K
Khách
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4 tháng 7 2017
Đề này nhá : \(\frac{2x+5}{45}+\frac{2x+6}{44}=\frac{2x-3}{53}+\frac{2x}{50}\)
\(\Leftrightarrow\left(\frac{2x+5}{45}+1\right)+\left(\frac{2x+6}{44}+1\right)=\left(\frac{2x-3}{53}+1\right)+\left(\frac{2x}{50}+1\right)\)
\(\Leftrightarrow\frac{2x+50}{45}+\frac{2x+50}{44}=\frac{2x+50}{53}+\frac{2x+50}{50}\)
\(\Leftrightarrow\frac{2x+50}{45}+\frac{2x+50}{44}-\frac{2x+50}{53}-\frac{2x+50}{50}=0\)
\(\Leftrightarrow\left(2x+50\right)\left(\frac{1}{45}+\frac{1}{44}-\frac{1}{53}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{45}+\frac{1}{44}-\frac{1}{53}-\frac{1}{50}\right)\ne0\)
Nên 2x + 50 = 0
=> 2x = -50
=> x = -25
KP
2
8 tháng 10 2020
Ta có: \(3^{x+3}\cdot3^{2x-1}+3^{2x}\cdot3^{x+1}=324\)
\(\Leftrightarrow3^{3x+2}+3^{3x+1}=324\)
\(\Leftrightarrow3^{3x+1}\cdot\left(3+1\right)=324\)
\(\Leftrightarrow3^{3x+1}\cdot4=324\)
\(\Leftrightarrow3^{3x+1}=81=3^4\)
\(\Rightarrow3x+1=4\)
\(\Leftrightarrow x=1\)
8 tháng 10 2020
\(3^{x+3}\cdot3^{2x-1}+3^{2x}\cdot3^{x+1}=324\)
\(3^{x+3+2x-1}+3^{2x+x+1}=324\)
\(3^{3x+2}+3^{3x+1}=324\)
\(3^{3x+1}\cdot\left(3+1\right)=324\)
\(3^{3x+1}\cdot4=324\)
\(3^{3x+1}=324:4\)
\(3^{3x+1}=81\)
\(3^{3x+1}=3^4\)
\(\Rightarrow3x+1=4\)
\(3x=4-1\)
\(3x=3\)
\(x=3:3\)
\(x=1\)
13 tháng 4 2019
\(a,\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+...+\frac{1}{\left[2x-2\right]\cdot2x}=\frac{1}{8}\)
\(\Rightarrow\frac{1}{2}\left[\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+...+\frac{2}{\left[2x-2\right]\cdot2x}\right]=\frac{1}{8}\)
\(\Rightarrow\frac{1}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}\right]=\frac{1}{8}\)
\(\Rightarrow\frac{1}{2}\left[\frac{1}{2}-\frac{1}{2x}\right]=\frac{1}{8}\)
\(\Rightarrow\left[\frac{1}{2}-\frac{1}{2x}\right]=\frac{1}{8}:\frac{1}{2}\)
\(\Rightarrow\left[\frac{1}{2}-\frac{1}{2x}\right]=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2x}=\frac{1}{2}-\frac{1}{4}\)
\(\Rightarrow\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow2x=4\Leftrightarrow x=2\)
Vậy x = 2
13 tháng 4 2019
Mun ảnh đại diện cute
<3
À tk mk nhé. giờ mk tk bn trước
3 tháng 11 2015
\(1\frac{1}{2}\)( x+1) -3 = \(\frac{1}{3}\)
\(\frac{3}{2}\)(x+1)-3 =\(\frac{1}{3}\)
\(\frac{3}{2}\)(x+1) = \(\frac{1}{3}\)+3
\(\frac{3}{2}\)(x+1) = \(\frac{10}{3}\)
(x+1) = \(\frac{10}{3}\):\(\frac{3}{2}\)
(x+1) = \(\frac{20}{9}\)
x = \(\frac{20}{9}\)-1
x = \(\frac{11}{9}\)
1x+\(\frac{1}{2}\)=1
1x = 1 - \(\frac{1}{2}\)
1x = \(\frac{1}{2}\)
x = \(\frac{1}{2}\):1
x = \(\frac{1}{2}\)
13 tháng 3 2020
a) Ta có:
\(\frac{3}{x+2}=\frac{5}{2x+1}\)
\(\Rightarrow3\left(2x+1\right)=5\left(x+2\right)\)
\(\Rightarrow6x+3=5x+10\)
\(\Rightarrow6x-5x=10-3\)
\(\Rightarrow x=7\)
b)Ta có:
\(\frac{5}{8x-2}=\frac{-4}{7-x}\)
\(\Rightarrow5\left(7-x\right)=-4\left(8x-2\right)\)
\(\Rightarrow35-5x=-32x+8\)
\(\Rightarrow-5x+32x=8-35\)
\(\Rightarrow27x=-27\)
\(\Rightarrow x=-1\)
c) Ta có:
\(\frac{4}{3}=\frac{2x-1}{x}\)
\(\Rightarrow4x=3\left(2x-1\right)\)
\(\Rightarrow4x=6x-3\)
\(\Rightarrow3=6x-4x=2x\)
\(\Rightarrow x=\frac{3}{2}\)
13 tháng 3 2020
d)Ta có:
\(\frac{2x-1}{3}=\frac{3x+1}{4}\)
\(\Rightarrow4\left(2x-1\right)=3\left(3x+1\right)\)
\(\Rightarrow8x-4=9x+3\)
\(\Rightarrow8x-9x=3+4\)
\(\Rightarrow-x=7\Rightarrow x=-7\)
e)Ta có:
\(\frac{4}{x+2}=\frac{7}{3x+1}\)
\(\Rightarrow4\left(3x+1\right)=7\left(x+2\right)\)
\(\Rightarrow12x+4=7x+14\)
\(\Rightarrow12x-7x=14-4\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
f)Ta có:
\(\frac{-3}{x+1}=\frac{4}{2-2x}\)
\(\Rightarrow-3\left(2-2x\right)=4\left(x+1\right)\)
\(\Rightarrow-6+6x=4x+4\)
\(\Rightarrow6x-4x=4+6\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\)
22 tháng 11 2016
a) \(\left(2x+1\right)^3=125\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
b) \(1999^{2x-6}=1\)
\(\Rightarrow1999^{2x-1}=1999^0\)
\(\Rightarrow2x-1=0\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
c) \(x^{2002}=x\)
\(\Rightarrow x^{2002}-x=0\)
\(\Rightarrow x.\left(x^{2001}-1\right)=0\)
\(\Rightarrow x=0\) hoặc \(x^{2001}-1=0\)
+) \(x=0\)
+) \(x^{2001}-1=0\Rightarrow x^{2001}=1\Rightarrow x=1\)
Vậy \(x\in\left\{0;1\right\}\)
d) \(\left(x-1\right)^2=9\)
\(\Rightarrow x-1=\pm3\)
+) \(x-1=3\Rightarrow x=4\)
+) \(x-1=-3\Rightarrow x=-2\)
Vậy \(x\in\left\{4;-2\right\}\)
e) \(\left(2x-3\right)^2=81\)
\(\Rightarrow2x-3=\pm9\)
+) \(2x-3=9\Rightarrow2x=12\Rightarrow x=6\)
+) \(2x-3=-9\Rightarrow2x=-6\Rightarrow x=-3\)
Vậy \(x\in\left\{6;-3\right\}\)
Các phần khác làm tương tự
2x + 1x= 3x
vậy x là bất kì số nào cũng dc
3x
k cho minh