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4. ( x - 250 ) : 6 = 64 - 12
( x- 250 ) : 6 = 52
x - 250 = 312
x = 562
5. 10x = 1030
=> x = 103
6. 30x = 120
x = 4
7. \(x=2023\)
\(8.165-\left(35:x+3\right).19=13\)
\(\left(35:x+3\right).19=152\)
\(35:x+3=8\)
\(35:x=5\)
\(x=7\)
4) \(\left(x-250\right)\div6=4^3-2^2\times3\)
\(\left(x-250\right)\div6=64-4\times3\)
\(\left(x-250\right)\div6=64-12=52\)
\(x-250=52\times6=312\)
\(x=312+250\)
\(x=562\)
5) \(2x+3x+5x=1030\)
\(x\left(2+3+5\right)=1030\)
\(10x=1030\)
\(x=1030\div10\)
\(x=103\)
6) \(15x-35x+50x=120\)
\(x\left(15-35+50\right)=120\)
\(30x=120\)
\(x=120\div30\)
\(x=4\)
7) \(\dfrac{1}{2}x+\dfrac{1}{6}x+\dfrac{1}{3}x=2023\)
\(x\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\right)=2023\)
\(x\times1=2023\)
\(x=2023\)
8) \(165-\left(35\div x+3\right)\times19=13\)
\(\left(35\div x+3\right)\times19=165-13\)
\(\left(35\div x+3\right)\times19=152\)
\(35\div x+3=152\div19=8\)
\(35\div x=8-3=5\)
\(x=35\div5\)
\(x=7\)
\(3x\left(2x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=-1\\x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=0\end{cases}}\)
\(\frac{\frac{6}{5}+\frac{6}{35}-\frac{6}{125}-\frac{6}{2009}-\frac{6}{2011}}{\frac{7}{5}+\frac{7}{35}-\frac{7}{125}-\frac{7}{2009}-\frac{7}{2011}}\)
\(=\frac{6.(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011})}{7.(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011})}\)
\(=\frac{6}{7}\)
Tìm x
\(a,3x(2x+1)=0\)
\(\Rightarrow\hept{\begin{cases}3x=0\\2x+1=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=0\\x=\frac{-1}{2}\end{cases}}\)
Vậy \(x=0\)hoặc \(x=\frac{-1}{2}\)
\(b.\frac{2}{3}-\frac{1}{3}(x-\frac{3}{2})-\frac{1}{2}(2x+1)=5\)
\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}-x(\frac{1}{3}+1)=5\)
\(\frac{4}{3}x=\frac{2}{3}-5\)
\(\frac{4}{3}x=\frac{-13}{3}\)
\(x=\frac{-13}{3}\div\frac{4}{3}\)
\(x=\frac{-13}{4}\)
Chúc ban học tốt
1: Ta có: \(20-2\left(x+4\right)=4\)
\(\Leftrightarrow2\left(x+4\right)=16\)
\(\Leftrightarrow x+4=8\)
hay x=4
5: Ta có: \(\left(x+1\right)^3=27\)
\(\Leftrightarrow x+1=3\)
hay x=2
a, 2x + 35 -x+27=0
x +62=0
x=-62
b, 2x -41 -3x + 23 =0
-x -18=0
-x=18
x=-18
c, 4x -12-3x-15= -124
x -27=-124
x= -97
d, Suy ra x+3 =0 hoặc 2x-18=0
x=-3 hoặc 2x=18 => x=9
vậy x=-3 hoặc x=9
5)
để \(\frac{5x-3}{x+1}\)là số nguyên
\(5x-3⋮x+1\)
\(x+1⋮x+1\)
\(\Rightarrow5\left(x+1\right)⋮x+1\)
\(5x-3-\left(5x-5\right)⋮x+1\)
\(-2⋮x+1\)
\(\Rightarrow x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
x+1 | 1 | -1 | 2 | -2 |
x | 0 | -2 | 1 | -3 |
Vậy \(x\in\left\{0;-2;1;-3\right\}\)
1) -16 + 23 + x = - 16
ó x = -23
2) 2x – 35 = 15
ó 2x = 50
ó x = 25
3) 3x + 17 = 12
ó 3x = -5
ó x = -5/3
4) │x - 1│= 0
=> x – 1 = 0 ó x = 1
5) -13 .│x│ = -26
ó |x| = 2
ó x = 2 hoặc x = -2
\(\left(x+2\right)-2=0\)
\(\Rightarrow x+2-2=0\)
\(\Rightarrow x=0\)
\(\left(x+3\right)+1=7\)
\(\Rightarrow x+3+1=7\)
\(\Rightarrow x+4=7\)
\(\Rightarrow x=3\)
\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)
\(\Rightarrow3x=12\)
\(\Rightarrow x=4\)
\(\left(5x+4\right)-1=13\)
\(\Rightarrow5x+4-1=13\)
\(\Rightarrow5x+3=13\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
\(\left(4x-8\right)-3=5\)
\(\Rightarrow4x-8-3=5\)
\(\Rightarrow4x-11=5\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
\(8-\left(2x+4\right)=2\)
\(\Rightarrow8-2x-4=2\)
\(\Rightarrow4-2x=2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
\(7+\left(5x+2\right)=14\)
\(\Rightarrow7+5x+2=14\)
\(\Rightarrow9+5x=14\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=1\)
\(5-\left(3x-11\right)=1\)
\(\Rightarrow5-3x+11=1\)
\(\Rightarrow16-3x=1\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=5\)
c: \(\Leftrightarrow2x+1=3\)
hay x=1