Tìm x,y biết:
a) (x-5) + |4-2y| = 0
b)|x-3| + |x-1/2|=0
Giúp mình với nha.
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9 tháng 12 2021
\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
NN
2
TP
25 tháng 12 2022
\((2x-1)^2+(x+3)^2-5(x+7)(x-7)=0\)
\(< =>4x^2-4x+1+x^2+6x+9-5\left(x^2-7^2\right)=0\\ < =>4x^2-4x+1+x^2+6x+9-5x^2+245=0\\ < =>2x+255=0\\ < =>2x=-255=>x=\dfrac{-255}{2}\)
Vậy \(x=\dfrac{-255}{2}\)
DX
25 tháng 12 2022
\(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Rightarrow2x+255=0\Rightarrow2x=-255\Rightarrow x=-\dfrac{255}{2}\)
13 tháng 1 2022
Bài 2:
a: =>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
29 tháng 1 2023
a: x/2=-5/y
=>xy=-10
=>\(\left(x,y\right)\in\left\{\left(1;-10\right);\left(-10;1\right);\left(-1;10\right);\left(10;-1\right);\left(2;-5\right);\left(-5;2\right);\left(-2;5\right);\left(5;-2\right)\right\}\)
b: =>xy=12
mà x>y>0
nên \(\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)
c: =>(x-1)(y+1)=3
=>\(\left(x-1;y+1\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(2;2\right);\left(4;0\right);\left(0;-4\right);\left(-2;-2\right)\right\}\)
d: =>y(x+2)=5
=>\(\left(x+2;y\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(-1;5\right);\left(3;1\right);\left(-3;-5\right);\left(-7;-1\right)\right\}\)
11 tháng 1 2021
a) Ta có: \(7x^2-28=0\)
\(\Leftrightarrow7\left(x^2-4\right)=0\)
\(\Leftrightarrow7\left(x-2\right)\left(x+2\right)=0\)
mà 7>0
nên (x-2)(x+2)=0
hay \(\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-2\right\}\)
b) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)
\(\Leftrightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)
mà \(\dfrac{2}{3}>0\)
nên x(x-2)(x+2)=0
hay \(\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-2;2\right\}\)
c) Ta có: \(2x\left(3x-5\right)-\left(5-3x\right)=0\)
\(\Leftrightarrow2x\left(3x-5\right)+\left(3x-5\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=5\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{5}{3};-\dfrac{1}{2}\right\}\)
d) Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-1-5\right)\left(2x-1+5\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-2\right\}\)
VT
2
4 tháng 10 2021
\(a,\Rightarrow4x\left(x^2-9\right)=0\\ \Rightarrow4x\left(x-3\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,\Rightarrow\left(3x-5-x-1\right)\left(3x-5+x+1\right)=0\\ \Rightarrow\left(2x-6\right)\left(4x-4\right)=0\\ \Rightarrow2\left(x-3\right)4\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
4 tháng 10 2021
a) \(\Rightarrow4x\left(x^2-9\right)=0\)
\(\Rightarrow4x\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow\left(3x-5-x-1\right)\left(3x-5+x+1\right)=0\)
\(\Rightarrow\left(2x-6\right)\left(4x-4\right)=0\)
\(\Rightarrow8\left(x-3\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
12 tháng 10 2021
Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
HP
21 tháng 9 2021
a. 9x2 - 6x - 3 = 0
<=> 3(3x2 - 2x - 1) = 0
<=> 3(3x2 - 3x + x - 1) = 0
<=> \(3\left[3x\left(x-1\right)+\left(x-1\right)\right]=0\)
<=> 3(3x + 1)(x - 1) = 0
<=> \(\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\)
b. (2x + 1)2 - 4(x + 2)2 = 9
<=> (2x + 1)2 - \(\left[2\left(x+2\right)\right]^2=9\)
<=> (2x + 1 - 2x - 4)(2x + 1 + 2x + 4) = 9
<=> -3(4x + 5) = 9
<=> 4x + 5 = -3
<=> 5 + 3 = -4x
<=> -4x = 8
<=> -x = 2
<=> x = -2
21 tháng 9 2021
a) \(\Leftrightarrow\left(9x^2-6x+1\right)-4=0\)
\(\Leftrightarrow\left(3x-1\right)^2-4=0\)
\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)
\(\Leftrightarrow12x=-24\Leftrightarrow x=-2\)
c) \(\Leftrightarrow3x^2-6x+3-3x^2+15x=21\)
\(\Leftrightarrow9x=18\Leftrightarrow x=2\)
d) \(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)
\(\Leftrightarrow2x=-40\Leftrightarrow x=-20\)
6 tháng 11 2021
\(a,\Leftrightarrow\left(x+3\right)\left(x+3-x+3\right)=0\Leftrightarrow x=-3\\ b,\Leftrightarrow x=0\left(x^2+4>0\right)\)
NH
6 tháng 11 2021
\(a,x^2+2.x.3+3^2-\left(x^2-3^2\right)=0\)
\(x^2+6x+9-x^2+9=0\)
\(6x+18=0\)
\(6x=-18\)
\(x=-3\)
Vậy x=-3
\(b,5x^3+20x=0\)
\(5x\left(x^2+4\right)=0\)
\(Th1:5x=0=>x=0\)
\(Th2:x^2+4=0\)
\(x^2=-4\)(vô lý)
Vậy x=0
\(\left|x-3\right|+\left|x-\dfrac{1}{2}\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)( vô lý)
Vậy \(S=\varnothing\)
b: \(\left|x-3\right|+\left|x-\dfrac{1}{2}\right|\ge0\forall x\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\in\varnothing\)