\(\dfrac{2}{x}=\dfrac{x}{8}\Rightarrow x^2=16=4^2=\left\{{}\begin{matrix}+4\\-4\end{matrix}\right.\)\(A=\dfrac{4116-14}{10290-35}=\dfrac{4102}{10255}=\dfrac{2}{3}\)

a, (x² - 5)(x² - 24) < 0

=> x² - 5 và x² - 24 trái dấu

Mà x² - 5 > x² - 24 => \(\hept{\begin{cases}x^2-5>0\\x^2-24>0\end{cases}\Rightarrow5< x^2< 24}\)

Vì x \(\in\)Z nên x² = 9;16

+) x² = 9 => x = 3 hoặc x = -3

+) x² = 16 => x = 4 hoặc x = -4

Vậy...

b,

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Mà \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)

=> x + 1 = 0 => x = 0 - 1 => x = -1

\(\frac{x+1}{14}+\frac{x+2}{13}=\frac{x+3}{12}+\frac{x+4}{11}\)

\(\Rightarrow\left(\frac{x+1}{14}+1\right)+\left(\frac{x+2}{13}+1\right)=\left(\frac{x+3}{12}+1\right)+\left(\frac{x+4}{11}+1\right)\)

\(\Rightarrow\frac{x+15}{14}+\frac{x+15}{13}=\frac{x+15}{12}+\frac{x+15}{11}\)

\(\Rightarrow\frac{x+15}{14}+\frac{x+15}{13}-\frac{x+15}{12}-\frac{x+15}{11}=0\)

\(\Rightarrow\left(x+15\right)\left(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\right)=0\)

Mà \(\left(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\right)\ne0\)

=> x + 15 = 0 => x = 0 - 15 => x = -15

Áp dụng công thức \(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\) ta được:

    \(\frac{x+2}{x+6}=\frac{3}{x+1}\)

      \(\Rightarrow\left(x+1\right)\left(x+2\right)=3\left(x+6\right)\)

       \(\Leftrightarrow x^2+3x+2=3x+18\)

        \(\Leftrightarrow x^2=16\)

 Vậy \(x\in\left\{4;-4\right\}\)

(x+2)/(x+6)=3/(x+1)

<=>  (x+2)(x+1)/(x+6)(x+1)=3(x+6)/(x+6)(x+1)

=>(x+2)(x+1)=3(x+6)

<=> x^2+x+2x+2=3x+18

<=> x^2=16

<=>x^2=4^2 hoặc (-4)^2

<=> x=4 hoặc x=-4

Vậy......... 

Ta có:

\(\frac{x+2}{x+6}=\frac{3}{x+1}\)

Áp dụng tỉ lệ thức ta có:

\(\left(x+2\right).\left(x+1\right)=\left(x+6\right).3\)

\(\Rightarrow x.\left(x+2\right)+2.\left(x+1\right)=x.3+6.3\)

\(\Rightarrow x^2+x+2x+2=3x+18\)

\(\Rightarrow x^2+3x+2=3x+18\)

\(\Leftrightarrow x^2+2=18\)

\(\Leftrightarrow x=\sqrt{18-2}=4\)

\(\frac{x-3}{7-5x}=\frac{1}{x-2}\)

\(\Rightarrow\left(x-3\right)\left(x-2\right)=7-5x\)

\(\Rightarrow x^2-2x-3x+6=7-5x\)

\(\Rightarrow x^2-2x-3x+5x=7-6\)

\(\Rightarrow x^2=1\Rightarrow\orbr{\begin{cases}x=-1\\x=1\end{cases}}\)

k nhé,Vy Nguyễn Đặng Khánh !

nhân tích chéo

\(\frac{x-3}{7-5x}=\frac{1}{x-2}\)

\(\Rightarrow\left(x-3\right)\left(x-2\right)=1\left(7-5x\right)\)

\(\Leftrightarrow x^2-3x-2x+6=7-5x\)

\(\Leftrightarrow x^2-1=0\)

\(\Leftrightarrow x^2=1\Leftrightarrow x=1\)

vậy x=1

<=>(x-3)(x-2)=7-5x

<=>\(x^2-2x-3x+6=7-5x\)

<=>\(x^2-2x-3x+5x=7-6\)

<=>\(x^2=1\)

Vậy \(x\in\left\{1;-1\right\}\)

Ta có:

4116 / 10290 = 7.7.7.712 - 7.2 / 7.7.7.7 30-7.5

                   = 12-2 / 30-5

                  = 10 / 25 

                = 2 / 5

  nhớ k cho mình nha

câu 

này =2/5 

k mk nha

7.7.7.7.12 -2.7 /  7.7.7.7.30- 7.5 nha mình quên dấu nhân

A=4102/10255<=>2051*2/2051*5<=>A=2/5

B=2828/4242<=>1414*2/1414*3<=>B=2/3

ủng hộ nha mn
 

\(\frac{x}{5}\le\frac{12}{x}\Rightarrow x^2\le60\left(1\right)\)

\(\frac{12}{x}\le\frac{x}{3}\Rightarrow x^2\ge36\left(2\right)\)

Từ (1) và (2)

\(\Rightarrow36\le x^2\le60\) và \(x\in N\)

\(\Rightarrow6\le x\le7,75\)

Vậy \(x=6;7\)

Ta có:-2/4=x/10=>4.x=-2.10=>4.x=-20=>x=-20:4=>x=-5

Thay x=-5 ta có:-5/10=-7/y=>10.(-7)=y.(-5)=>y.(-5)=-70=>y=-70:(-5)=>y=14

Thay y=14 ta có:-7/14=z/-24=>14.z=-24.(-7)=>14.z=168=>z=168:14=>z=12

Vậy x=-5;y=14;z=12

Đúng 100% luôn nha!

$M=(\frac{1}{x+1}+\frac{2}{1-x}+\frac{x}{x^2-1})\div \frac{1}{x+1}\\=-\frac{3}{(x-1)(x+1)}\times (x+1)\\=-\ ...