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\(\frac{3}{x-2}=\frac{-2}{\frac{1}{8}}\)
\(\Rightarrow\frac{3}{8}=-2\left(x-2\right)\)
\(\frac{3}{8}=-2x-\left(-4\right)\)
\(\frac{3}{8}=-2x+4\)
\(-2x=\frac{3}{8}-4=\frac{-29}{8}\)
\(x=\frac{-29}{8}.\frac{1}{-2}=\frac{29}{16}\)
\(\frac{3}{x-2}=-\frac{2}{\frac{1}{8}}\)
\(\Rightarrow\frac{3}{x-2}=\frac{\left(-2\right).8}{1}\)
\(\Rightarrow\frac{3}{x-2}=-16\)
\(\Rightarrow x-2=3:\left(-16\right)\)
\(\Rightarrow x-2=-\frac{3}{16}\)
\(\Rightarrow x=-\frac{3}{16}+2\)
\(\Rightarrow x=\frac{29}{16}\)
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
\(\frac{x+2}{x+6}=\frac{3}{x+1}\)
\(\Rightarrow\left(x+2\right)\left(x+1\right)=3\left(x+6\right)\)
\(\Rightarrow x^2+x+2x+2=3x+18\)
\(\Rightarrow x^2+x+2x-3x=18-2\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x=\pm4\)
các phần còn lại tương tự :)
a)\(\frac{x+2}{x+6}\) =\(\frac{3}{x+1}\)
<=>\(\frac{\left(x+2\right)\left(x+1\right)}{\left(x+6\right)\left(x+1\right)}\) =\(\frac{3\left(x+6\right)}{\left(x+1\right)\left(x+6\right)}\)
=> ( x+2) ( x+1) = 3(x+6)
<=> x2 +3x +3 = 3x +18
<=> x2 +3x -3x = 18 -3
<=> x2 = 15
=> x = \(\sqrt{15}\)
Vậy x=\(\sqrt{15}\)
b)
I x - 1/2 I + I x - 1/3 I + I x - 1/6 I = x
I 3x I - ( 1/2 + 1/3 + 1/6 ) = x
I 3x I - 1 = x
=> 2x = 1
x = 1 : 2
x = 0,5
\(\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x+2\right)\left(x-2\right)\)
\(\Rightarrow x^2+2x-3=x^2-4\)
\(\Rightarrow2x=-1\Rightarrow x=-\frac{1}{2}\)
\(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
=> (x-1)(x+3) = (x+2)(x-2) ( tích chéo bạn nhé :)))))
=> x^2 - x + 3x - 3 = x^2 + 2x - 2x - 4 ( bước này là mình nhân tung ra bạn nhé ^.^)
=> x^2 + 2x - 3 = x^2 - 4
=> x^2 +2x = x^2 - 4 +3
=> x^2 + 2x = x^2 - 1
=> 2x = -1 ( bớt cả 2 vế đi x^2)
=> x = \(\frac{-1}{2}\)
đúng 100% nha
ủng hộ mk đi bạn
a)Ta có : 2x+2y-z-7=0 => 2x+2y-z=7
Ta có : \(x=\frac{y}{2}=>\frac{x}{2}=\frac{y}{4}\)
Mà \(\frac{y}{4}=\frac{z}{5}\)nên \(\frac{x}{2}=\frac{y}{4}=\frac{z}{5}=\frac{2x}{4}=\frac{2y}{8}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x}{2}=\frac{y}{4}=\frac{z}{5}=\frac{2x}{4}=\frac{2y}{8}=\frac{2x+2y-z}{4+8-5}=\frac{7}{7}=1\)
Từ \(\frac{x}{2}=1=>x=2\)
Từ\(\frac{y}{4}=1=>y=4\)
Từ \(\frac{z}{5}=1=>z=5\)
\(\frac{x}{2}=\frac{y}{4}=\frac{z}{5}=\frac{2x}{4}=\frac{2y}{8}\)
a, x/-3=-27/y
=> x.x=(-27).(-3)
=> x^2=81
=x thuộc {9;-9}
Mấy câu sau làm tương tự nhé!
a) \(\frac{x}{-3}=\frac{-27}{x}\Leftrightarrow x^2=\left(-3\right).\left(-27\right)\)
x2=81 <=> x2=92
Vậy x=9 hoặc x=-9
b) \(\frac{5}{-x}=\frac{x}{-20}\Leftrightarrow5.\left(-20\right)=x.\left(-x\right)\)
-100=-x2 <=>-(100)=-(x2)=-(52)=-(x2)
=> x=5 hoặc x=-5
c) \(\frac{-x}{3}=\frac{48}{-x}\Leftrightarrow\left(-x\right)\left(-x\right)=3.48\)
x2=144 <=>x2=122
Vậy x=12 hoặc x=-12
d) \(\frac{-2}{x}=\frac{-x}{18}\Leftrightarrow\left(-2\right).18=x.\left(-x\right)\)
-36=-x2<=>-(36)=-(x2)<=>-(62)=-(x2)
=>x=6 hoặc x-6
ta có:\(\frac{x-1}{x+2}=\frac{x-2}{x+3}\Rightarrow\left(x-1\right)\times\left(x+3\right)=\left(x+2\right)\times\left(x-2\right)\)
\(\Rightarrow\) \(x^2-x+3x-3=x^2+2x-2x-4\)
\(\Rightarrow\) \(x^2+2x-3=x^2-4\)
\(\Rightarrow\) \(x^2-x^2+2x=3-4\)
\(\Rightarrow\) \(2x=-1\)
\(\Rightarrow x=-1:2\)
\(\Rightarrow x=\frac{-1}{2}\)