Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có: \(\left(2x-4\right)\left(3x+1\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)\left[2\left(3x+1\right)+\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(6x+2+x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\cdot7x=0\)
Vì 7≠0
nên \(\left[{}\begin{matrix}x-2=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
Vậy: x∈{0;2}
b) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\cdot3x=0\)
Vì 3≠0
nên \(\left[{}\begin{matrix}x+2=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)
Vậy: x∈{0;-2}
c) Ta có: \(2x^2-x=0\)
\(\Leftrightarrow x\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{1}{2}\right\}\)
d) Ta có: \(x^3-6x^2+9x=0\)
\(\Leftrightarrow x\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow x\left(x-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy: x∈{0;3}
k) Ta có: \(x^3+3x^2+x+3=0\)
\(\Leftrightarrow x^2\left(x+3\right)+\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+1\right)=0\)(1)
Ta có: \(x^2+1\ge1>0\forall x\)(2)
Từ (1) và (2) suy ra x+3=0
hay x=-3
Vậy: x=-3
cái bài a) thì số 2 đâu ra thế bạn?
<=>(x−2)[2(3x+1)+(x−2)]=0
a) \(\left(y-1\right)^2=9\)
\(\Rightarrow\left(y-1\right)^2=3^2=\left(-3\right)^2\)
\(\Rightarrow x-1=3\Rightarrow x=4\)
\(\Rightarrow x-1=-3\Rightarrow x=-2\)
Vậy: \(x=4\) hoặc \(-2\)
Ik mk nha, hôm nay ngày mai, ngày kia mk ik 3 lần lại cho bạn (thành 9 lần)
Nhớ kb với mìn lun nha!! Mk rất vui đc làm quen vs bạn, cảm ơn mn nhìu lắm
b) \(x^3+6x^2+9x=0\)
\(\Leftrightarrow x^3+3x^2+3x^2+9x=0\)
\(\Leftrightarrow x^2\left(x+3\right)+3x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+3x\right)=0\)
\(\Leftrightarrow\left(x+3\right)x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)^2x=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=0\\x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=0\end{cases}}}\)
Vậy \(x\in\left\{-3;0\right\}\)
a) \(2x\left(x-2\right)+x^2=4\)
\(\Leftrightarrow2x\left(x-2\right)+x^2-4=0\)
\(\Leftrightarrow2x\left(x-2\right)+\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{-2}{3}\end{cases}}}\)
Vậy \(x\in\left\{\frac{-2}{3};2\right\}\)
a. x.(x+3)-x2+15=0
=> x^2 + 3x - x^2 + 15 = 0
=> 3x + 15 = 0
=> 3x = -15
=> x = -5
vậy_
b. (2x-1)(x+3) - x(2x-6) =15
=> 2x^2 + 6x - x - 3 - 2x^2 + 6x = 15
=> x - 3 = 15
=> x = 18
vậy_
c. x3 -36x = 0
=> x(x^2 - 36) = 0
=> x = 0 hoặc x^2 - 36 = 0
=> x = 0 hoặc x^2 = 36
=> x = 0 hoặc x = 6 hoặc x = -6
vậy_
d. 6x2 + 6x =x2+2x+1
=> 6x(x + 1) = (x + 1)^2
=> 6x(x + 1) - (x + 1)^2 = 0
=> (x + 1)(6x - x - 1) = 0
=> (x + 1)(5x - 1) = 0
=> x = -1 hoặc 5x = 1
=> x = -1 hoặc x = 1/5
vậy_
e. x(3x+1)=1-9x2
=> x(3x + 1) = (1 - 3x)(1 + 3x)
=> x(3x + 1) - (1 - 3x)(1 + 3x) = 0
=> (3x + 1)(x - 1 + 3x) = 0
=> (3x + 1)(4x - 1) = 0
=> 3x + 1 = 0 hoặc 4x - 1 = 0
=> 3x = -1 hoặc 4x = 1
=> x = -1/3 hoặc x = 1/4
vậy_
a) 2x (x-5) -(x2-10x +25)=0
\(\Leftrightarrow\)2x(x-5)-(x-5)2=0
\(\Leftrightarrow\)(x-5)(2x-x+5)=0
\(\Leftrightarrow\)(x-5)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
b) x2 - 9 +3x(x+3) = 0
\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0
\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0
\(\Leftrightarrow\)(x+3)(x-3+3x)=0
\(\Leftrightarrow\)(x+3)(4x-3)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)
c) x3 - 16x = 0
\(\Leftrightarrow\)x(x2-16)=0
\(\Leftrightarrow\)x(x-4)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) (2x+3)(x-2) - (x2 -4x+4) = 0
\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0
\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0
\(\Leftrightarrow\)(x-2)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
e) 9x2 -(x2 -2x +1)=0
\(\Leftrightarrow\)(3x)2-(x-1)2=0
\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0
\(\Leftrightarrow\)(2x+1)(4x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
f)x3-4x2 -9x +36 = 0
\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0
\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0
\(\Leftrightarrow\)(x-4)(x2-9)=0
\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
g) 3x - 6 = (x-1).(x-2)
\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)
\(\Leftrightarrow\)x-1=3
\(\Leftrightarrow\)x=4
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)
k) x2 -1 = (x-1).(2x+3)
\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)
\(\Leftrightarrow\)x+1=2x+3
\(\Leftrightarrow\)x-2x=3-1
\(\Leftrightarrow\)-x=2
\(\Leftrightarrow\)x=-2
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6
\(\Leftrightarrow\)6x-8=6
\(\Leftrightarrow\)6x=14
\(\Leftrightarrow\)x=\(\frac{7}{3}\)
a) x4 - 16x2 = 0
<=> ( x2 )2 - ( 4x )2 = 0
<=> ( x2 - 4x )( x2 + 4x ) = 0
<=> [ x( x - 4 ) ][ x( x + 4 ) ] = 0
<=> x( x - 4 )x( x + 4 ) = 0
<=> x2( x - 4 )( x + 4 ) = 0
<=> \(\hept{\begin{cases}x^2=0\\x-4=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)( thay bằng dấu hoặc hộ mình nhé )
b) 9x2 + 6x + 1 = 0
<=> ( 3x )2 + 2.3x.1 + 12 = 0
<=> ( 3x + 1 )2 = 0
<=> 3x + 1 = 0
<=> 3x = -1
<=> x = -1/3
c) x2 - 6x = 16
<=> x2 - 6x - 16 = 0
<=> x2 + 2x - 8x - 16 = 0
<=> x( x + 2 ) - 8( x + 2 ) = 0
<=> ( x + 2 )( x - 8 ) = 0
<=> \(\orbr{\begin{cases}x+2=0\\x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}\)
d) 9x2 + 6x = 80
<=> 9x2 + 6x - 80 = 0
<=> 9x2 + 30x - 24x - 80 = 0
<=> 9x( x + 10/3 ) - 24( x + 10/3 ) = 0
<=> ( x + 10/3 )( 9x - 24 ) = 0
<=> \(\orbr{\begin{cases}x+\frac{10}{3}=0\\9x-24=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{10}{3}\\x=\frac{8}{3}\end{cases}}\)
e) Áp dụng công thức an.bn = ( ab )n ta có :
25( 2x - 1 )2 - 9( x + 1 )2 = 0
<=> 52( 2x - 1 )2 - 32( x + 1 )2 = 0
<=> [ 5( 2x - 1 ) ]2 - [ 3( x + 1 ) ]2 = 0
<=> ( 10x - 5 )2 - ( 3x + 3 )2 = 0
<=> [ ( 10x - 5 ) - ( 3x + 3 ) ][ ( 10x - 5 ) + ( 3x + 3 ) ] = 0
<=> ( 10x - 5 - 3x - 3 )( 10x - 5 + 3x + 3 ) = 0
<=> ( 7x - 8 )( 13x - 2 ) = 0
<=> \(\orbr{\begin{cases}7x-8=0\\13x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{7}\\x=\frac{2}{13}\end{cases}}\)
Bài làm :
a) x4 - 16x2 = 0
<=> ( x2 )2 - ( 4x )2 = 0
<=> ( x2 - 4x )( x2 + 4x ) = 0
<=> [ x( x - 4 ) ][ x( x + 4 ) ] = 0
<=> x( x - 4 )x( x + 4 ) = 0
<=> x2( x - 4 )( x + 4 ) = 0
Vậy x=0 hoặc x=±4
b) 9x2 + 6x + 1 = 0
<=> ( 3x )2 + 2.3x.1 + 12 = 0
<=> ( 3x + 1 )2 = 0
<=> 3x + 1 = 0
<=> 3x = -1
<=> x = -1/3
c) x2 - 6x = 16
<=> x2 - 6x - 16 = 0
<=> x2 + 2x - 8x - 16 = 0
<=> x( x + 2 ) - 8( x + 2 ) = 0
<=> ( x + 2 )( x - 8 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}\)
d) 9x2 + 6x = 80
<=> 9x2 + 6x - 80 = 0
<=> 9x2 + 30x - 24x - 80 = 0
<=> 9x( x + 10/3 ) - 24( x + 10/3 ) = 0
<=> ( x + 10/3 )( 9x - 24 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{10}{3}=0\\9x-24=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{10}{3}\\x=\frac{8}{3}\end{cases}}\)
e) 25( 2x - 1 )2 - 9( x + 1 )2 = 0
<=> 52( 2x - 1 )2 - 32( x + 1 )2 = 0
<=> [ 5( 2x - 1 ) ]2 - [ 3( x + 1 ) ]2 = 0
<=> ( 10x - 5 )2 - ( 3x + 3 )2 = 0
<=> [ ( 10x - 5 ) - ( 3x + 3 ) ][ ( 10x - 5 ) + ( 3x + 3 ) ] = 0
<=> ( 10x - 5 - 3x - 3 )( 10x - 5 + 3x + 3 ) = 0
<=> ( 7x - 8 )( 13x - 2 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}7x-8=0\\13x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{7}\\x=\frac{2}{13}\end{cases}}\)
a) Ta có : x4 - 16x2 = 0
=> x4 - 8x2 - 8x2 + 64 = 64
=> x2(x2 - 8) - 8(x2 - 8) = 64
=> (x2 - 8)2 = 64
=> \(\orbr{\begin{cases}x^2-8=8\\x^2-8=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=16\\x^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\pm4\\x=0\end{cases}}\Rightarrow x\in\left\{4;-4;0\right\}\)
b) Ta có 9x2 + 6x + 1 = 0
=> 9x2 + 3x + 3x + 1 = 0
=> 3x(3x + 1) + (3x + 1) = 0
=> (3x + 1)2 = 0
=> 3x + 1 = 0
=> x = -1/3
c) Ta có x2 - 6x = 16
=> x2 - 6x + 9 = 25
=> (x - 3)2 = 25
=> \(\orbr{\begin{cases}x-3=5\\x-3=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=-2\end{cases}}\Rightarrow x\in\left\{8;-2\right\}\)
d) 9x2 + 6x = 80
=> 9x2 + 6x + 1 = 81
=> (3x + 1)2 = 81
=> \(\orbr{\begin{cases}3x+1=9\\3x+1=-9\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=-\frac{10}{3}\end{cases}\Rightarrow x\in}\left\{\frac{8}{3};\frac{-10}{3}\right\}\)
e) 25(2x - 1)2 - 9(x + 1)2 = 0
=> [5(2x - 1)]2 - [3(x + 1)]2 = 0
=> (10x - 5)2 - (3x + 3)2 = 0
=> (10x - 5 - 3x - 3)(10x - 5 + 3x + 3) = 0
=> (7x - 8)(13x - 2) = 0
=> \(\orbr{\begin{cases}7x=8\\13x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{8}{7}\\x=\frac{2}{13}\end{cases}}\)
a)\(\left(x-2\right)^2-\left(2x+3\right)^2=0\Rightarrow\left(x-2+2x+3\right)\left(x-2-2x-3\right)=0\)
\(\Rightarrow\left(3x+1\right)\left(-x-5\right)=0\Rightarrow\left[{}\begin{matrix}3x+1=0\\-x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
b)\(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\Rightarrow\left[3\left(2x+1\right)+2\left(x+1\right)\right]\left[3\left(2x+1\right)-2\left(x+1\right)\right]=0\)
\(\Rightarrow\left[8x+5\right]\left[4x+1\right]=0\Rightarrow\left[{}\begin{matrix}8x+5=0\\4x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)
c)\(x^3-6x^2+9x=0\Rightarrow x\left(x^2-6x+9\right)=0\Rightarrow x\left(x-3\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
d) \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x^2-1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x-1\right)\left(x+1\right)+x\left(x-1\right)=0\)
\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)\left(x+1\right)+1\right]=0\)
\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)^2+1\right]=0\)
Do \(\left(x+1\right)^2+1>0\)
\(\Rightarrow x\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)