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VT = 101x + 1+2+3+...+100 = 101x + 1/2*100*101
PT <=> 101x + 50*101 = 5353
<=> x + 50 = 53
<=> x = 3
Ta có: x+(x+1)+(x+2)+...+(x+100)=5353
<=> 101x + (1 + 2 + ....+ 100) = 5353
<=> 101x + 2550 = 5353
=> 101x = 5353 - 2550
=> 101x = 2803
=> x =2803 : 101
=> x =????????????
a) \(3\left(x-4\right)-\left(8-x\right)=12\)
\(3x-12-8+x=12\)
\(4x=12+12+8\)
\(4x=32\)
\(x=8\)
b) \(4\left(x-5\right)-\left(x-7\right)=-19\)
\(4x-20-x+7=-19\)
\(3x=-19+20-7\)
\(3x=-6\)
\(x=-2\)
c) \(7\left(x-3\right)-5\left(3-x\right)=11x-5\)
\(7\left(x-3\right)+5\left(x-3\right)=11x-5\)
\(\left(x-3\right).12=11x-5\)
\(12x-36-11x+5=0\)
\(x-31=0\)
\(x=31\)
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: \(\left|2x-3\right|-1=2\)
\(\Leftrightarrow\left|2x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
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
a) 2.(x-1/3) - (x-1/2) = 1/2.x
2.x - 2/3 - x + 1/2 = 1/2.x
=> 2.x-x - 1/2.x = 2/3 -1/2
1/2.x = 1/6
x = 1/3
bài b bn làm tương tự nha
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
\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{5}{4}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}:\frac{5}{3}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}.\frac{3}{5}\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)
Đề bài có vấn đề nhé bạn! Chỗ cuối x+? thì mình mới làm được