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LL
27 tháng 6 2019
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
27 tháng 6 2019
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
20 tháng 9 2021
\(1,\\ \left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\\ \Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)
\(2,\\ a,\left|2x-3\right|>5\Leftrightarrow\left[{}\begin{matrix}2x-3< -5\\2x-3>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -1\\x>4\end{matrix}\right.\\ b,\left|3x-1\right|\le7\Leftrightarrow\left[{}\begin{matrix}3x-1\le7\\1-3x\le7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{8}{3}\\x\ge-2\end{matrix}\right.\\ c,\cdot x< -\dfrac{3}{2}\\ \Leftrightarrow5-3x+\left(-2x-3\right)=7\Leftrightarrow2-5x=7\Leftrightarrow x=-1\left(ktm\right)\\ \cdot-\dfrac{3}{2}\le x\le\dfrac{5}{3}\\ \Leftrightarrow\left(5-3x\right)+\left(2x+3\right)=7\Leftrightarrow8-x=7\Leftrightarrow x=1\left(tm\right)\\ \cdot x>\dfrac{5}{3}\\ \Leftrightarrow\left(3x-5\right)+\left(2x+3\right)=7\Leftrightarrow5x-2=7\Leftrightarrow x=\dfrac{9}{5}\left(tm\right)\\ \Leftrightarrow S=\left\{1;\dfrac{9}{5}\right\}\)
30 tháng 10 2016
a) 2x = 16 b) 3x + 1 = 9x
2x = 24 3x + 1 = 32x
x = 4 x + 1 = 2x
x = 1
c) 23x + 2 = 4x + 2
23x + 2 = 22(x + 2)
3x + 2 = 2(x + 2)
3x + 2 = 2x + 4
x = 2
d) 32x - 1 = 243
32x - 1 = 35
2x - 1 = 5
2x = 6
x = 3
PA
8 tháng 7 2016
a.
\(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(6x^2+21x-2x-7-6x^2+5x-6x+5=16\)
\(\left(6x^2-6x^2\right)+\left(21x-2x+5x-6x\right)-\left(7-5\right)=16\)
\(18x-2=16\)
\(18x=16+2\)
\(18x=18\)
\(x=\frac{18}{18}\)
\(x=1\)
b.
\(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)
\(10x^2+9x-10x^2-15x+2x+3=8\)
\(\left(10x^2-10x^2\right)-\left(15x-9x-2x\right)+3=8\)
\(-4x=8-3\)
\(-4x=5\)
\(x=-\frac{5}{4}\)
c.
\(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)-2=0\)
\(21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)
\(\left(15x^2-15x^2\right)+\left(25x+21x-10x+6x\right)-\left(35+4+2\right)=0\)
\(42x=41\)
\(x=\frac{41}{42}\)
TT
1 tháng 10 2017
a. \(5.\left(x-2\right)+3.\left(x-2\right)=0\)
\(\Rightarrow8.\left(x-2\right)=0\)
\(\Rightarrow x-2=0:8\)
\(\Rightarrow x-2=0\)
\(\Rightarrow x=2\)
Vậy...
b. \(\dfrac{2}{3}+\dfrac{5}{2}:x=\dfrac{2}{4}\)
\(\Rightarrow\dfrac{5}{2}:x=\dfrac{2}{4}-\dfrac{2}{3}\)
\(\Rightarrow\dfrac{5}{2}:x=\dfrac{-1}{6}\)
\(\Rightarrow x=\dfrac{5}{2}:\dfrac{-1}{6}=-15\)
Vậy...
c. \(2.\left(x-\dfrac{1}{7}\right)=0\)
\(\Rightarrow x-\dfrac{1}{7}=0:2\)
\(\Rightarrow x-\dfrac{1}{7}=0\)
\(\Rightarrow x=\dfrac{1}{7}\)
Vậy...
d. \(\dfrac{11}{20}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\Rightarrow\dfrac{2}{5}+x=\dfrac{11}{12}:\dfrac{2}{3}\)
\(\Rightarrow\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{1}{4}-\dfrac{2}{5}=\dfrac{-3}{20}\)
Vậy...
e. \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(\Rightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}=\dfrac{-5}{7}\)
Vậy...
g. \(\dfrac{2}{3}x+\dfrac{5}{7}=\dfrac{3}{10}\)
\(\Rightarrow\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{5}{7}\)
\(\Rightarrow\dfrac{2}{3}x=\dfrac{-29}{70}\)
\(\Rightarrow x=\dfrac{-29}{70}:\dfrac{2}{3}=\dfrac{-87}{140}\)
Vậy...
a) \(\left(2x-3\right)^2=16\)
\(\left(2x-3\right)^2=4^2\)
\(2x-3=4\)
\(2x=7\)
\(x=\dfrac{7}{2}=3,5\)
b) \(\left(3x-2\right)^5=-243\)
\(\left(3x-2\right)^5=-3^5\)
\(3x-2=-3\)
\(3x=-1\)
\(3x=-\dfrac{1}{3}\)
c) \(\left(x-7\right)^{x+1}=\left(x-7\right)^{x+11}\)
\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\left(x-7\right)^{x+1}\times\left[1-\left(x-7\right)^{10}\right]=0\)
\(\left(x-7\right)^{x+1}=0\) ; \(1-\left(x-7\right)^{10}=0\)
\(x-7=0;\left(x-7\right)^{10}=1\)
\(x=7;\left(x-7=1;x-7=-1\right)\)
\(x=7;x=8;x=6\)
a, (2\(x\) - 3)2 = 16
\(\left[{}\begin{matrix}2x-3=-4\\2x-3=4\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-1\\2x=7\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy \(x\in\){ - \(\dfrac{1}{2}\); \(\dfrac{7}{2}\)}
b, (3\(x\) - 2)5 = -243
( 3\(x\) - 2)5 = (-3)5
3\(x\) - 2 = -3
3 \(x\) = -1
\(x\) = - \(\dfrac{1}{3}\)
Vậy \(x\) = -\(\dfrac{1}{3}\)
c, \(\left(x-7\right)\)\(x+1\) = (\(x-7\))\(x+11\)
(\(x-7\))\(^{x+1}\).( \(\left(x-7\right)^{10}\) - 1 ) = 0
\(\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=7\\x-7=-1\\x-7=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=7\\x=6\\x=8\end{matrix}\right.\)
Vậy \(x\in\){ 6; 7; 8}