mn làm hộ mk vs
a) 27x^3y - a^3b^3y
b) (xy+4)^2 - 4(x+y)^2
c) x^2 - xz - 9y^3 +3yz
d) 36 - 4x^2 - 20xy -25y^2
Những câu hỏi liên quan
Thực hiện phép tính:
a, 7x.(2x-5)+(4x-3).(x+2)-16x2
b, (2x+3y)2+(2x-3y)2-2.(4x2-9y2)
c, (27x3y6+15x2y4-6xy3):3xy
d, (x2-10xy+25y2):(x-5y)
e, (x+2).(x2-2x+4)+(1-x).(1+x+x2)+19
13 tháng 8 2023 lúc 18:18
Phân tích các đa thức sau đây thành nhân tử
a, 36x^2 - ( 3x -2 ) ^2
b, 16(4x+5)^5 - 25 (2x+2)^2
c, ( x - y + 4 )^2
d, (x+1)^4 - (x-1)^4
e, 16x^2 - 24xy + 9y^2
f, -x^4/4 + 2x^2y^3 - 4y^6
g , 64x^3 +1
h, x^3y^6z^9 - 125
k, 27x^6 - 8x^3
I , x^6 - y^6
m, 27x^3 - 54x^2y + 36xy^2 - 8y^3
n, y^9 - 9x^2y^6 + 27x^4y^3 - 27x^6
làm ơn giải chi tiết giúp mik vs ạ , cảm ơn
a: =(6x)^2-(3x-2)^2
=(6x-3x+2)(6x+3x-2)
=(9x-2)(3x+2)
d: \(=\left[\left(x+1\right)^2-\left(x-1\right)^2\right]\left[\left(x+1\right)^2+\left(x-1\right)^2\right]\)
\(=4x\cdot\left[x^2+2x+1+x^2-2x+1\right]\)
=8x(x^2+1)
e: =(4x)^2-2*4x*3y+(3y)^2
=(4x-3y)^2
f: \(=-\left(\dfrac{1}{4}x^4-2\cdot\dfrac{1}{2}x^2\cdot2y^3+4y^6\right)\)
\(=-\left(\dfrac{1}{2}x^2-2y^3\right)^2\)
g: =(4x)^3+1^3
=(4x+1)(16x^2-4x+1)
k: =x^3(27x^3-8)
=x^3(3x-2)(9x^2+6x+4)
l: =(x^3-y^3)(x^3+y^3)
=(x-y)(x+y)(x^2-xy+y^2)(x^2+xy+y^2)
Đúng 0
Bình luận (0)
Thực hiện phép tính
a) 6 xy^2 : 3y
b) 62x^4y^3 :2x^3y^2
c) 18x^4y^3 : (-6x^2y)
d) 27x^5y^6 : 9x^3y^3
e) 18x^3y^4 : 12xy^3
a: \(=\dfrac{6}{3}\cdot x\cdot\dfrac{y^2}{y}=2xy\)
b: \(=\dfrac{62}{2}\cdot\dfrac{x^4}{x^3}\cdot\dfrac{y^3}{y^2}=31xy\)
c: \(=\dfrac{-18}{6}\cdot\dfrac{x^4}{x^2}\cdot\dfrac{y^3}{y}=-3x^2y^2\)
d: \(=\dfrac{27}{9}\cdot\dfrac{x^5}{x^3}\cdot\dfrac{y^6}{y^3}=3x^2y^3\)
e: \(=\dfrac{18}{12}\cdot\dfrac{x^3}{x}\cdot\dfrac{y^4}{y^3}=\dfrac{3}{2}x^2y\)
Đúng 1
Bình luận (0)
A.5x^2y^3-25x^3y^4+10x^3y^3
B.12x^2y-18xy^2-30y^2
C.5(x-y)-y(x-y)
D.y(x-z)+7(z-x)
E.27x^2(y-1)-9x^3(1-y)
F.36-12x+x^2
G.x^2+2xy+y^2-xz-yz
H.x^4+64
I.27x^2(y-1)-9x^3(1-y)
K.36-12x+x^2
M.-4x^2+4x-1
N.x^2+5x+6
P.x^2-x-6
Q.x^4-5x^2+4
Rút gọn biểu thức :
a, A = ( x - y ) ( x^2 + xy + y^2 ) - ( x + y ) ( x^2 - xy + y^2 )
b, B = ( a^2b^2 - 5a ) ( a^4 + b^4 + 5a^3b^2 + 25d^2 )
c, C = ( 2x + 3y ) ( 4x^2 - 6xy + 9y^2 )
d, D = ( y + 2 ) ( y^2 - 2y + 4 )
1) x^2-4xy-x+3y^2+3y
2) 6x^2+xy -7x-2y^2+7y-5
3) 2a^2+5ab-3b^2-7b-2
4) 6x^2-xy-2y^2+3x-2y
5) 2x^2 - 3xy-4x-9y^2-6y
Giúp mk với mk đang cần gấp
1: \(=\left(x-3y\right)\left(x-y\right)-\left(x-3y\right)=\left(x-3y\right)\left(x-y-1\right)\)
4: \(=6x^2-4xy+3xy-2y^2+3x-2y\)
\(=\left(3x-2y\right)\left(2x+y\right)+3x-2y=\left(3x-2y\right)\left(2x+y+1\right)\)
Đúng 0
Bình luận (0)
22 tháng 8 2023 lúc 10:04
Bài 8: Phân tích đa thức thành nhân tử.
a, x^4 - y^4
b, x^2 - 3y^2
c, (3x - 2y)^2 - (2x - 3y)^2
d, 9(x -y)^2 - 4(x + y)^2
e, (4x^2 - 4x + 1) - (x+1)^2
f, x^3 + 27
g, 27x^3 - 0,001
h, 125x^3 - 1
a) \(x^4-y^4\)
\(=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)
b) \(x^2-3y^2\)
\(=x^2-\left(y\sqrt{3}\right)^2\)
\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)
c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)
\(=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)
\(=\left(5x-5y\right)\left(x+y\right)\)
\(=5\left(x-y\right)\left(x+y\right)\)
d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)
\(=\left[3\left(x-y\right)+2\left(x+y\right)\right]\left[3\left(x-y\right)-2\left(x+y\right)\right]\)
\(=\left(3x-3y+2x+2y\right)\left(3x-3y-2x-2y\right)\)
\(=\left(5x-y\right)\left(x-5y\right)\)
e) \(\left(4x^2-4x+1\right)-\left(x+1\right)^2\)
\(=\left(2x-1\right)^2-\left(x+1\right)\)
\(=\left(2x-1+x+1\right)\left(2x-1-x-1\right)\)
\(=3x\left(x-2\right)\)
f) \(x^3+27\)
\(=x^3+3^3\)
\(=\left(x+3\right)\left(x^2-3x+9\right)\)
g) \(27x^3-0,001\)
\(=\left(3x\right)^3-\left(0,1\right)^3\)
\(=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)
h) \(125x^3-1\)
\(=\left(5x\right)^3-1^3\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)
Đúng 2
Bình luận (0)
Giải hệ phương trình :
a) \(\hept{\begin{cases}x^2+y^5+x-9y=2\\x^4+4=-4x-25y^2\end{cases}}\)
b) \(\hept{\begin{cases}x^2-4x+3=0\\x^2+xy+y^2=3\end{cases}}\)
b) \(\hept{\begin{cases}x^2-4x+3=0\left(1\right)\\x^2+xy+y^2=3\left(2\right)\end{cases}}\)
Từ (1) <=> (x - 1)(x - 3) = 0 \(\Leftrightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
Với x = 3 => (2) <=> 32 + 3y + y2 = 3
<=> y2 + 3y + 6 = 0
<=> \(\left(2y+3\right)^2=-15\)<=> PT vô nghiệm
Với x = 3 => (1) <=> 12 + y + y2 = 3
<=> (y - 1)(y + 2) = 0
<=> \(\orbr{\begin{cases}y=1\\y=-2\end{cases}}\)
=> Hệ có 2 nghiệm (x ; y) = (1;1) ; (1 ; - 2)
27x^3 - 27 x^2 +3x - 1
1/27 + x^3
x^3- 3x^2+3x-1
0,001-1000x^3
12/5 x^2y^2-9x^4 - 4/25y^4
a^2y^2+b^2x^2-2axby
100-(3x-y)^2
64x^2-(8a+b)^2
27x^3-a^3b^3
b: \(x^3+\dfrac{1}{27}=\left(x+\dfrac{1}{3}\right)\left(x^2-\dfrac{1}{3}x+\dfrac{1}{9}\right)\)
c: \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
e: \(a^2y^2-2axby+b^2x^2\)
\(=\left(ay\right)^2-2\cdot ay\cdot bx+\left(bx\right)^2\)
\(=\left(ay-bx\right)^2\)
f: \(100-\left(3x-y\right)^2\)
\(=\left(10-3x+y\right)\left(10+3x-y\right)\)
g: \(64x^2-\left(8a+b\right)^2\)
\(=\left(8x\right)^2-\left(8a+b\right)^2\)
\(=\left(8x-8a-b\right)\left(8x+8a+b\right)\)
Đúng 0
Bình luận (0)