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Lời giải:
a.
$\frac{2}{3}+\frac{1}{3}:3\times x=20\text{%}$
$\frac{2}{3}+\frac{1}{9}\times x=\frac{1}{5}$
$\frac{1}{9}\times x=\frac{1}{5}-\frac{2}{3}=\frac{-7}{15}$
$x=\frac{-7}{15}: \frac{1}{9}=\frac{-21}{5}$
b.
$\frac{3-x}{5-x}=\frac{6}{11}$
$\Rightarrow 6(5-x)=11(3-x)$
$\Rightarrow 30-6x=33-11x$
$\Rightarrow 5x=3$
$\Rightarrow x=\frac{3}{5}$
Bài 3:
a: Ta có: 60-3(x-2)=51
\(\Leftrightarrow x-2=3\)
hay x=5
b: Ta có: \(4x-20=25:2^2\)
\(\Leftrightarrow4x=\dfrac{25}{4}+20=\dfrac{105}{4}\)
hay \(x=\dfrac{105}{16}\)
c: Ta có: \(8\cdot6+288:\left(x-3\right)^2=50\)
\(\Leftrightarrow288:\left(x-3\right)^2=50-48=2\)
\(\Leftrightarrow\left(x-3\right)^2=144\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
a: =>10+3x-3=6x+10
=>3x-3=6x
=>-3x=3
=>x=-1
b: =>x+1=0 hoặc x-2=0
=>x=-1 hoặc x=2
a) \(10+3\left(x-1\right)=10+6x\)
\(\Rightarrow10+3x-3=10+6x\)
\(\Rightarrow3x-6x=10-10+3\)
\(\Rightarrow-3x=3\)
\(\Rightarrow x=-\dfrac{3}{3}\)
\(\Rightarrow x=-1\)
b) \(\left(x+1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
`a,`
`x*3+7=16`
`=>3x = 16 - 7`
`=> 3x = 9`
`=> x = 9 \div 3`
`=> x = 3`
Vậy, `x = 3`
`b,`
` x- 152 \div 2 = 46?`
`=> x - 76 = 46`
`=> x = 46 + 76`
`=> x = 122`
Vậy, `x = 122.`
`c,`
` 74-2*(x+3)=34`
`=> 2(x + 3) = 74 - 34`
`=> 2(x+2) = 40`
`=> x + 2 = 40 \div 2`
`=> x + 2 = 20`
`=> x = 20 - 2`
`=> x = 18`
Vậy, `x = 18.`
a/ x - 7 = 12
=> x = 12 + 7 = 19
b/ 9 + 4(x - 5) = 13
=> 4x - 20 = 4
=> 4x = 24
=> x = 6
c/ (x+2)3 = 64
=> x + 2 = 4
=> x = 2
a) x-7=12
x=12+7
x=19
b) 9+4.(x-5)=13
4.(x-5)=13-9
4.(x-5)=4
(x-5)=4:4
(x-5)=1
x=5+1
x=6
cau cuoi mik ko bt lam nha!
chuc ban hoc tot
b) \(3^x\cdot3^2+3^x=7290\)
\(3^x\left(3^2+1\right)=720\)
\(3^x\cdot10=7290\)
\(=>3^x=729=3^6\)
=> \(x=6\)
\(a,\dfrac{x}{3}=-\dfrac{1}{2}\)
\(\Leftrightarrow2x=-1.3\)
\(\Leftrightarrow2x=-3\)
\(\Leftrightarrow x=-\dfrac{3}{2}\)
\(b,\dfrac{x}{3}=\dfrac{3}{x}\)
\(\Leftrightarrow x.x=3.3\)
\(\Leftrightarrow x^2=9\)
\(\Leftrightarrow x=\pm3\)
a)
=> \(\left[{}\begin{matrix}x-2< 3\\x-2< -3\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
TH1
X-2<3
x<5
Th2
x-2<-3
x<-1
Vậy no là x<5 , x<-1