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\(\dfrac{x}{3}=\dfrac{-10}{6}\)
\(\Leftrightarrow6x=-10.3\)
\(\Leftrightarrow6x=-30\)
\(\Leftrightarrow x=-30:6\)
\(\Leftrightarrow x=-5\)
Vậy \(x=-5\)
a) 2/7 : x = 11/6 : 7/12
2/7 : x = 22/7
x = 2/7 : 22/7
x = 1/11
b) (2 - x)/3 = -3/(x - 2)
(2 - x)(x - 2) = -3.3
-(x - 2)² = -9
(x - 2)² = 9
x - 2 = 3 hoặc x - 2 = -3
*) x - 2 = 3
x = 3 + 2
x = 5
*) x - 2 = -3
x = -3 + 2
x = -1
Vậy x = -1; x = 5
c) (x - 1)/(x + 2) = 2/3
3(x - 1) = 2(x + 2)
3x - 3 = 2x + 4
3x - 2x = 4 + 3
x = 7
a) 2/7 : x = 11/6 : 7/12
2/7 : x = 22/7
x = 2/7 : 22/7
x = 1/11
b) (2 - x)/3 = -3/(x - 2)
(2 - x)(x - 2) = -3.3
-(x - 2)² = -9
(x - 2)² = 9
x - 2 = 3 hoặc x - 2 = -3
*) x - 2 = 3
x = 3 + 2
x = 5
*) x - 2 = -3
x = -3 + 2
x = -1
Vậy x = -1; x = 5
c) (x - 1)/(x + 2) = 2/3
3(x - 1) = 2(x + 2)
3x - 3 = 2x + 4
3x - 2x = 4 + 3
x = 7
Bài 2:
a:
1: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)
hay \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
P/s : Vì bạn lớp 7 nên mình giải chi tiết nha!
Có \(\frac{x-1}{x+2}=\frac{x-2}{x+3}\Leftrightarrow\left(x-1\right)\left(x+3\right)=\left(x+2\right)\left(x-2\right)\)
\(\Leftrightarrow x\left(x+3\right)-1\left(x+3\right)=x\left(x-2\right)+2\left(x-2\right)\)
\(\Leftrightarrow x^2+3x-x-3=x^2-2x+2x-4\)
\(\Leftrightarrow x^2+\left(3x-x\right)-3=x^2+\left(2x-2x\right)-4\)
\(\Leftrightarrow x^2+2x-3=x^2-4\Leftrightarrow x^2-x^2+2x=3-4=-1\)
\(=\left(x^2-x^2\right)+2x=-1\Leftrightarrow2x=1\Leftrightarrow x=1\div2=\frac{1}{2}\)
a, \(\frac{x+2}{5}=\frac{1}{x-2}\Rightarrow\left(x+2\right)\left(x-2\right)=5\Rightarrow x^2-2x+2x-4=5\Rightarrow x^2=9\Rightarrow x=\pm3\)
b, \(\frac{3}{x-4}=\frac{x+4}{3}\Rightarrow\left(x+4\right)\left(x-4\right)=9\Rightarrow x^2-4x+4x-16=9\Rightarrow x^2=25\Rightarrow x=\pm5\)
c, \(\frac{x+2}{2}=\frac{1}{1-x}\Rightarrow\left(x+2\right)\left(1-x\right)=2\Rightarrow x-x^2+2-2x=2\Rightarrow-x^2-x=0\Rightarrow-x\left(x+1\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
\(\frac{7}{x-1}=\frac{x+1}{9}\)
\(\Rightarrow\left(x-1\right)\left(x+1\right)=7.9\)
\(\Rightarrow x^2-1=63\)
\(\Rightarrow x^2=64\)
\(\Rightarrow\orbr{\begin{cases}x=8\\x=-8\end{cases}}\)
\(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
\(\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x+2\right)\left(x-2\right)\)
\(\Rightarrow x^2+2x-3=x^2-4\)
\(\Rightarrow2x-3=4\)
\(\Rightarrow2x=7\)
\(\Rightarrow x=\frac{7}{2}\)
Ta có :\(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
=> (x + 2)(x - 2) = (x - 1)(x + 3)
=> x2 - 2x + 2x - 4 = x2 - x + 3x - 3
=> x2 - 4 = x2 + 2x - 3
=> 2x - 3 = -4
=> 2x = -1
=> x = -1/2
Vậy x = -1/2 là giá trị cần tìm