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câu E
\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )
3, Tìm x, biết
\(d,\dfrac{-16}{x}=\dfrac{x}{-4}=>x^2=\left(-16\right).\left(-4\right)=>x^2=64\)
\(=>x=8\) hay \(x=-8\)
\(e,\dfrac{x}{-2}=\dfrac{\dfrac{8}{25}}{-x}=>-x^2=-2.\dfrac{8}{5}=\dfrac{-16}{25}\)
\(=>-x^2=0,64=>x=0,8\)
\(g,\dfrac{x}{-15}=\dfrac{-60}{x}\)
\(=>x^2=\left(-15\right).\left(-60\right)\)\(=>x^2=900=>x=30\) hay \(x=-30\)
d) \(\dfrac{-16}{x}=\dfrac{x}{-4}\)
= 16 . 4 = x.x
= 64 = \(x^2\)
= \(8^2=x^2\)
vậy x = 8
e)\(\dfrac{x}{-2}=\dfrac{8}{\dfrac{25}{-x}}\)
= -2 . \(\dfrac{8}{25}\) = -x . x
= -0,64 = \(-x^2\)
= 0,64 = \(x^2\)
0,8\(^2=x^2\)
vậy x = 0,8
g) \(\dfrac{x}{-15}=\dfrac{-60}{x}\)
= -15 . -60 = x.x
= 900 = \(x^2\)
30 \(^2=x^2\)
vậy x = 30
a) \(x^2=\left(-15\right).\left(-60\right)=900=>x=\)\(\pm\)\(30\)
b) \(-x^2=\dfrac{-16}{25}=>x^2=\dfrac{16}{25}=>x=\)\(\pm\)\(\dfrac{4}{5}\)
a)\(\dfrac{x}{-15}\)= \(-\dfrac{60}{x}\)
=> x . x = -15 . (-60)
=> \(^{x^2}\) = 900
x = 30
b) \(-\dfrac{2}{x}\) = \(-\dfrac{x}{\dfrac{8}{25}}\)
=> -2 . \(\dfrac{8}{25}\) = x . (-x)
=> \(\dfrac{-16}{25}\) = \(^{x^2}\)
=> x = \(\dfrac{4}{5}\)và \(-\dfrac{4}{5}\)
nhớ tích cho mk vs nha >_<
bài 1) ta có : \(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow2\left(x+y\right)=3\left(2x-y\right)\)
\(\Leftrightarrow2x+2y=6x-3y\Leftrightarrow4x=5y\Leftrightarrow\dfrac{x}{y}=\dfrac{5}{4}\)
vậy \(\dfrac{x}{y}=\dfrac{5}{4}\)
bài 1
\(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow\dfrac{2.\dfrac{x}{y}-1}{\dfrac{x}{y}+1}=\dfrac{2.\dfrac{x}{y}+2-3}{\dfrac{x}{y}+1}=2-\dfrac{3}{\dfrac{x}{y}+1}=\dfrac{2}{3}\)
\(2-\dfrac{2}{3}=\dfrac{4}{3}=\dfrac{3}{\dfrac{x}{y}+1}\)
\(\left(\dfrac{x}{y}+1\right)=\dfrac{9}{4}\Rightarrow\dfrac{x}{y}=\dfrac{9}{4}-\dfrac{4}{4}=\dfrac{5}{4}\)
a: Đặt A=0
=>-2/3x=5/9
hay x=-5/6
b: Đặt B(x)=0
=>(x-2/5)(x+2/5)=0
=>x=2/5 hoặc x=-2/5
c: Đặt C(X)=0
\(\Leftrightarrow x^3\cdot\dfrac{1}{2}=-\dfrac{4}{27}\)
\(\Leftrightarrow x^3=-\dfrac{8}{27}\)
hay x=-2/3
a: =>\(-\dfrac{6+x}{2}-\dfrac{3}{2}=2\)
=>-x-6-3=4
=>-x-9=4
=>-x=5
hay x=-5
b: =>(x+1)2=16
=>x+1=4 hoặc x+1=-4
=>x=3 hoặc x=-5
c: \(\Leftrightarrow\left(\dfrac{x-2}{27}-1\right)+\left(\dfrac{x-3}{26}-1\right)+\left(\dfrac{x-4}{25}-1\right)+\left(\dfrac{x-5}{24}-1\right)+\left(\dfrac{x-44}{5}+3\right)=0\)
=>x-29=0
hay x=29
Bài 2:
a: =>x^2=60
=>\(x=\pm2\sqrt{15}\)
b: =>2^2x+3=2^3x
=>3x=2x+3
=>x=3
c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)
\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)
=>1/2x-2=4
=>1/2x=6
=>x=12
\(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\left(ĐK:x\ne0\right)\)
\(\Rightarrow\dfrac{2}{x}=\dfrac{x}{\dfrac{8}{25}}\)
\(\Rightarrow\dfrac{2}{x}=x\cdot\dfrac{25}{8}\)
\(\Rightarrow\dfrac{2}{x}=\dfrac{25x}{8}\)
\(\Rightarrow25x\cdot x=2\cdot8\)
\(\Rightarrow25x^2=16\)
\(\Rightarrow x^2=\dfrac{16}{25}\)
\(\Rightarrow x^2=\left(\dfrac{4}{5}\right)^2\)
\(\Rightarrow x=\pm\dfrac{4}{5}\)
Vậy:\(x\in\left\{-\dfrac{4}{5};\dfrac{4}{5}\right\}\)
-2/x = -x/(8/25)
x.(-x) = -2.(8/25)
-x² = -16/25
x² = 16/25
x = -4/5 hoặc x = 4/5