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DN
17 tháng 12 2021
a.\(x-\dfrac{2}{3}=\dfrac{8}{7}\)
\(x=\dfrac{8}{7}+\dfrac{2}{3}\)
x=\(\dfrac{38}{21}\)
b.\(\left(x+\dfrac{1}{3}\right)=\dfrac{4}{25}
\)
x=\(\dfrac{4}{25}-\dfrac{1}{3}\)
x=\(-\dfrac{13}{75}\)
c.\(-\dfrac{2}{3}:x+\dfrac{5}{8}=-\dfrac{7}{12}\)
\(-\dfrac{2}{3}:x=-\dfrac{29}{24}\)
x=\(\dfrac{16}{29}\)
7 tháng 2 2021
a) Ta có: \(\left|2.5-x\right|=1.3\)
\(\Leftrightarrow\left|x-2.5\right|=1.3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2.5=1.3\\x-2.5=-1.3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3.8\\x=-1.3+2.5=1.2\end{matrix}\right.\)
Vậy: \(x\in\left\{3.8;1.2\right\}\)
b) Ta có: \(1.6-\left|x-0.2\right|=0\)
\(\Leftrightarrow\left|x-0.2\right|=1.6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-0.2=1.6\\x-0.2=-1.6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1.8\\x=-1.4\end{matrix}\right.\)
Vậy: \(x\in\left\{1.8;-1.4\right\}\)
c) Ta có: \(13^x=169\)
\(\Leftrightarrow13^x=13^2\)
\(\Leftrightarrow x=2\)
Vậy: x=2
d) Ta có: \(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\)
\(\Leftrightarrow\dfrac{2}{x}=\dfrac{x}{\dfrac{8}{25}}\)
\(\Leftrightarrow x^2=\dfrac{16}{25}\)
hay \(x\in\left\{\dfrac{4}{5};-\dfrac{4}{5}\right\}\)
Vậy: \(x\in\left\{\dfrac{4}{5};-\dfrac{4}{5}\right\}\)
7 tháng 2 2021
\(|2,5-x|=1,3\)
\(\Rightarrow2,5-x=1,3\) hoặc -(2,5-x)=1,3
=>x=2,5-1,3 x-2,5=1,3
=>x=1,2 x=1,3+2,5=3,8
Vậy \(x\in\left\{1,2;3,8\right\}\)
\(1,6-|x-0,2|=0\Rightarrow|x-0,2|=1,6\)
=> x-0,2=1,6 hoặc -(x-0,2)=1,6
=>x=1,6+0,2 0,2-x=1,6
=>x=3,8 x=0,2-1,6=-1,4
Vậy \(x\in\left\{3,8;-1,4\right\}\)
\(13^x=169\Rightarrow13^x=13^2\Rightarrow x=2\)
\(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\Rightarrow-2\times\dfrac{8}{25}=-x\times x\Rightarrow\dfrac{-16}{25}=-x^2\Rightarrow\dfrac{16}{25}=x^2\Rightarrow x^2=\left(\dfrac{4}{5}\right)^2=\left(-\dfrac{4}{5}\right)^2\)
\(\Rightarrow x=\dfrac{4}{5}\) hoặc \(x=-\dfrac{4}{5}\)
H9
HT.Phong (9A5)
CTVHS
12 tháng 1
\(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\left(ĐK:x\ne0\right)\)
\(\Rightarrow\dfrac{2}{x}=\dfrac{x}{\dfrac{8}{25}}\)
\(\Rightarrow\dfrac{2}{x}=x\cdot\dfrac{25}{8}\)
\(\Rightarrow\dfrac{2}{x}=\dfrac{25x}{8}\)
\(\Rightarrow25x\cdot x=2\cdot8\)
\(\Rightarrow25x^2=16\)
\(\Rightarrow x^2=\dfrac{16}{25}\)
\(\Rightarrow x^2=\left(\dfrac{4}{5}\right)^2\)
\(\Rightarrow x=\pm\dfrac{4}{5}\)
Vậy:\(x\in\left\{-\dfrac{4}{5};\dfrac{4}{5}\right\}\)
KV
12 tháng 1
-2/x = -x/(8/25)
x.(-x) = -2.(8/25)
-x² = -16/25
x² = 16/25
x = -4/5 hoặc x = 4/5
2 tháng 10 2017
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. )
22 tháng 9 2018
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 \(x=-30\)
22 tháng 9 2018
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
21 tháng 7 2022
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
10 tháng 6 2017
a) \(x^2=\left(-15\right).\left(-60\right)=900=>x=\)\(\pm\)\(30\)
b) \(-x^2=\dfrac{-16}{25}=>x^2=\dfrac{16}{25}=>x=\)\(\pm\)\(\dfrac{4}{5}\)
HV
10 tháng 10 2017
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 >_<
Điều kiện : x khác 0
\(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\)
\(\Leftrightarrow-x^2=-2.\dfrac{8}{25}\)
\(\Leftrightarrow-x^2=-\dfrac{16}{25}\)
\(\Leftrightarrow x^2=\dfrac{16}{25}\)
\(\Leftrightarrow x=\dfrac{4}{5}\)(tmđk)
Vậy x = \(\dfrac{4}{5}\)
\(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\left(x\ne0\right)\)
\(\Rightarrow x.\left(-x\right)=-2.\dfrac{8}{25}\)
\(\Rightarrow-x^2=-\dfrac{16}{25}\)
\(\Rightarrow x^2=\dfrac{16}{25}\)
\(\Rightarrow x^2=\left(\pm\dfrac{4}{5}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{4}{5};-\dfrac{4}{5}\right\}\)