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1
\(\left(x-2\right):2.3=6\)
\(\Leftrightarrow\left(x-2\right):2=2\)
\(\Leftrightarrow\left(x-2\right)=4\)
\(\Leftrightarrow x=4+2=6\)
c) ta có
\(\left[\left(2x+1\right)+1\right]m:2=625\)
\(\Leftrightarrow\left[\left(2x+1\right)+1\right]\left\{\left[\left(2x+1\right)-1\right]:2+1\right\}=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-1:2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2=1249\)
\(\Leftrightarrow\left(2x+1\right)^2+1=1251\)
\(\Leftrightarrow\left(2x+1\right)^2=1250\)
...
2
\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{5}{4}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}:\frac{5}{3}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}.\frac{3}{5}\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)
Bài 1:
Ta có: \(4-2\left(x+1\right)=2\)
\(\Leftrightarrow2\left(x+1\right)=2\)
\(\Leftrightarrow x+1=1\)
hay x=0
Bài 2:
Ta có: \(\left|2x-3\right|-1=2\)
\(\Leftrightarrow\left|2x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
a/ 99x+(1+2+...+99)=0
99x+(1+99)x49,5=0
x=5
b/3x-1-2-3+1+2+3+4+...+10=0
3x+4+5+...+10=0
x=-49/3
Bài 2:
a: =>x=0 hoặc x+3=0
=>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
3(x+5) -3 = 2(x+1) +7
⇒3x+15-3=2x+2+7
⇒3x+12=2x+9
⇒x=-3
15-(x+2)² = -1
⇒(x+2)2=16
⇒\(\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\)
⇒\(\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
75-(2x+1)³ =11
⇒(2x+1)³=64⇒\(\left[{}\begin{matrix}2x+1=8\\2x+1=-8\end{matrix}\right.\)⇒\(\left[{}\begin{matrix}x=3,5\\x=-4,5\end{matrix}\right.\)a) ta có: 3(x+5)-3=2(x+1)+7
nên 3x-2x=2+7-15+3
hay x=-3
b) Ta có: \(15-\left(x+2\right)^2=-1\)
nên \(\left(x+2\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
a) Ta có: \(3\left(x+5\right)-3=2\left(x+1\right)+7\)
\(\Leftrightarrow3x+15-3=2x+2+7\)
\(\Leftrightarrow3x+12=2x+9\)
hay x=-3
b) Ta có: \(15-\left(x+2\right)^2=-1\)
\(\Leftrightarrow\left(x+2\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
c) Ta có: \(75-\left(2x+1\right)^3=11\)
\(\Leftrightarrow\left(2x+1\right)^3=64\)
\(\Leftrightarrow2x+1=4\)
hay \(x=\dfrac{3}{2}\)
`3(x+5)-3=2(x+1)+7`
`3x+15-3=2x+2+7`
`x=-3`
.
`15-(x+2)^2=-1`
`(x+2)^2=16`
`(x+2)^2=4^2=(-4)^2`
`[(x+2=4),(x+2=-4):}`
`[(x=2),(x=-6):}`
.
`75-(2x+1)^3=11`
`(2x+1)^3=64`
`(2x+1)^2=4^3`
`2x+1=4`
`x=3/2`