Tìm x
a) | x| +|x-2| =2
b) |2x-1| +|9-2x|=8
c) |3x+7|+|3x-20|=27
d) |10-x| +|x+34|=40
K
Khách
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25 tháng 8 2020
a) ( 3x + 2 )( x - 1 ) - ( x + 2 )( 3x + 1 ) = 7
<=> 3x2 - x - 2 - ( 3x2 + 7x + 2 ) = 7
<=> 3x2 - x - 2 - 3x2 - 7x - 2 = 7
<=> -8x - 4 = 7
<=> -8x = 11
<=> x = -11/8
b) ( 6x + 5 )( 2x + 3 ) - ( 4x + 3 )( 3x - 2 ) = 8
<=> 12x2 + 28x + 15 - ( 12x2 + x - 6 ) = 8
<=> 12x2 + 28x + 15 - 12x2 - x + 6 = 8
<=> 27x + 21 = 8
<=> 27x = -13
<=> x = -13/27
c) 2x( x + 3 ) - ( x + 1 )( 2x + 1 ) - 5 = 9
<=> 2x2 + 6x - ( 2x2 + 3x + 1 ) - 5 = 9
<=> 2x2 + 6x - 2x2 - 3x - 1 - 5 = 9
<=> 3x - 6 = 9
<=> 3x = 15
<=> x = 5
d) ( 5x + 3 )( 4x - 7 ) - ( 10x + 9 )( 2x - 3 ) = 10
<=> 20x2 - 23x - 21 - ( 20x2 - 12x - 27 ) = 10
<=> 20x2 - 23x - 21 - 20x2 + 12x + 27 = 10
<=> -11x + 6 = 10
<=> -11x = 4
<=> x = -4/11
A
25 tháng 8 2020
a, \(\left(3x+2\right)\left(x-1\right)-\left(x+2\right)\left(3x+1\right)=7\Leftrightarrow-8x-4=7\Leftrightarrow x=-\frac{11}{8}\)
b, \(\left(6x+5\right)\left(2x+3\right)-\left(4x+3\right)\left(3x-2\right)=8\Leftrightarrow27x+21=8\Leftrightarrow x=-\frac{13}{27}\)
c, \(2x\left(x+3\right)-\left(x+1\right)\left(2x+1\right)-5=9\Leftrightarrow3x-6=9\Leftrightarrow x=5\)
d, \(\left(5x+3\right)\left(4x-7\right)-\left(10x+9\right)\left(2x-3\right)=10\Leftrightarrow-11x+6=10\Leftrightarrow x=-\frac{4}{11}\)
3T
2
31 tháng 10 2021
\(a,\Rightarrow x=3\\ b,\Rightarrow2x-1=2\Rightarrow x=\dfrac{3}{2}\\ c,\Rightarrow\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\\ d,\Rightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\\ e,\Rightarrow2x+5=3^2=9\Rightarrow x=2\)
18 tháng 1 2022
a: =>-2x=90/91
hay x=-45/91
b: =>2x=-7
hay x=-7/2
c: ->-3x=-12
hay x=4
KM
16 tháng 8 2019
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(=>2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
KM
16 tháng 8 2019
b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)
=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.
\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)
24 tháng 9 2021
\(1,A=\left(3x+7\right)\left(2x+3\right)-\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =6x^2+23x+21-2x-3-6x^2-23x+55\\ =73-2x\left(đề.sai\right)\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ 2,\\ a,\Leftrightarrow30x^2+18x+3x-30x^2=7\\ \Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\\ \Leftrightarrow79x=79\Leftrightarrow x=1\\ c,\Leftrightarrow\left(x+5\right)\left(x^2+3x+2\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\\ \Leftrightarrow17x=17\Leftrightarrow x=1\)
\(d,\Leftrightarrow7x-2x^2-3+x^2+x-6=-x^2-x+2\\ \Leftrightarrow9x=11\Leftrightarrow x=\dfrac{11}{9}\)
a/ Ta có :
\(\left|x-2\right|=\left|2-x\right|\)
\(\Leftrightarrow\left|x\right|+\left|x-2\right|=\left|x\right|+\left|2-x\right|\)
\(\Leftrightarrow\left|x\right|+\left|x-2\right|\ge\left|x+2-x\right|\)
\(\Leftrightarrow\left|x\right|+\left|x-2\right|\ge2\)
Dấu "=" xảy ra khi :
\(x\left(x-2\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x-2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x-2\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\ge2\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x\le2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow2\le x\le2\Leftrightarrow x=2\)
Vậy \(x=2\)
b/ \(\left|2x-1\right|+\left|9-2x\right|\ge\left|2x-1+9-2x\right|\)
\(\Leftrightarrow\left|2x-1\right|+\left|9-2x\right|\ge8\)
Dấu "=" xảy ra khi :
\(\left(2x-1\right)\left(9-2x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1\ge0\\9-2x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1\le0\\9-2x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\le\dfrac{9}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\x\ge\dfrac{9}{2}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\dfrac{1}{2}\le x\le\dfrac{9}{2}\)
Vậy ....
c/ Ta có : \(\left|3x-20\right|=\left|20-3x\right|\)
\(\Leftrightarrow\left|3x+7\right|+\left|3x-20\right|=\left|3x+7\right|+\left|20-3x\right|\)
\(\Leftrightarrow\left|3x+7\right|+\left|3x-20\right|\ge\left|3x+7+20-3x\right|\)
\(\Leftrightarrow\left|3x+7\right|+\left|3x-20\right|\ge27\)
Dấu "=" xảy ra khi :
\(\left(3x+7\right)\left(20-3x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x+7\ge0\\20-3x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}3x+7\le0\\20-3x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-\dfrac{7}{3}\\x\le\dfrac{20}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-\dfrac{7}{3}\\x\ge\dfrac{20}{3}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\dfrac{20}{3}\le x\le-\dfrac{7}{3}\)
Vậy...
d/ \(\left|10-x\right|+\left|x+30\right|\ge\left|10-x+x+30\right|\)
\(\Leftrightarrow\left|10-x\right|+\left|x+30\right|\ge40\)
Dấu "=" xảy ra khi :
\(\left(10-x\right)\left(x+30\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}10-x\ge0\\x+30\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}10-x\le0\\x+30\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}10\ge x\\x\ge-30\end{matrix}\right.\\\left\{{}\begin{matrix}10\le x\\x\le-30\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow10\ge x\ge-30\)
Vậy...