tìm x
a, (2x - 3)\(^2\) = |3 - 2x|
b, (x - 1)\(^2\) + (2x - 1)\(^2\) - 0
c, x - 2\(\sqrt{x}\) = 0
d, (x - 1)\(^2\) + 1/7 = 0
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TK
1
KV
16 tháng 10 2023
a) \(6x^2-72x=0\)
\(6x\left(x-12\right)=0\)
\(6x=0\) hoặc \(x-72=0\)
*) \(6x=0\)
\(x=0\)
*) \(x-12=0\)
\(x=12\)
Vậy \(x=0;x=12\)
b) \(-2x^4+16x=0\)
\(-2x\left(x^3-8\right)=0\)
\(-2x=0\) hoặc \(x^3-8=0\)
*) \(-2x=0\)
\(x=0\)
*) \(x^3-8=0\)
\(x^3=8\)
\(x=2\)
Vậy \(x=0;x=2\)
c) \(x\left(x-5\right)-\left(x-3\right)^2=0\)
\(x^2-5x-x^2+6x-9=0\)
\(x-9=0\)
\(x=9\)
d) \(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)=0\)
\(x^3-6x^2+12x-8-x^3+8=0\)
\(-6x^2+12x=0\)
\(-6x\left(x-2\right)=0\)
\(-6x=0\) hoặc \(x-2=0\)
*) \(-6x=0\)
\(x=0\)
*) \(x-2=0\)
\(x=2\)
Vậy \(x=0;x=2\)
TA
1
12 tháng 10 2021
\(a,\Rightarrow3x^2-3x+6-2x-3x^2=0\\ \Rightarrow-5x=-6\Rightarrow x=\dfrac{6}{5}\\ b,\Rightarrow\left(x-1\right)\left(x-1+x+2\right)=0\\ \Rightarrow\left(x-2\right)\left(2x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{2}\end{matrix}\right.\\ c,\Rightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\\ \Rightarrow\left(x^2+1\right)\left(2x+3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\2x+3=0\end{matrix}\right.\\ \Rightarrow x=-\dfrac{3}{2}\\ d,\Rightarrow2x^2+x-6=0\\ \Rightarrow2x^2+4x-3x-6=0\\ \Rightarrow2x\left(x+2\right)-3\left(x+2\right)=0\\ \Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
30 tháng 7 2021
a) \(\left(2x-1\right)^2-25=0\)
⇔ \(\left(2x-1\right)^2-5^2=0\)
⇔ \(\left(2x-1-5\right)\left(2x-1+5\right)=0\)
⇒ \(2x-1-5=0\) hoặc \(2x-1+5=0\)
⇔ \(x=3\) hoặc \(x=-2\)
TC
30 tháng 7 2021
Bài 1: Tìm x
a) (2x-1) ² - 25 = 0
<=> (2x-1)2 = 25
<=> 2x-1 = 5 hay 2x-1 =-5
<=> 2x= 6 hay 2x=-4
<=> x=3 hay x= -2
Vậy S={3; -2}
b) 3x (x-1) + x - 1 = 0
<=> (x-1)(3x+1)=0
<=> x-1=0 hay 3x+1=0
<=> x=1 hay 3x=-1
<=> x=1 hay x=\(\dfrac{-1}{3}\)
Vậy S={1;\(\dfrac{-1}{3}\)}
c) 2(x+3) - x ² - 3x = 0
<=> 2(x+3)- x(x+3)=0
<=> (x+3)(2-x)=0
<=> x+3=0 hay 2-x=0
<=> x=-3 hay x=2
Vậy S={-3;2}
d) x(x - 2) + 3x - 6 = 0
<=> x(x-2)+3(x-2)=0
<=> (x-2)(x+3)=0
<=> x-2=0 hay x+3=0
<=> x=2 hay x=-3
Vậy S={2;-3}
e) 4x ² - 4x +1 = 0
<=> (2x-1)2=0
<=> 2x-1=0
<=> 2x=1
<=> x=\(\dfrac{1}{2}\)
Vậy S={\(\dfrac{1}{2}\)}
f) x +5x2 = 0
<=> x(1+5x)=0
<=>x=0 hay 1+5x=0
<=> x=0 hay 5x=-1
<=> x=0 hay x= \(\dfrac{-1}{5}\)
Vậy S={0;\(\dfrac{-1}{5}\)}
g) x ²+ 2x -3 = 0
<=> x2-x+3x-3=0
<=> x(x-1)+3(x-1)=0
<=> (x-1)(x+3)=0
<=> x-1=0 hay x+3=0
<=> x=1 hay x=-3
Vậy S={1;-3}
26 tháng 4 2022
a. 5 - 3(x + 4) = -1
⇔ 5 - 3x - 12 = -1
⇔ 3x = -1 - 5 + 12
⇔ 3x = 6
⇔ x = 2
TT
26 tháng 4 2022
\(d,2x^2-3=5\)
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x=\pm2\)
\(e,x\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)
24 tháng 12 2023
`#3107.101107`
`1.`
`a,`
`(2x - 3)^2 = |3 - 2x|`
`=> (2x - 3)^2 = |2x - 3|`
`=>`\(\left[{}\begin{matrix}2x-3=\left(2x-3\right)^2\\2x-3=-\left(2x-3\right)^2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x-3-\left(2x-3\right)^2=0\\2x-3+\left(2x-3\right)^2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}\left(2x-3\right)\left(1-2x+3\right)=0\\\left(2x-3\right)\left(1+2x-3\right)=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x-3=0\\4-2x=0\\2x-2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=1\end{matrix}\right.\)
Vậy, `x \in {3/2; 2; 1}`
`b,`
`(x - 1)^2 + (2x - 1)^2 = 0`
`=>`\(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(2x-1\right)^2=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-1=0\\2x-1=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy, `x \in {1; 1/2}`
`c,`
`5 - x^2 = 1`
`=> x^2 = 4`
`=> x^2 = (+-2)^2`
`=> x = +-2`
Vậy, `x \in {-2; 2}`
`d,`
`x - 2\sqrt{x} = 0`
`=> x^2 - (2\sqrt{x})^2 = 0`
`=> x^2 - 4x = 0`
`=> x(x - 4) = 0`
`=>`\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy, `x \in {0; 4}`
`g,`
`(x - 1) + 1/7 = 0`
`=> x - 1 + 1/7 = 0`
`=> x - 6/7 = 0`
`=> x = 6/7`
Vậy, `x = 6/7.`
N
10 tháng 10 2021
\(a.\left(x^2+4x+4\right)+\left(x^2-6x+9\right)=2x^2+14x\)
\(x^2+4x+4+x^2-6x+9-2x^2-14x=0\)
\(-18x+13=0\)
\(x=\dfrac{13}{18}\)
Vậy \(S=\left\{\dfrac{13}{18}\right\}\)
\(b.\left(x-1\right)^3-125=0\)
\(\left(x-1\right)^3=125\)
\(x-1=5\)
\(x=6\)
Vậy \(S=\left\{6\right\}\)
\(c.\left(x-1\right)^2+\left(y +2\right)^2=0\)
\(Do\left(x-1\right)^2\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)
Mà \(\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy \(S=\left\{1;-2\right\}\)
\(d.x^2-4x+4+x^2-2xy+y^2=0\)
\(\left(x-2\right)^2+\left(x-y\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x-y\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
Vậy \(S=\left\{2;2\right\}\)
HM
2
19 tháng 10 2021
\(b,\Rightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Rightarrow5\left(x+2\right)=0\\ \Rightarrow x=-2\\ c,\Rightarrow2x\left(x^2-2x+1\right)=0\\ \Rightarrow2x\left(x-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ d,\Rightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Rightarrow3x\left(-x-2\right)=0\\ \Rightarrow-3x\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3x=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
D
1
25 tháng 10 2021
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(x^2-1=0\Rightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c) \(x^2-9=0\Rightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
d) \(\Rightarrow\left(2x-4\right)\left(2x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
2) \(\Rightarrow\left(5x-3\right)\left(5x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
a: \(\left(2x-3\right)^2=\left|3-2x\right|\)
=>\(\left\{{}\begin{matrix}\left|2x-3\right|>=0\\\left(2x-3\right)^2=\left(2x-3\right)\end{matrix}\right.\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)=0\)
=>\(\left(2x-3\right)\left(2x-3-1\right)=0\)
=>\(\left(2x-3\right)\left(2x-4\right)=0\)
=>\(\left[{}\begin{matrix}2x-3=0\\2x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)
b: \(\left(x-1\right)^2+\left(2x-1\right)^2=0\)
=>\(x^2-2x+1+4x^2-4x+1=0\)
=>\(5x^2-6x+2=0\)
\(\Delta=\left(-6\right)^2-4\cdot5\cdot2=36-20\cdot2=-4< 0\)
=>Phương trình vô nghiệm
c: ĐKXĐ: x>=0
\(x-2\sqrt{x}=0\)
=>\(\sqrt{x}\cdot\sqrt{x}-2\cdot\sqrt{x}=0\)
=>\(\sqrt{x}\left(\sqrt{x}-2\right)=0\)
=>\(\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
d: \(\left(x-1\right)^2+\dfrac{1}{7}=0\)
mà \(\left(x-1\right)^2+\dfrac{1}{7}>=\dfrac{1}{7}>0\forall x\)
nên \(x\in\varnothing\)