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câu c) mang tính mua vui hay gì hả bn
mếu thật thì x=0,x=số nào cx đc(câu trả lời này mang tính mua vui thôi nhé)
\(\dfrac{4-x}{-5}=\dfrac{-5}{4-x}\)
\(\left(4-x\right)^2=25=5^2=\left(-5\right)^2\)
4-x=5 hoặc 4-x=-5
x=-1 hoặc x=9
`|7/5 x+2/3| = |4/3 x-1/4|`
\(\left[{}\begin{matrix}\dfrac{7}{5}x+\dfrac{2}{3}=\dfrac{4}{3}x-\dfrac{1}{4}\\\dfrac{7}{5}x+\dfrac{2}{3}=-\dfrac{4}{3}x+\dfrac{1}{4}\end{matrix}\right.\\ \left[{}\begin{matrix}x=-\dfrac{55}{4}\\x=-\dfrac{25}{164}\end{matrix}\right.\)
\(\dfrac{5}{4}+\left(2x-\dfrac{1}{2}\right)=\dfrac{5}{6}\\ =>2x-\dfrac{1}{2}=\dfrac{5}{6}-\dfrac{5}{4}\\ =>2x-\dfrac{1}{2}=\dfrac{10}{12}-\dfrac{15}{12}\\ =>2x-\dfrac{1}{2}=-\dfrac{5}{12}\\ =>2x=-\dfrac{5}{12}+\dfrac{1}{2}\\ =>2x=-\dfrac{5}{12}+\dfrac{6}{12}\\ =>2x=\dfrac{1}{12}\\ =>x=\dfrac{1}{12}:2\\ =>x=\dfrac{1}{12}\cdot\dfrac{1}{2}\\ =>x=\dfrac{1}{24}\)
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\(\dfrac{3}{2}-\left(x+\dfrac{1}{4}\right)=\dfrac{5}{8}\\ =>x+\dfrac{1}{4}=\dfrac{3}{2}-\dfrac{5}{8}\\ =>x+\dfrac{1}{4}=\dfrac{12}{8}-\dfrac{5}{8}\\ =>x+\dfrac{1}{4}=\dfrac{7}{8}\\ =>x=\dfrac{7}{8}-\dfrac{1}{4}\\ =>x=\dfrac{7}{8}-\dfrac{2}{8}\\ =>x=\dfrac{5}{8}\)
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\(\dfrac{x}{3}=\dfrac{12}{x}\\ =>x^2=3\cdot12\\ =>x^2=36\\ =>x^2=6^2\\ =>x=\pm6\)
Tìm x:
a) \(\dfrac{5}{4}+\left(2x-\dfrac{1}{2}\right)=\dfrac{5}{6}\)
\(=>2x-\dfrac{1}{2}=\dfrac{5}{6}-\dfrac{5}{4}\)
\(=>2x-\dfrac{1}{2}=\dfrac{-5}{12}\)
\(=>2x=\dfrac{-5}{12}+\dfrac{1}{2}\)
\(=>2x=\dfrac{1}{12}\)
\(=>x=\dfrac{1}{12}:2\)
\(=>x=\dfrac{1}{24}\)
b) \(\dfrac{3}{2}-\left(x+\dfrac{1}{4}\right)=\dfrac{5}{8}\)
\(=>x+\dfrac{1}{4}=\dfrac{3}{2}-\dfrac{5}{8}\)
\(=>x+\dfrac{1}{4}=\dfrac{7}{8}\)
\(=>x=\dfrac{7}{8}-\dfrac{1}{4}\)
\(=>x=\dfrac{5}{8}\)
c) \(\dfrac{x}{3}=\dfrac{12}{x}\)
Ta có: \(x.x=3.12\)
\(\Rightarrow x^2=36\)
Vậy x = 6 hoặc x = -6
Chúc bạn học tốt
Ta có: \(20\%x-x+\dfrac{1}{5}=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-4}{5}x=\dfrac{3}{4}-\dfrac{1}{5}=\dfrac{15}{20}-\dfrac{4}{20}=\dfrac{11}{20}\)
\(\Leftrightarrow x=\dfrac{11}{20}:\dfrac{-4}{5}=\dfrac{11}{20}\cdot\dfrac{5}{-4}=\dfrac{-55}{80}=\dfrac{-11}{16}\)
Vậy: \(x=-\dfrac{11}{16}\)
a: =>1/3(2x-5)=-2/3-3/2=-4/6-9/6=-13/6
=>2x-5=-13/6*3=-13/2
=>2x=-3/2
=>x=-3/4
b: =>2/5x=-3/4-1/2=-5/4
=>x=-5/4:2/5=-5/4*5/2=-25/8
a)
\(-\dfrac{2}{3}-\dfrac{1}{3}\left(2x-5\right)=\dfrac{3}{2}\\ \Rightarrow\dfrac{1}{3}\left(2x-5\right)=-\dfrac{2}{3}-\dfrac{3}{2}\\ \Rightarrow\dfrac{1}{3}\left(2x-5\right)=-\dfrac{13}{6}\\ \Rightarrow2x-5=-\dfrac{13}{6}:\dfrac{1}{3}=-\dfrac{13}{2}\\ \Rightarrow2x=-\dfrac{13}{2}+5\\ \Rightarrow2x=-\dfrac{3}{2}\\ \Rightarrow x=-\dfrac{3}{2}:2\\ \Rightarrow x=-\dfrac{3}{4}\)
b)
\(\dfrac{2}{5}x+\dfrac{1}{2}=-\dfrac{3}{4}\\ \Rightarrow\dfrac{2}{5}x=-\dfrac{3}{4}-\dfrac{1}{2}\\ \Rightarrow\dfrac{2}{5}x=-\dfrac{5}{4}\\ \Rightarrow x=-\dfrac{5}{4}:\dfrac{2}{5}=-\dfrac{25}{8}\)
`|5/4 x-7/2| -|5/8 x +3/5|=0`
`|5/4 x-7/2|=|5/8 x+3/5|`
\(\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{7}{2}=\dfrac{5}{8}x+\dfrac{3}{5}\\\dfrac{5}{4}x-\dfrac{7}{2}=-\dfrac{5}{8}x-\dfrac{3}{5}\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{164}{25}\\x=\dfrac{116}{75}\end{matrix}\right.\)
Vậy....
a \(\Rightarrow\left|x\right|=\dfrac{19}{20}\Rightarrow\left[{}\begin{matrix}x=\dfrac{19}{20}\\x=-\dfrac{19}{20}\end{matrix}\right.\)
b \(\Rightarrow\left|x-5\right|=-\dfrac{17}{12}\) vô lí vì\(VT=\left|x-5\right|\ge0\) mà \(VP=-\dfrac{17}{20}< 0\)
\(\Rightarrow\) ko có x
a/ /x/= \(\dfrac{1}{5}+\dfrac{3}{4}\)
/x/= 0.95
=> x= \(\pm0.95\)
Vạy....
b/ /x-5/ =\(-\dfrac{5}{3}+\dfrac{1}{4}\)
/x-5/ =\(-\dfrac{17}{12}\) (vô lý)
Vậy...