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

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

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

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

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

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 \(x\in\left\{1;-1\right\}\)

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:

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

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

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

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

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

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

vừa chị làm rồi em ko hiểu à?

Ta có: \(\dfrac{-3}{7}-\dfrac{4}{7}:x=-2\)

\(\Leftrightarrow\dfrac{4}{7}:x=\dfrac{-3}{7}+2=\dfrac{11}{7}\)

hay \(x=\dfrac{11}{7}:\dfrac{4}{7}=\dfrac{11}{7}\cdot\dfrac{7}{4}=\dfrac{11}{4}\)

Vậy: \(x=\dfrac{11}{4}\)

KO CẦN ĐỌC TIẾP NX

đag thi à?

a, \(x-1\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)

b, \(2x-1\inƯ\left(-4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

c, \(\dfrac{3\left(x-1\right)+10}{x-1}=3+\dfrac{10}{x-1}\Rightarrow x-1\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

d, \(\dfrac{4\left(x-3\right)+3}{-\left(x-3\right)}=-4-\dfrac{3}{x+3}\Rightarrow x+3\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)

1200-31/40=47969/40

câu 2 =45/56+25/56=(45+25)/56=70/56=5/4

\(\Rightarrow\left(x+1\right)\times\left(x-1+1\right):2=1225\\ \Rightarrow\left(x+1\right)\times x=1225\times2=2450=49\times50\\ \Rightarrow x=49\)

a \(\dfrac{6}{15}+\dfrac{11}{15}=\dfrac{17}{15}\)

b \(\dfrac{22}{77}-\dfrac{14}{77}=\dfrac{8}{77}\)

c \(\dfrac{11}{13}\times\dfrac{26}{31}=11\times\dfrac{2}{31}=\dfrac{22}{31}\)

d \(\dfrac{1}{2}\times3\times\dfrac{2}{5}=\dfrac{3}{5}\)

a, 2/5 + 11/15 = 6/15 + 11/15 = 17/15

b,2/7 - 2/11 = 22/77 - 14/77 = 8/77

c, 11/13 x 26/31= 11 x 2/31 =22/31

d, 1/2 : 1/3 x 2/5

=  ( 1/2 x 2/5 ) : 1/3

=     1/5        : 1/3

=         3/5

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

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

Từ (1) và (2)

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

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

Vậy \(x=6;7\)