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Ai chẳng biết chuyển vế đổi dấu :v
a) \(x-7=4x+10\)
\(x-4x=10+7\)
\(-3x=17\)
\(x=\dfrac{17}{-3}\)
Vậy \(x=\dfrac{17}{-3}\)
b) \(2x+5=-3x+7\)
\(2x+3x=7-5\)
\(5x=2\)
\(x=\dfrac{2}{5}\)
Vậy \(x=\dfrac{2}{5}\)
c) \(x-\left(3x+7\right)=6x-1\)
\(x-3x-7=6x-1\)
\(-2x-7=6x+1\)
\(-7-1=6x+2x\)
\(-8=8x\)
\(x=\dfrac{-8}{8}=-1\)
Vậy \(x=-1\)
d) \(x+\left(5x-1\right)=15\)
\(x+5x-1=15\)
\(6x=15+1\)
\(6x=16\)
\(x=\dfrac{16}{6}=\dfrac{8}{3}\)
Vậy \(x=\dfrac{8}{3}\)
1 , x - 7 = 4x + 10
x - 4x = 10 + 7
- 3x = 17
x = 17 : ( - 3 )
x = \(\dfrac{-17}{3}\)
2 , 2x + 5 = -3x + 7
2x + 3x = 7 -5
5x = 2
x = 2 : 5
x =\(\dfrac{2}{5}\)
3 , x - ( 3x + 7 ) = 6x - 1
x - 3x - 7 = 6x - 1
x - 3x -6x = -1 +7
-8x = 6
x = 6 : ( -8 )
x = \(\dfrac{-3}{4}\)
4 , x + ( 5x -1 ) = 15
x + 5x - 1 = 15
x + 5x = 15 + 1
6x = 16
x = 16 : 6
x = \(\dfrac{8}{3}\)
5 , / x + 1 / = / 2x - 5 /
TH 1 : x + 1 = 2x - 5
x - 2x = -5 -1
- x = -4
= > x = 4
TH 2 : -x -1 = -2x + 5
-x + 2x = 5 + 1
x = 6
6 , / 3x + 8 / - / x -10 / = 0
3x + 8 - x + 10 = 0
3x - x = 0 - 10 - 8
2 x = -18
x = -18 : 2
x = - 9
\(-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\\ \Rightarrow-x+2x=\dfrac{3}{5}+\dfrac{4}{7}\\ \Rightarrow x=\dfrac{21}{35}+\dfrac{20}{35}\\ \Rightarrow x=\dfrac{41}{35}\)
Vậy `x=41/35`
__
\(\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\\ \Rightarrow\left(\dfrac{3}{7}-\dfrac{2}{3}\right)x=\dfrac{10}{21}\\ \Rightarrow\left(\dfrac{9}{21}-\dfrac{14}{21}\right)x=\dfrac{10}{21}\\ \Rightarrow\dfrac{-5}{21}x=\dfrac{10}{21}\\ \Rightarrow x=\dfrac{10}{21}:\left(-\dfrac{5}{21}\right)\\ \Rightarrow x=-2\)
Vậy `x=-2`
a)
-4/7 - x = 3/5 - 2x
2x - x = 3/5 + 4/7
x = 41/35
Vậy x = 41/35
b)
3/7.x - 2/3.x = 10/21
x(3/7 - 2/3) = 10/21
x.(-5/21) = 10/21
x = 10/21 : (-5/21) = -2
Vậy x = -2
\(1)\)\(-\dfrac{10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
\(=\dfrac{10}{11}\left(-\dfrac{8}{9}+\dfrac{7}{18}\right)\)
\(=\dfrac{10}{11}\left(\dfrac{-16}{18}+\dfrac{7}{18}\right)\)
\(=\dfrac{10}{11}.\left(-\dfrac{1}{2}\right)=-\dfrac{5}{11}\)
\(2)\)\(\dfrac{12}{25}.\dfrac{23}{7}-\dfrac{12}{7}.\dfrac{13}{25}\)
\(=\dfrac{12}{7}.\dfrac{23}{25}-\dfrac{12}{7}.\dfrac{13}{25}\)
\(=\dfrac{12}{7}.\left(\dfrac{23}{25}-\dfrac{13}{25}\right)\)
\(=\dfrac{12}{7}.\dfrac{2}{5}=\dfrac{24}{35}\)
\(3)\)\(\dfrac{3}{7}.\dfrac{16}{15}-\dfrac{2}{15}.\dfrac{-3}{7}\)
\(=\dfrac{3}{7}.\dfrac{16}{15}-\dfrac{3}{7}.\dfrac{-2}{15}\)
\(=\dfrac{3}{7}.\left(\dfrac{16}{15}+\dfrac{2}{15}\right)\)
\(=\dfrac{3}{7}.\dfrac{18}{15}=\dfrac{18}{35}\)
\(4)\)\(-\dfrac{4}{13}.\dfrac{5}{17}+\dfrac{-12}{13}.\dfrac{4}{17}\)
\(=-\dfrac{4}{13}.\dfrac{5}{17}+\dfrac{-4}{13}.\dfrac{12}{17}\)
\(=-\dfrac{4}{13}.\left(\dfrac{5}{17}+\dfrac{12}{17}\right)\)
\(=-\dfrac{4}{13}.\dfrac{17}{17}=-\dfrac{4}{13}\)
`#040911`
`1)`
`-10/11 * 8/9 + 7/18 . 10/11`
`= 10/11 * (-8/9 + 7/18)`
`= 10/11 * (-1/2)`
`= -5/11`
`2)`
`12/25 * 23/7 - 12/7 *13/25`
`= 12/7 * 23/25 - 12/7 * 13/25`
`= 12/7 * (23/25 - 13/25)`
`= 12/7 * 2/5`
`= 24/35`
`3)`
`3/7 * 16/15 - 2/15 * (-3)/7`
`= 3/7 * (16/15 + 2/15)`
`= 3/7 * 6/5`
`= 18/35`
`4)`
`-4/13 * 5/17 + (-12)/13 * 4/17`
`= -4/17 * 5/13 + (-12)/13 * 4/17`
`= 4/17 * (-5/13 - 12/13)`
`= 4/17 * (-17)/13`
`= -4/13`
\(B=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^{10}}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}=\dfrac{2^{10}\cdot3^8\left(1-3^2\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{1-9}{6}=\dfrac{-8}{6}=-\dfrac{4}{3}\)
a: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
=>\(\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
b: \(\Leftrightarrow-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-8}-\dfrac{1}{x-8}+\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
=>1/x-1=3/4
=>x-1=4/3
=>x=7/3
=\(\dfrac{2^{19}.27^3-15.\left(-4\right)^9.9^4}{6^9.2^{10}+\left(-12\right)^{10}}\)
=\(\dfrac{2^{19}.3^9-2^{18}.3^9.5}{6^9.2^{10}+6^{10}.2^{10}}\)
=\(\dfrac{2^{18}.3^9\left(2-2.5\right)}{6^9.2^{10}\left(1+6\right)}\)
=\(\dfrac{2^{18}.3^9.\left(-8\right)}{3^9.2^{19}.7}\)
=\(\dfrac{-8}{14}=\dfrac{-4}{7}\)
bạn làm sai rùi kết quả là \(\dfrac{1}{2}\) mình hỏi thầy rùi
\(A=1+2+2^2+2^3+...+2^{2020}\)
\(2A=2+2^2+2^3+2^4+...+2^{2021}\)
\(2A-A=\left(2+2^2+2^3+2^4+....+2^{2021}\right)-\left(1+2+2^2+2^3+...+2^{2020}\right)\)
\(A=2^{2021}-1\)
a) Ta có: \(\left(x-1\right)^{x+2}-\left(x-1\right)^{x+4}=0\)
\(\Leftrightarrow\left(x-1\right)^x\cdot\left(x-1\right)^2-\left(x-1\right)^x\cdot\left(x-1\right)^4=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\cdot\left[1-\left(x-1\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
b) Ta có: \(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{15}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\)
\(\Leftrightarrow2x=\dfrac{1}{64}\)
hay \(x=\dfrac{1}{128}\)