Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) 5xy ( x - y ) - 2x + 2y
= 5xy ( x - y ) - 2 ( x - y )
= ( x - y ) ( 5xy - 2 )
b) 6x-2y-x(y-3x)
= 2 ( y - 3x ) - x ( y - 3x )
= ( y - 3x ( ( 2 - x )
c) x2 + 4x - xy-4y
= x ( x + 4 ) - y ( x + 4 )
( x + 4 ) ( x - y )
d) 3xy + 2z - 6y - xz
= ( 3xy - 6y ) + ( 2z - xz )
= 3y ( x - 2 ) + z ( x - 2 )
= ( x - 2 ) ( 3y + z )
a,5xy(x-y)-2x+2y=5xy(x-y)-2(x-y)=(x-y)(5xy-2)
b,6x-2y-x(y-3x)=-2(y-3x)-x(y-3x)=(y-3x)(-2-x)
c,x^2+4x-xy-4y=x(x+4)-y(x+4)=(x+4)(x-y)
d,3xy+2z-6y-xz=(3xy-6y)+(2z-xz)=3y(x-2)+z(2-x)=3y(x-2)-z(x-2)=(x-2)(3y-z)
11)
a,4-9x^2=0
(2-3x)(2+3x)=0
2-3x=0=>x=2/3 hoặc 2+3x=0=>x=-2/3
b,x^2 +x+1/4=0
(x+1/2)^2 =0
x+1/2=0
x=-1/2
c,2x(x-3)+(x-3)=0
(x-3)(2x+1)=0
x-3=0=>x=3 hoặc 2x+1=0=>x=-1/2
d,3x(x-4)-x+4=0
3x(x-4)-(x-4)=0
(x-4)(3x-1)=0
x-4=0=>x=4 hoặc 3x-1=0=>x=1/3
e,x^3-1/9x=0
x(x^2-1/9)=0
x(x+1/3)(x-1/3)=0
x=0 hoặc x+1/3=0=>x=-1/3 hoặc x-1/3=0=>x=1/3
f,(3x-y)^2-(x-y)^2 =0
(3x-y-x+y)(3x-y+x-y)=0
2x(4x-2y)=0
4x(2x-y)=0
x=0hoặc 2x-y=0=>x=y/2
Bài 1 : Ta có : x3 + 2x2 + x
= x3 + x2 + x2 + x
= x2(x + 1) + x(x + 1)
= (x2 + x)(x + 1)
= x(x + 1)2
Bài : 2 :
a) Ta có : \(\frac{2}{3}x\left(x^2-4\right)=0\)
\(\Rightarrow\frac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)
=> x = 0
x - 2 = 0
x + 2 = 0
=> x = 0
x = 2
x = -2
a) \(x\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
Vậy tập nghiệm của pt là \(S=\left\{4;5\right\}\)
b) \(x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-6\end{cases}}\)
Vậy tập nghiệm của pt là \(S=\left\{-6;7\right\}\)
A) \(\left(x-3\right)^2-\left(x+2\right)^2\)
\(=\left(x-3-x-2\right)\left(x-3+x+2\right)\)
\(=-5.\left(2x-1\right)\)
B) \(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
\(=\left(2x\right)^3-y^3-\left[\left(2x\right)^3+y^3\right]\)
\(=8x^3-y^3-8x^3-y^3\)
\(=-2y^3\)
C) \(x^2+6x+8\)
\(=x^2+6x+9-1\)
\(=\left(x+3\right)^2-1\)
\(=\left(x+3-1\right)\left(x+3+1\right)\)
\(=\left(x+2\right)\left(x+4\right)\)
bài 3 A) \(x^2-16=0\)
\(\left(x-4\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
B) \(x^4-2x^3+10x^2-20x=0\)
\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\left(x^3+10x\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Bài 10 :
Câu a :
\(5xy\left(x-y\right)-2x+2y\)
\(=5xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(5xy-2\right)\)
Câu b :
\(6x-2y-x\left(y-3x\right)\)
\(=2\left(3x-y\right)+x\left(3x-y\right)\)
\(=\left(3x-2y\right)\left(2+x\right)\)
Câu c :
\(x^2+4x-xy-4y\)
\(=x\left(x+4\right)-y\left(x+4\right)\)
\(=\left(x+4\right)\left(x-y\right)\)
Câu d :
\(3xy+2z-6y-xz\)
\(=\left(3xy-6y\right)-\left(xz-2z\right)\)
\(=3y\left(x-2\right)-z\left(x-2\right)\)
\(=\left(x-2\right)\left(3y-z\right)\)
Bài 11 :
Câu a :
\(4-9x^2=0\)
\(\Leftrightarrow\left(2-3x\right)\left(2+3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2-3x=0\\2+3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy ........................
Câu b :
\(x^2+x+\dfrac{1}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x+\dfrac{1}{2}=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy........................
Câu c :
\(2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy..................
Câu d :
\(3x\left(x-4\right)-x+4=0\)
\(\Leftrightarrow3x\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy................................
Câu e :
\(x^3-\dfrac{1}{9}x=0\)
\(\Leftrightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Leftrightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{3}=0\\x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
Vậy........................
Câu f :
\(\left(3x-y\right)^2-\left(x-y\right)^2=0\)
\(\Leftrightarrow\left(3x-y-x+y\right)\left(3x-y+x-y\right)=0\)
\(\Leftrightarrow2x\left(4x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\4x-2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy..........................
Bài 1.
a) -2x( -3x + 2 ) - ( x + 2 )2
= 6x2 - 4x - ( x2 + 4x + 4 )
= 6x2 - 4x - x2 - 4x - 4
= 5x2 - 8x - 4
b) ( x + 2 )( x2 - 2x + 4 ) - 2( x + 1 )( 1 - x )
= x3 + 8 + 2( x + 1 )( x - 1 )
= x3 + 8 + 2( x2 - 1 )
= x3 + 8 + 2x2 - 2
= x3 + 2x2 + 6
c) ( 2x - 1 )2 - 2( 4x2 - 1 ) + ( 2x + 1 )2
= 4x2 - 4x + 1 - 8x2 + 2 + 4x2 + 4x + 1
= 4
d) x2 - 3x + xy - 3y
= x( x - 3 ) + y( x - 3 )
= ( x - 3 )( x + y )
Bài 2.
a) 4x2 - 4xy + y2 = ( 2x - y )2
b) 9x3 - 9x2y - 4x + 4y
= 9x2( x - y ) - 4( x - y )
= ( x - y )( 9x2 - 4 )
= ( x - y )( 3x - 2 )( 3x + 2 )
c) x3 + 2 + 3( x3 - 2 )
= x3 + 2 + 3x3 - 6
= 4x3 - 4
= 4( x3 - 1 )
= 4( x - 1 )( x2 + x + 1 )
Bài 3.
2( x - 2 ) = x2 - 4x + 4
⇔ ( x - 2 )2 - 2( x - 2 ) = 0
⇔ ( x - 2 )( x - 2 - 2 ) = 0
⇔ ( x - 2 )( x - 4 ) = 0
⇔ x = 2 hoặc x = 4
1
a, 2x2+4x+2-2y2 = 2(x2+2x+1-y2)= 2[(x+1)2-y2 ] = 2(x-y+1)(x+y+1)
b, 2x - 2y - x2 + 2xy - y2= 2(x -y) - (x2 - 2xy + y2) = 2(x-y)-(x-y)2=(x-y)(2-x+y)
c, x2-y2-2y-1=x2-(y2+2y+1)=x2-(y+1)2=(x-y-1)(x+y+1)
d, x2-4x-2xy-4y+y2= x2-2xy+y2-4x-4y=(x-y)
2.
a, x2-3x+2=x2-x-2x+2=x(x-1)-2(x-1)=(x-2)(x-1)
b, x2+5x+6=x2+2x+3x+6=x(x+2)+3(x+2)=(x+3)(x+2)
c, x2+6x-6=
a) \(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
Vậy tập nghiệm \(S=\left\{-4;0;4\right\}\)
b) \(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^3+10x\right)\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x^2+10\right)\left(x-2\right)=0\)
Mà \(x^2+10>0\)nên \(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy tập nghiệm S = { 0;2}
bài 2) a) \(x^2+3x+2=0\)
ta có : \(a-b+c=1-3+2=0\)
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(x_1=-1\) ; \(x_2=\dfrac{-c}{a}=\dfrac{-2}{1}=-2\)
vậy phương trình có 2 nghiệm \(x=-1;x=-2\)
b) \(x^2-x-6=0\)
\(\Delta=\left(-1\right)^2-4.1.\left(-6\right)=1+24=25>0\)
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(x_1=\dfrac{1+5}{2}=3\)
\(x_2=\dfrac{1-5}{2}=-2\)
vậy phương trình có 2 nghiệm \(x=3;x=-2\)