> (72 - 84 )  - (20x + 36x ) = (30x - 6x ) - 240 + 84

=> -12 - 56x = 24x - 156

=> -12 + 156 = 24x + 56x 

=> 144 = 80x

=> x = 144  : 80

=> x = 9/5

đk: \(x\ge0\)

Ta có: \(\sqrt{x}+2\sqrt{x+3}=x+4\)

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

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

\(\Leftrightarrow\left|\sqrt{x-3}-1\right|=\sqrt{x}-1\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-3}-1=\sqrt{x}-1\\\sqrt{x-3}-1=1-\sqrt{x}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-3}=\sqrt{x}\left(ktm\right)\\\sqrt{x-3}+\sqrt{x}=2\end{cases}}\)

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

\(\Leftrightarrow2\sqrt{x^2-3x}=7-2x\)

\(\Leftrightarrow4\left(x^2-3x\right)=\left(7-2x\right)^2\)

\(\Leftrightarrow4x^2-12x=49-28x+4x^2\)

\(\Leftrightarrow16x=49\)

\(\Rightarrow x=\frac{49}{16}\)

\(( \sqrt{x+3}-1)^2\) chứ bạn.

với cả là \(\sqrt{x+3}\)  mà   có phải \(\sqrt{x-3} \)  đâu

Đặt A=2/3+2/6+2/12+...+2/768

=2/3(1+1/2+1/4+...+1/256)

Đặt B=1+1/2+1/4+...+1/256

=>2B=2+1+1/2+...+1/128

=>B=2-1/256=511/256

=>\(A=\dfrac{2}{3}\cdot\dfrac{511}{256}=\dfrac{511}{128\cdot3}=\dfrac{511}{384}\)