Tìm x
\(a,\left|x.\left(x-7\right)\right|=x\)
\(b,\left|x-1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|=5x\)
\(c,7^{x+2}+2.7^{x-1}=345\)
K
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Những câu hỏi liên quan
3 tháng 2 2019
Ta có : 2xy - x - y = 2
<=> 2xy - x = 2 + y
<=> x(2y - 1) = y + 2
=> x = \(\frac{y+2}{2y-1}\)
Vì x nguyên nên \(\frac{y+2}{2y-1}\) nguyên
Ta có ; \(\frac{y+2}{2y-1}=\frac{2y+4}{2y-1}=\frac{\left(2y-1\right)+5}{2y-1}=\frac{2y-1}{2y-1}+\frac{5}{2y-1}=1+\frac{5}{2y-1}\)
Để \(\frac{y+2}{2y-1}\) nguyên thì \(\frac{5}{2y-1}\) nguyên
Suy ra : 2y - 1 \(\in\) Ư(5) = {-5;-1;1;5}
Ta có bảng :
| 2y - 1 | -5 | -1 | 1 | 5 |
| 2y | -4 | 0 | 2 | 6 |
| y | -2 | 0 | 1 | 3 |
| x | 0 | -2 | 3 | 1 |
3 tháng 2 2019
Cảm ơn bạn nhá 
Nhưng mà \(\dfrac{y+2}{2y-1}\) làm sao mà bằng \(\dfrac{2y+4}{2y-1}\)
Phải \(2x\) mới bằng \(\dfrac{2y+4}{2y-1}\) được chứ 
HT
22 tháng 7 2017
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
HT
22 tháng 7 2017
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
10 tháng 7 2023
\(a,\left(x+2\right)^2-9=0\\ \Leftrightarrow\left(x+2-3\right)\left(x+2+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{1;-5\right\}\)
\(b,x^2-2x+1=25\\ \Leftrightarrow\left(x-1\right)^2=25\\ \Leftrightarrow\left(x-1\right)^2-25=0\\ \Leftrightarrow\left(x-1-5\right)\left(x-1+5\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ Vậy\dfrac{ }{ }S=\left\{6;-4\right\}\)
\(c,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\\ \Leftrightarrow25x^2+10x+1-25x^2+9=30\\ \Leftrightarrow25x^2+10x-25x^2=30-1-9\\ \Leftrightarrow10x=20\\ \Leftrightarrow x=2\\ Vậy\dfrac{ }{ }S=\left\{2\right\}\)
\(d,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\\ \Leftrightarrow x^3-1-x\left(x^2-4\right)=5\\ \Leftrightarrow x^3-1-x^3+4x=5\\ \Leftrightarrow x^3-x^3+4x=5+1\\ \Leftrightarrow4x=6\\ \Leftrightarrow x=\dfrac{3}{2}\\ Vậy\dfrac{ }{ }S=\left\{\dfrac{3}{2}\right\}\)
10 tháng 7 2023
a: =>(x+2-3)(x+2+3)=0
=>(x-1)(x+5)=0
=>x=1 hoặc x=-5
b: =>(x-1)^2=25
=>x-1=5 hoặc x-1=-5
=>x=-4 hoặc x=6
c: =>25x^2+10x+1-25x^2+9=30
=>10x+10=30
=>x+1=3
=>x=2
d: =>x^3-1-x(x^2-4)=5
=>x^3-1-x^3+4x=5
=>4x=6
=>x=3/2
19 tháng 3 2020
\(5x\left(x-3\right)\left(x-1\right)-4x\left(x^2-2x\right)\)
\(5x^3-5x^2-15x^2+15x-4x^3+8x^2\)
\(x^3-12x^2+15x\)
\(-4x\left(x+3\right)\left(x-4\right)-3x\left(x^2-x+1\right)\)
\(-4x^3+16x^2-12x^2+48x-3x^3+3x^2-3x\)
\(-7x^3+7x^2+45x\)
28 tháng 9 2017
a/ \(\left|2x-1,6\right|-2,3=1,4\)
\(\Leftrightarrow\left|2x-1,6\right|=3,7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1,6=3,7\\2x-1,6=-3,7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5,3\\2x=-2,1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2,65\\x=-1,05\end{matrix}\right.\)
Vậy ....
b/ \(5,4-\left|3x-1,2\right|=5,5\)
\(\Leftrightarrow\left|3x-1,2\right|=-0,1\)
Mà \(\left|3x-1,2\right|\ge0\)
\(\Leftrightarrow x\in\varnothing\)
c/ \(\left|x+1,3\right|+\left|x+2,4\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+1,3\right|\ge0\\\left|x+2,4\right|\ge0\end{matrix}\right.\) \(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+1,3+x+2,4=4x\)
\(\Leftrightarrow2x+3,7=4x\)
\(\Leftrightarrow3,7=4x-2x\)
\(\Leftrightarrow2x=3,7\)
\(\Leftrightarrow x=1,85\)
Vậy ....
d/ \(\left|x-1,2\right|+\left|2,5-x\right|=0\)
Mà \(\left\{{}\begin{matrix}\left|x-1,2\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x-1,2\right|=0\\\left|2,5-x\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1,2=0\\2,5-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=2,5\end{matrix}\right.\) (loại)
Vậy ..
S
28 tháng 9 2017
a, \(\left|2x-1,6\right|-2,3=1,4\)
\(\Rightarrow\left|2x-1,6\right|=3,7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1,6=3,7\\2x-1,6=-3,7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2,65\\x=-1,05\end{matrix}\right.\)
b,\(5,4-\left|3x-1,2\right|=5,5\)
\(\Rightarrow\left|3x-1,2\right|=-0,1\) (vô lí)
Vì \(\left|x\right|\ge0\) mà \(\left|3x-1,2\right|< 0\)
Vậy, không có giá trị của x thỏa mãn.
c, \(\left|x+1,3\right|+\left|x+2,4\right|=4x\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+1,3\right|\ge0\\\left|x+2,4\right|\ge0\end{matrix}\right.\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+1,3+x+2,4=4x\)
\(\Leftrightarrow x+x+1,3+2,4=4x\)
\(\Leftrightarrow2x+3,7=4x\)
\(\Leftrightarrow2x-4x=-3,7\)
\(\Leftrightarrow-2x=-3,7\)
\(\Leftrightarrow x=\dfrac{3,7}{2}\)
d, \(\left|x-1,2\right|+\left|2,5-x\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-1,2\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x-1,2\right|=0\\\left|2,5-x\right|=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1,2=0\\2,5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=2,5\end{matrix}\right.\)
a) \(\left|x\left(x-7\right)\right|=x\)
\(\Rightarrow\orbr{\begin{cases}x\left(x-7\right)=x\\x\left(x-7\right)=-x\end{cases}\Leftrightarrow\orbr{\begin{cases}x-7=1\\x-7=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=8\\x=6\end{cases}}}\)
b) \(\left|x-1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|=5x\)
\(\Rightarrow x-1,1+x+1,2+x+1,3+x+1,4=5x\)
\(\Leftrightarrow4x+2,8=5x\)
\(\Leftrightarrow x=2,8\)
\(a.\)\(\left|x.\left(x-7\right)\right|=x\)( Đk: \(x\ge0\))
\(\Leftrightarrow\orbr{\begin{cases}x.\left(x-7\right)=x\\x.\left(x-7\right)=-x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=x:x\\x-7=-x:x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=1\\x-7=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1+7\\x=-1+7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=8\\x=6\end{cases}}\)
\(b.\)\(\left|x-1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|=5x\)( Đk: \(5x\ge0\Leftrightarrow x\ge0\))
\(\Rightarrow x-1,1+x+1,2+x+1,3+x+1,4=5x\)
\(\Leftrightarrow\left(x+x+x+x\right)+\left(-1,1+1,2+1,3+1,4\right)=5x\)
\(\Leftrightarrow4x+2,8=5x\)
\(\Leftrightarrow2,8=5x-4x\)
\(\Leftrightarrow x=2,8\)
\(c.\)\(7^{x+2}+2.7^{x-1}=345\)
\(\Leftrightarrow7^{x-1}.7^{x+3}+2.7^{x-1}=345\)
\(\Leftrightarrow7^{x-1}.\left(7^{x+3}+2\right)=345\)
\(......................\)
Đến đây mk ko bt làm nữa, tự lm nhé !