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\(\frac{x-20}{x-10}=\frac{x+40}{x+70}=\frac{-20-40}{-10-70}=\frac{6}{8}=\frac{3}{4}.\)
\(\Rightarrow4\cdot\left(x-20\right)=3\cdot\left(x-10\right)\Leftrightarrow4x-80=3x-30\Leftrightarrow x=50.\)
bn dinh thuy linh gioi that, toàn dung ngôn ngữ toán hoc nên bai lam rat ấn tuong
\(\frac{x-20}{x-10}=\frac{x+40}{x+70}\)
\(\Leftrightarrow\left(x-20\right)\left(x+70\right)=\left(x-10\right)\left(x+40\right)\)
\(\Leftrightarrow x^2+70x-20x-140=x^2+40x-10x-140\)
\(\Leftrightarrow50x=30x\)
\(\Leftrightarrow50x-30x=0\)
\(\Leftrightarrow20x=0\)
\(\Leftrightarrow x=0\)
Chết , sửa chút !!! Nhân ngu wá
\(\frac{x-20}{x-10}=\frac{x+40}{x+70}\)
\(\Leftrightarrow\left(x-20\right)\left(x+70\right)=\left(x-10\right)\left(x+40\right)\)
\(\Leftrightarrow x^2+70x-20x-140=x^2+40x-10x-400\)
\(\Leftrightarrow20x+260=0\)
\(\Leftrightarrow x=-13\)
a, \(\left(7,5-x\right):\left(3,5+x\right)=5:6\)
\(\Leftrightarrow\frac{7,5-x}{3,5+x}=\frac{5}{6}\)
\(\Leftrightarrow6.\left(7,5-x\right)=5.\left(3,5+x\right)\)
\(\Leftrightarrow45-6x=17,5+5x\)
\(\Leftrightarrow45-17,5=5x+6x\)
\(\Leftrightarrow27,5=11x\)
\(\Leftrightarrow x=\frac{27,5}{11}=2,5\)
Vậy : \(x=2,5\)
b) Tương tự như câu a.
Chúc bạn học tốt nhé !!
a)\(\frac{x+7}{x}=\frac{4}{5}\)
=>5(x+7)=4x
5x+35=4x
5x-4x=-35
x=-35
Vậy x=-35
b)(x-20)/(x+40)=(x+70)/(x-10)
(x-20)*(x-10)=(x+40)*(x+70)
x(x-10)-20(x-10)=x(x+70)+40(x+70)
x2-10x-20x+200=x2+70x+40x+2800
x2-x2-10x-20x-70x-40x=2800-200
0-(10x+20x+70x+40x)=2600
-140x=2600
x=2600/(-140)
x=\(-\frac{130}{7}\)
Vậy x=-130/7
a) (x + 15) : x = 4 : 3
=> x : x + 15 : x = \(\frac{4}{3}\)
=> 1 + 15 : x = \(\frac{4}{3}\)
=> 15 : x = \(\frac{4}{3}\)- 1 = \(\frac{1}{3}\)
=> x = 15 : \(\frac{1}{3}\)
=> x = 45
\(\frac{x-20}{x-10}=\frac{x+40}{x+70}\)
\(\Rightarrow\) (x - 20)(x + 70) = (x - 10)(x + 40)
\(\Rightarrow\) x2 + 70x - 20x - 140 = x2 + 40x - 10x - 400
\(\Rightarrow\) x2 + 70x - 20x - x2 - 40x + 10x = - 400 + 140
\(\Rightarrow\) 20x = 260
\(\Rightarrow\) x = 130
đề như vậy hả bạn?
\(\frac{x-20}{x-10}=\frac{x+40}{x+70}\)
=> \(\left(x-20\right)\left(x+70\right)=\left(x+40\right)\left(x-10\right)\)
=> \(x^2+70x-20x-1400=x^2-10x+40x-400\)
=> \(\left(x^2-x^2\right)+\left(70x-20x+10x-40x\right)=-400+1400\)
=> \(-43x=1000\Rightarrow x=\frac{-1000}{43}\)