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Câu 1 :
\(\left(x-2\right)^2=x^2-4x+4\)
Câu 2:
\(2x^2\left(4x-5x^3\right)+10x^5-5x^3\)
\(=8x^3-10x^5+10x^5-5x^3\)
\(=3x^3\)
\(\left(x-2\right)\left(x^2-2x+4\right)+\left(x-4\right)\left(x-2\right)\)
\(=x^3-4x^2+8x-8+x^2-6x+8\)
\(=x^3-3x^2+2x\)
Còn lại tự làm nha dài lắm
(2x-1)*(y-1)=10
suy ra 2x-1=10/(y-1)
suy ra (y-1) thuộc ước của 10.ta có bảng sau:
|
y-1 |
1 |
-1 |
2 |
-2 |
5 |
-5 |
10 |
-10 |
|
y |
2 |
0 |
3 |
-1 |
6 |
-4 |
11 |
-9 |
|
x |
3 |
-4,5 |
13/6 |
-2 |
1/5 |
-0,5 |
1 |
0 |
|
Kết quả |
Nhận |
Loại |
Loại |
Nhận |
Loại |
Loại |
Nhận |
nhận |
vậy...........................
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 2:
\(\Leftrightarrow4x^3-4x-3x+3=0\)
=>4x(x-1)(x+1)-3(x-1)=0
=>(x-1)(4x^2+4x-3)=0
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{3}{2}\right\}\)
a)B=3x3 -2y3-6x2y2+xy
B=(3x3-6x2y2)+(xy-2y3)
B=3x2(x-2y2)+y(x-2y2)
B=(x-2y2)(3x2+y)
tại x=\(\frac{2}{3}\)và y=\(\frac{1}{2}\)ta có B=(x-2y2)(3x2+y)=(\(\frac{2}{3}\)-2*\(\frac{1}{2}\)^2 )(3*\(\frac{2}{3}\)^2+\(\frac{1}{2}\))=\(\frac{1}{6}\)*\(\frac{11}{6}\)=\(\frac{11}{36}\)
b)C= 2x+xy2-x2y-2y
C=(2x-2y)+(xy2-x2y)
C=2(x-y)-xy(x-y)
C=(2-xy)(x-y)
tại x=\(-\frac{1}{2}\)và y=\(-\frac{1}{3}\)ta có C=(2-xy)(x-y)=(2-\(-\frac{1}{2}\)*\(-\frac{1}{3}\))(\(-\frac{1}{2}\)+\(\frac{1}{3}\))=\(\frac{-11}{36}\)
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:
\(A=x^2y-y+xy^2-x=\left(x^2y+xy^2\right)-\left(x+y\right)\\ =xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)
Voqis x=-1;y=3 ta có:
\(A=\left(-1+3\right)\left(-1\cdot3-1\right)=2\cdot\left(-4\right)=-8\)
b) \(B=x^2y^2+xy+x^3+y^3=\left(x^2y^2+x^3\right)+\left(xy+y^3\right)\\ =x^2\left(y^2+x\right)+y\left(x+y^2\right)=\left(x+y^2\right)\left(x^2+y\right)\)
Với x=-1;y=3 ta có:
\(B=\left(-1+3^2\right)\left(-1^2+3\right)=8\cdot2=16\)
c) \(C=2x+xy^2-x^2y-2y=\left(2x-2y\right)+\left(xy^2-x^2y\right)\\ =2\left(x-y\right)+xy\left(y-x\right)=\left(x-y\right)\left(2-xy\right)\)
Với x=-1;y=3 ta có:
\(C=\left(-1-3\right)\left(2-\left(-1\right)\cdot3\right)=-4\cdot5=-20\)
d) phân tích tt