(2-x)^3=(2-x)^5
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`x/2+x+x/3+x+x+x/4=5 3/4`
`=>3x+x/2+x/3+x/4=23/4`
`=>49/12x=23/4`
`=>x=69/49`
Vậy `x=69/49`
5(x-1)=125-25
5(x-1)=100
x-1=100:5
x-1=20
x=20+1
x=21
12(x-1):3=64+8
12(x-1):3=72
12(x-1)=72.3
12(x-1)=216
x-1=216:12
x-1=18
x=18+1
x=19
(x-1)^3=5^3
=>x-1=5
x=5+1
x=6
\(\dfrac{1}{2}-\dfrac{5}{12}x=\dfrac{2}{3}\)
\(\dfrac{5}{12}x=\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{3}{6}-\dfrac{4}{6}\)
\(\dfrac{5}{12}x=\dfrac{-1}{6}\)
\(x=\dfrac{-1}{6}:\dfrac{5}{12}=\dfrac{-1}{6}.\dfrac{12}{5}\)
\(x=\dfrac{-2}{5}\)
\(-\left|1,7-x\right|-\dfrac{5}{3}=\dfrac{2}{3}\\ \Rightarrow\left|1,7-x\right|=-\dfrac{5}{3}-\dfrac{2}{3}=-\dfrac{7}{3}\left(l\right)\)
Vậy không có giá trị x thoả mãn
`5/2 -3(1/3-x)=1/4-7x`
`=> 5/2 - 1 + 3x=1/4 -7x`
`=>3x+7x= 1/4 - 5/2 +1`
`=> 10x= 1/4 - 10/4 +4/4`
`=>10x= -5/4`
`=>x=-5/4 :10`
`=>x=-5/4 xx1/10`
`=>x= -5/40=-1/8`
| x-2/3| = 1/3+2/5
= 11/15
=> x-2/3=11/15 hoặc x-2/3=-11/15
x= 7/5 x = -1/15
k cho mk nha
\(\left|x-\frac{2}{3}\right|-\frac{2}{5}=\frac{1}{3}\)
\(\Rightarrow\left|x-\frac{2}{3}\right|=\frac{1}{3}+\frac{2}{5}=\frac{11}{15}\)
\(\Rightarrow\orbr{\begin{cases}\Rightarrow x-\frac{2}{3}=\frac{11}{15}\\x-\frac{2}{3}=\frac{-11}{15}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\Rightarrow x=\frac{7}{15}\\x=\frac{-1}{15}\end{cases}}\)
a: x/3-1/6=1/5
=>x/3=11/30
hay x=11/90
b: =>1/2x=2
hay x=4
c: =>2/3:x=-7-1/3=-22/3
=>x=-1/11
A = 1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ... + 99 x 100
A = 2 + 6 + 12 + 20 + ... 9900
A = [2+9900] rồi bn nhân tổng số từ số 2 - 9900
\(3A=1.2.3+2.3.\left(4-1\right)+...+99.100.\left(101-98\right)\)
\(3A=1.2.3+2.3.4-1.2.3+...+99.100.101-98.99.100\)
\(3A=99.100.101\)
\(\Rightarrow A=\frac{99.100.101}{3}\)
(x+1)+(x+2)+(x+3)=4x
x+1+x+2+x+3=4x
(x+x+x)+(1+2+3)=4x
x*3+6=4x
6=1*x(bớt cả hai vế đi 3*x)
x=6/1(Tìm thừa số)
x=6
\(\left(2-x\right)^3=\left(2-x\right)^5\)
\(\Leftrightarrow\left(2-x\right)^5-\left(2-x\right)^3=0\)
\(\Leftrightarrow\left(2-x\right)^3\left[\left(2-x\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2-x\right)^3=0\\\left(2-x\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\2-x=1\\2-x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=3\end{matrix}\right.\)
(2 - \(x\))3 = (2 - \(x\))5
(2 - \(x\))3 - (2 - \(x\))5 = 0
(2 - \(x\))3.[1 - (2 - \(x\))2] = 0
\(\left[{}\begin{matrix}\left(2-x\right)^3=0\\1-\left(2-x\right)^2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}2-x=0\\\left(2-x\right)^2=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\2-x=-1\\2-x=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=3\\x=1\end{matrix}\right.\)
Vậy \(x\) \(\in\) {1; 2; 3}