1/2x2 + 4x
2/15x3 + 5x2 - 10x
3) 5x2(x-2y) + 15x (x -2y)
4) 3(x - y) + 5x(y - x)
5/ 5x2 - 10x
6) 3x-6y
7)25x2+5x3+x2y
8)14x2y-21xy2+28x2y2
9)x(y-1)- y(y-1)
10) 10x(x-y)-8y(y-x)
giúp mình vs mn cảm ơn mn rất nhiều ạ
RÚT GỌN BIỂU THỨC
A) (x+y)-(x-y)
B) 3x(5x2-2x-1)-15x3
C) ( 5x-2y) (x2-xy+1)+7x2y
Lời giải:
a. $(x+y)-(x-y)=x+y-x+y=(x-x)+y+y=0+2y=2y$
b. $3x(5x^2-2x-1)-15x^3=15x^3-6x^2-3x-15x^3=-6x^2-3x$
c. $(5x-2y)(x^2-xy+1)+7x^2y=5x^3-5x^2y+5x-2x^2y+2xy^2-2y+7x^2y$
$=5x^3+(-5x^3y-2x^2y+7x^2y)+5x+2xy^2-2y$
$=5x^3+5x+2xy^2-2y$
a) x2 – x;
b) 5x2(x – 2y) – 15x(x – 2y);
c) 3(x – y) – 5x(y – x).
\(a,x^2-x=x\left(x-1\right)\\ b,5x^2\left(x-2y\right)-15x\left(x-2y\right)=\left(5x^2-15x\right)\left(x-2y\right)=5x\left(x-3\right)\left(x-2y\right)\\ c,3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)=\left(5x+3\right)\left(x-y\right)\)
Bài 1: Pân tích đa thức thành nhân tử
a, 3x-6y
b, 14x2y-21xy2+28x2y2
c, 10x (x-y)-8y(y-x)
Bài 2: Tìm x
a,5x ( x-2000 )-x+2000
b, x3-13x=0
MONG MỌI NGƯỜI GIÚP MIK :3
Bài 1:
a: \(3x-6y=3\cdot x-3\cdot2y=3\left(x-2y\right)\)
b: \(14x^2y-21xy^2+28x^2y^2\)
\(=7xy\cdot2x-7xy\cdot3y+7xy\cdot4xy\)
\(=7xy\left(2x-3y+4xy\right)\)
c: \(10x\left(x-y\right)-8y\cdot\left(y-x\right)\)
\(=10x\left(x-y\right)+8y\left(x-y\right)\)
\(=\left(x-y\right)\left(10x+8y\right)\)
\(=\left(2\cdot5x+2\cdot4y\right)\left(x-y\right)\)
\(=2\left(5x+4y\right)\left(x-y\right)\)
bài 2:
a: Đề thiếu vế phải rồi bạn
b: \(x^3-13x=0\)
=>\(x\left(x^2-13\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=13\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x=\pm\sqrt{13}\end{matrix}\right.\)
Bài 1:
a, $3x-6y$
$=3(x-2y)$
b, $14x^2y-21xy^2+28x^2y^2$
$=7xy(2x-3y+4xy)$
c, $10x(x-y)-8y(y-x)$
$=10x(x-y)-8y[-(x-y)]$
$=10x(x-y)+8y(x-y)$
$=(x-y)(10x+8y)$
$=2(x-y)(5x+4y)$
Bài 2:
a, Đề thiếu rồi bạn nhé.
b, \(x^3-13x=0\)
\(\Rightarrow x\left(x^2-13\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=13\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{matrix}\right.\)
1. (x3 – 3x2 + x – 3) : (x – 3) 2. (2x4 – 5x2 + x3 – 3 – 3x) : (x2 – 3) 3. (x – y – z)5 : (x – y – z)3 4. (x2 + 2x + x2 – 4) : (x + 2) 5. (2x3 + 5x2 – 2x + 3) : (2x2 – x + 1) 6. (2x3 – 5x2 + 6x – 15) : (2x – 5)
1: \(=x^2+1\)
3: \(=\left(x-y-z\right)^2\)
Phân tích đa thức sau thành phân tử :
h. 3x^3(2y - 3z) - 15x(2y - 3z)^2
k. 3x(x + 2) + 5(-x - 2)
l. 18^2(3 + x) + 3(x + 3)
m. 14x^2y - 21xy^2 + 28x^2y^2
n. 10x(x - y) - 8y(y - x).
mn ơi,xin mn hãy giúp mik vs ạ. e đg cần gấp ah/cj ạ:<
h) \(=3x\left(2y-3z\right)\left[x^2-5\left(2y-3z\right)\right]=3x\left(2y-3z\right)\left(x^2-10y+15z\right)\)
k) \(=\left(x+2\right)\left(3x-5\right)\)
l) \(=\left(18^2+3\right)\left(x+3\right)=327\left(x+3\right)\)
m) \(=7xy\left(2x-3y+4xy\right)\)
n) \(=2\left(x-y\right)\left(5x-4y\right)\)
Cứu với ạ
Làm tính chia
1) (x3 – 3x2 + x – 3) : (x – 3) 2) (2x4 – 5x2 + x3 – 3 – 3x) : (x2 – 3)
3) (x – y – z)5 : (x – y – z)3 4) (x2 + 2x + x2 – 4) : (x + 2)
5) (2x3 + 5x2 – 2x + 3) : (2x2 – x + 1) | 6) (2x3 – 5x2 + 6x – 15):(2x – 5) |
a)5X2 - 2y = 8
2X + y = 5
b) 2X + y = 3
X – 2y = 4
c) 3X + 3y = 1
2X – y = -8
d) 4X + 5y = 3
X – 5y = 5
a) ( 10x3y - 5x2y2 - 25 x4y3) : ( -5xy)
Ta có : -5xy( -2x2 + xy + 5x3y2) : ( - 5xy)
Vậy , ta được thương là : -2x2 + xy + 5x3y2
b) ( 27x3 - y3) : ( 3x - y)
Ta có : ( 3x - y)( 9x2 + 3xy + y2) : ( 3x - y)
Vậy , ta được thương là : 9x2 + 3xy + y2
C,D chịu
Phân tích các đa thức sau thành nhân tử :
a/ 10x(x−y)−6y(y−x)10x(x−y)−6y(y−x)
b/ 14x2y−21xy2+28x3y214x2y−21xy2+28x2y2
c/ x2−4+(x−2)2x2−4+(x−2)2
d/ (x+1)2−25(x+1)2−25
d: \(=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)