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Bài 1:
a: =>13x+8=9x+20
=>4x=12
hay x=3
b: \(\Leftrightarrow5x-7=-8-11-3x\)
=>5x-7=-3x-19
=>8x=-12
hay x=-3/2
c: \(\Leftrightarrow\left[{}\begin{matrix}12x-7=5\\12x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{6}\end{matrix}\right.\)
e: =>3x+1=-5
=>3x=-6
hay x=-2
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
Câu 2:
a: \(\Leftrightarrow12x-60=7x-5\)
=>5x=55
=>x=11
b: \(\Leftrightarrow\left(2x-3\right)^{2010}\left[\left(2x-3\right)^2-1\right]=0\)
=>(2x-3)(2x-2)(2x-4)=0
hay \(x\in\left\{\dfrac{3}{2};1;2\right\}\)
1.Tính hợp lý:
a. 1152 - (374 + 1152) + (374 - 65) = 1152 - 374 - 1152 + 374 - 65 = ( 1152 - 1152 ) + ( -65) + ( 374 - 374 ) = 0 + ( - 65) + 0 = -65
Bài 1 : Tính hợp lý : c. \(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\) = \(\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\) = \(\dfrac{3^{29}.\left(11-3\right)}{2^2.3^{28}}\) = \(\dfrac{3^{29}.2^3}{2^2.3^{28}}\) = 6
a) \(4.5^2-32:2^5\)
\(=4.25-2^5:2^5\)
\(=100-1\)
\(=99.\)
b) \(9.8.14+6.\left(-17\right)\left(-12\right)+19.\left(-4\right).18\)
\(=9.2.4.14+6.3.\left(-4\right)\left(-17\right)+76.18\)
\(=18.56+18.68+18.76\)
\(=18\left(56+68+76\right)\)
\(=18\left(132+68\right)\)
\(=18.200\)
\(=3600.\)
c) \(\left(\dfrac{-1}{2}\right)^3-2.\left(\dfrac{-1}{2}\right)^2+3.\left(\dfrac{-1}{2}\right)+1\)
\(=\left(\dfrac{-1}{2}\right)\left[\left(\dfrac{-1}{2}\right)^2+2.\dfrac{-1}{2}+3\right]+1\)
\(=\left(\dfrac{-1}{2}\right)\left[\dfrac{1}{4}+\left(-1\right)+3\right]+1\)
\(\)\(=\left(\dfrac{-1}{2}\right)\left[\dfrac{1}{4}+2\right]+1\)
\(=\left(\dfrac{-1}{2}\right).\dfrac{9}{4}+1\)
\(=\dfrac{-9}{8}+1\)
\(=\dfrac{-1}{8}\)
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a) Ta có: \(\dfrac{x+12}{10-x}=-\dfrac{x-10+22}{x-10}=-1+\dfrac{22}{x-10}\)
Vì \(\left(x+12\right)⋮\left(10-x\right)\) nên \(22⋮\left(x-10\right)\)
Do đó ta có bảng:
| x-10 | -22 | -11 | -2 | -1 | 1 | 2 | 22 |
| x | -12 | -1 | 8 | 9 | 11 | 12 | 32 |
Vậy \(x\in\left\{-12;-1;8;9;11;12;32\right\}\)
c) \(\left(x-3\right)⋮\left(x+1\right)\)
=> \(\left(x-3\right)-\left(x+1\right)⋮\left(x+1\right)\)
=> \(\left(x-3-x-1\right)⋮\left(x+1\right)\)
=>\(-4⋮\left(x+1\right)\)
=> x+1\(\in\) ư(-4)= \(\left\{\pm1,\pm2,\pm4\right\}\)
ta có bảng sau
| x+1 | -4 | -2 | -1 | 1 | 2 | 4 |
| x | -5 | -3 | -2 | 0 | 1 | 3 |
vậy x\(\in\left\{-5,-3;-2;0;1;3\right\}\)
a) \(\left(x-1\right):3=2^3\) \(\Leftrightarrow\) \(\left(x-1\right):3=8\) \(x+1=24\) \(\Leftrightarrow\) \(x=23\) vậy \(x=23\)
b) \(12-2\left(x+5\right)=-10\) \(\Leftrightarrow\) \(12-2x-10=-10\)
\(\Leftrightarrow\) \(-2x=-12\) \(\Leftrightarrow\) \(x=6\) vậy \(x=6\)
c) \(x-12\left(x+5\right)=-10\) \(\Leftrightarrow\) \(x-12x-60=-10\)
\(\Leftrightarrow\) \(-11x=50\) \(\Leftrightarrow\) \(x=\dfrac{50}{-11}\) vậy \(x=\dfrac{50}{-11}\)
e) \(13-x:2=10\Leftrightarrow-x:2=-3\Leftrightarrow x=\dfrac{3}{2}\)
f) \(\left|12-x\right|-7=5\)
th1 : \(x\le12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(12-x-7=5\) \(\Leftrightarrow\) \(-x=0\Leftrightarrow x=0\)
th2 : \(x>12\) thì \(\left|12-x\right|-7=5\) \(\Leftrightarrow\) \(x-12-7=5\) \(\Leftrightarrow\) \(x=24\) vậy \(x=0;x=24\)
i) \(x^2-7=2\Leftrightarrow x^2=9\Leftrightarrow x=3\) vậy \(x=3\)
k) \(x^3-4=-12\) \(\Leftrightarrow\) \(x^3=-8\) \(\Leftrightarrow x=-2\) vậy \(x=-2\)
a)\(\left(x-1\right):3=2^3\Rightarrow x-1=2^3.3=24\Rightarrow x=25\)
b)\(12-2\left(x+5\right)=-10\Leftrightarrow12-2x-10=-10\Rightarrow2-2x=-10\Rightarrow2x=12\Rightarrow x=6\)c)\(x-12\left(x+5\right)=-10\Rightarrow x-12x-60=-10\Rightarrow-11x-60=-10\Rightarrow-11x=-70\Rightarrow x=\dfrac{70}{-11}\)d)\(6-\left|x\right|=5\Rightarrow\left|x\right|=1\Rightarrow x=\left\{\pm1\right\}\)
Làm nốt nha
D
\(\left(4^{-1}-3^{-1}\right)^{-1}=\left(\dfrac{1}{4}-\dfrac{1}{3}\right)^{-1}=\left(\dfrac{-1}{12}\right)^{-1}=\dfrac{1}{\dfrac{-1}{12}}=-12\)
Chọn A