Tìm X biết:
X x 1/4 + X x 1/3 =2
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\(4^{x+3}+4^{x+2}+4^{x+1}+4^x=5440\)
\(\Rightarrow4^x.4^3+4^x.4^2+4^x.4+4^x=5440\)
\(\Rightarrow4^x\left(4^3+4^2+4+1\right)=5440\)
\(\Rightarrow4^x.\left(64+16+4+1\right)=5440\)
\(\Rightarrow4^x.85=5440\)
\(\Rightarrow4^x=5440:85\)
\(\Rightarrow4^x=64=4^3\)
\(\Rightarrow x=3\)
dễ quá bạn ơi giải câu này nè mới chất
Q= 12 + 22 + 32 +...+ 1002
a: \(\dfrac{x}{6}=\dfrac{8}{3}\)
=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)
b: \(\dfrac{5}{x}=\dfrac{4}{9}\)
=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)
c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)
=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)
=>x=-1-3=-4
d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)
=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)
=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)
=>\(x=-\dfrac{69}{8}\)
f: ĐKXĐ: x<>1
\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)
=>\(\left(x-1\right)^2=3\cdot27=81\)
=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)
\(\frac{2}{3}\left(x-1\right)-x-\frac{3}{4}=1\)
<=> \(\frac{2}{3}x-\frac{2}{3}-x-\frac{3}{4}=1\)
<=> \(-\frac{1}{3}x-\frac{17}{12}=1\)
<=> \(-\frac{1}{3}x=\frac{29}{12}\)
<=> \(x=-\frac{29}{4}\)
\(\frac{5}{6}\left(x+2\right)-x-\frac{1}{2}=\frac{1}{3}\)
<=> \(\frac{5}{6}x+\frac{5}{3}-x-\frac{1}{2}=\frac{1}{3}\)
<=> \(-\frac{1}{6}x+\frac{7}{6}=\frac{1}{3}\)
<=> \(-\frac{1}{6}x=-\frac{5}{6}\)
<=> \(x=5\)
học tốt
a)\(\frac{1}{2}+\frac{3}{4}.x=\frac{1}{4}\)
\(\frac{3}{4}x=\frac{1}{4}-\frac{1}{2}\)
\(\frac{3}{4}.x=\frac{-1}{4}\)
\(x=\frac{-1}{4}:\frac{3}{4}\)
\(x=\frac{-1}{3}\)
Vậy \(x=\frac{-1}{3}\)
b)\(|x-5|-\frac{1}{3}=0,5\)
\(|x-5|=\frac{1}{2}+\frac{1}{3}\)
\(|x-5|=\frac{5}{6}\)
\(\Rightarrow\orbr{\begin{cases}x-5=\frac{5}{6}\\x-5=\frac{-5}{6}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{6}+5\\x=\frac{-5}{6}+5\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{35}{6}\\x=\frac{25}{6}\end{cases}}\)
Vậy\(x=\frac{35}{6}\)hoặc\(x=\frac{25}{6}\)
c, \(x\)(\(x\) - 2022) + 4.(2022 - \(x\)) = 0
(\(x\) - 2022).(\(x\) - 4) = 0
\(\left[{}\begin{matrix}x-2022=0\\x+4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2022\\x=4\end{matrix}\right.\)
`(3x-1)(x-3)-2(x-3)=9`
`-> 3x(x-3)-1(x-3)-2x+6=9`
`-> 3x^2-9x-x+3-2x+6=9`
`-> 3x^2-12x+9=9`
`-> 3x^2-12x=0`
`-> x(3x-12)=0`
`->`\(\left[{}\begin{matrix}x=0\\3x-12=0\end{matrix}\right.\)
`->`\(\left[{}\begin{matrix}x=0\\3x=12\end{matrix}\right.\)
`->`\(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy, `x={0 ; 4}`.
a) 1/4(x-3)+2=1/5
1/4.(x-3) = 1/5-2
1/4.(x-3) = -9/5
x-3 = (-9/5):1/4
x-3 = -36/5
x = -36/5+3
x= -21/5
`x xx1/4+x xx1/3=2`
`=> x xx (1/4+1/3)=2`
`=> x xx (3/12+4/12)=2`
`=> x xx 7/12=2`
`=> x=2:7/12`
`=>x=2xx12/7`
`=>x=24/7`
X x (1/4 +1/3)=2
X x 7/12=2
x= 2 : 7/12
x=24/7