1 Tìm x
a , ( 2 - x ) . ( 4 + x ) = 0
b , ( x - 6 ) . ( 3 - x ) = 0
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17 tháng 8 2021
a: Ta có: \(4\left(x+1\right)^2+\left(2x+1\right)^2-8\left(x-1\right)\left(x+1\right)-11=0\)
\(\Leftrightarrow4x^2+8x+4+4x^2+4x+1-8x^2+8-11=0\)
\(\Leftrightarrow12x=-2\)
hay \(x=-\dfrac{1}{6}\)
b: Ta có: \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)-1=0\)
\(\Leftrightarrow x^2+6x+9-x^2-4x+32-1=0\)
\(\Leftrightarrow2x=-40\)
hay x=-20
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 \(x=0;x=2\)
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}
HP
1
24 tháng 10 2021
\(a,Sửa:2021x-1+2022x\left(1-2021x\right)=0\\ \Leftrightarrow\left(2021x-1\right)\left(1-2022x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2021}\\x=\dfrac{1}{2022}\end{matrix}\right.\)
N
1
16 tháng 8 2023
`a,(5-x)(x-1) < 0`
`<=>5-x<0` hoặc `x-1<0`
`<=>5 <x` hoặc `x<1`
Vậy `S={x|5<x;x<1}`
`b,(x-4)(x+1/2) >= 0`
`<=>TH1 : {(x-4>=0),(x+1/2 >=0):}<=>{(x>=4(TM)),(x>= -1/2(L)):}`
`<=>TH2 :{(x-4<=0),(x+1/2 <= 0):} <=>{(x<=4(L)),(x<=-1/2(TM)):}`
`=>x<= -1/2` hoặc `x>=4`
Vậy `S={x|x<= -1/2 ; x>=4}`
KV
1 tháng 1
a) x² - 4 = 0
x² = 4
x = 2 hoặc x = -2
b) 2x(x + 5) - 3(5 + x) = 0
(x + 5)(2x - 3) = 0
X + 5 = 0 hoặc 2x - 3 = 0
*) x + 5 = 0
x = -5
*) 2x - 3 = 0
2x = 3
x = 3/2
c) x³ - 6x² + 11x - 6 = 0
x³ - x² - 5x² + 5x + 6x - 6 = 0
(x³ - x²) - (5x² - 5x) + (6x - 6) = 0
x²(x - 1) - 5x(x - 1) + 6(x - 1) = 0
(x - 1)(x² - 5x + 6) = 0
(x - 1)(x² - 2x - 3x + 6) = 0
(x - 1)[(x² - 2x) - (3x - 6)] = 0
(x - 1)[x(x - 2) - 3(x - 2)] = 0
(x - 1)(x - 2)(x - 3) = 0
x - 1 = 0 hoặc x - 2 = 0 hoặc x - 3 = 0
*) x - 1 = 0
x = 1
*) x - 2 = 0
x = 2
*) x - 3 = 0
x = 3
Vậy x = 1; x = 2; x = 3
K
30 tháng 6 2021
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
DT
1
28 tháng 12 2021
\(a,\Leftrightarrow x^2-2x-x^2+1=0\\ \Leftrightarrow-2x+1=0\Leftrightarrow x=\dfrac{1}{2}\\ b,\Leftrightarrow\left(2x-1-x-4\right)\left(2x-1+x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(3x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
NH
2
29 tháng 7 2021
`a)4x(x-2)+x-2=0`
`<=>(x-2)(4x+1)=0`
`<=>[(x-2=0),(4x+1=0):}`
`<=>[(x=2),(x=-1/4):}`
Vậy `S={2;-1/4}.`
`b)(3x-1)^3-9=0`
`<=>(3x-1-3)(3x-1+3)=0`
`<=>(3x-4)(3x+2)=0`
`<=>[(3x-4=0),(3x+2=0):}`
`<=>[(x=4/3),(x=-2/3):}`
Vậy `S={4/3;-2/3}.`
`c)x^3-8+(x-2)(x+1)=0`
`<=>(x-2)(x^2+2x+4)+(x-2)(x+1)=0`
`<=>(x-2)(x^2+3x+5)=0`
Mà `x^2+3x+5=(x+3/2)^2+11/4>=11/4>0`
`<=>x-2=0`
`<=>x=2`
Vậy `S={2}`
29 tháng 7 2021
a) Ta có: \(4x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{4}\end{matrix}\right.\)
b)Ta có: \(\left(3x-1\right)^2-9=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
c) Ta có: \(x^3-8+\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4+x+1\right)=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
NH
2
6 tháng 8 2021
a, \(4x\left(x-2\right)+x-2=0\Leftrightarrow\left(4x+1\right)\left(x-2\right)=0\Leftrightarrow x=-\dfrac{1}{4};x=2\)
b, \(\left(3x-1\right)^2-9=0\Leftrightarrow\left(3x-4\right)\left(3x+2\right)=0\Leftrightarrow x=\dfrac{4}{3};x=-\dfrac{2}{3}\)
c, \(x^3-8+\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)+\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+3x+5\ne0\right)=0\Leftrightarrow x=2\)
6 tháng 8 2021
a) Ta có: \(4x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{4}\end{matrix}\right.\)
b) Ta có: \(\left(3x-1\right)^2-9=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
a , ( 2 - x ) . ( 4 + x ) = 0
=> 2 - x = 0 hoặc 4 + x = 0
x = 2 - 0 x = 0 - 4
x = 2 x = -4
Vậy x = 2 hoặc x = -4
b , ( x - 6 ) . ( 3 - x ) = 0
=> x - 6 = 0 hoặc 3 - x = 0
x = 0 + 6 x = 3 - 0
x = 6 x = 3
Vậy x = 6 hoặc x = 3
a, (2-x) .( 4+x)=0
=> 2-x = 0 hoặc 4+x =0
x= 2 hoặc x= -4
b, (x-6).( 3-x) =0
=> x-6 =0 hoặc 3-x =0
x= 6 hoặc x=3