Bài 1: Thu gọn biểu thức
c) (x2-y)(3x+y2)-(6x4y-2xy4):2xy
Bài 2: phân tích thành nhân tử
a)10x2(2x-y)+6xy(y-2x) b) x2-2x+1-y2 c) x2-8x+12
PHÂN TÍCH ĐA THỨC THÀNH NHÂN TỬ
a, 3x(2x - y) + 5y(y - 2x)
b, (x - 5)2 - 9(x + y)2
c, y2 + 2yz + z2 - xy - xz
d, x2 - 9x2y2 + y2 + 2xy
e, x2 - 10x + 24
g, 6x2 + 7x - 5
h, x2 + 4xy - 12y2
k, a4 + 3a2 + 4
a) \(3x\left(2x-y\right)+5y\left(y-2x\right)\)
\(=3x\left(2x-y\right)-5y\left(2x-y\right)\)
\(=\left(3x-5y\right)\left(2x-y\right)\)
b) \(\left(x-5\right)^2-9\left(x+y\right)^2\)
\(=\left(x-5\right)^2-3^2\left(x+y\right)^2\)
\(=\left(x-5\right)^2-\left(3x+3y\right)^2\)
\(=\left(x-5+3x+3y\right)\left(x-5-3x-3y\right)\)
\(=\left(4x+3y-5\right)\left(-2x-3y-5\right)\)
a: \(3x\left(2x-y\right)+5y\left(y-2x\right)=\left(2x-y\right)\left(3x-5y\right)\)
e: \(x^2-10x+24=\left(x-4\right)\left(x-6\right)\)
g) \(6x^2+7x-5\)
=\(6x^2+10x-3x-5\)
=\(\left(6x^2+10x\right)-\left(3x+5\right)\)
=\(2x\left(3x+5\right)-\left(3x+5\right)\)
=\(\left(2x-1\right)\left(3x+5\right)\)
Bài 3:
b) Rút gọn biểu thức B=(x+y)(2x-y)+(xy4-x2y2):xy2
Bài 4: Phân tích thành nhân tử
a) 25x3-10x2+xx
b) x2-9x+9y-y2
c) 16-x2-4y2-4xy
Bài 5: Tìm x biết
a) 36-x3=00
b) (x+2)(x-2)-(x+1)2=7
Bài 3:
b. $B=(x+y)(2x-y)+(xy^4-x^2y^2):(xy^2)$
$=(2x^2-xy+2xy-y^2)+(y^2-x)$
$=2x^2+xy-y^2+y^2-x=2x^2+xy-x$
Bài 4:
a. $25x^3-10x^2+x=x(25x^2-10x+1)=x(5x-1)^2$
b. $x^2-9x+9y-y^2=(x^2-y^2)-(9x-9y)=(x-y)(x+y)-9(x-y)=(x-y)(x+y-9)$
c. $16-x^2-4y^2-4xy=16-(x^2+4y^2+4xy)$
$=4^2-(x+2y)^2=(4-x-2y)(4+x+2y)$
Bài 5:
a. $36-x^3=100$
$x^3=36-100=-64=(-4)^3$
$\Rightarrow x=-4$
b.
$(x+2)(x-2)-(x+1)^2=7$
$\Leftrightarrow (x^2-4)-(x^2+2x+1)=7$
$\Leftrightarrow -2x-5=7$
$\Leftrightarrow -2x=12$
$\Leftrightarrow x=-6$
Bài 1: Rút gọn các biểu thức:
a. (2x - 1)2 - 2(2x - 3)2 + 4
b. (3x + 2)2 + 2(2 + 3x)(1 - 2y) + (2y - 1)2
c. (x2 + 2xy)2 + 2(x2 + 2xy)y2 + y4
d. (x - 1)3 + 3x(x - 1)2 + 3x2(x -1) + x3
e. (2x + 3y)(4x2 - 6xy + 9y2)
f. (x - y)(x2 + xy + y2) - (x + y)(x2 - xy + y2)
g. (x2 - 2y)(x4 + 2x2y + 4y2) - x3(x – y)(x2 + xy + y2) + 8y3
a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)
Bài 2: Phân tích đa thức thành nhân tử
a) x2−xy+5y−25
b) xy−y2−3x+3y
c) x2(x−3)−4x+12
d) 2a(x+y)−x−y
e) 2x−4+5x2−10x
g) 10ax−5ay−2x+y
h) a2−2a+1−b2
a) x2-xy+5y-25
= x(2-y)+ 5(y-2)
= x(2-y)-5(2-y)
= (x-5)(2-y)
h: \(=\left(a-1-b\right)\left(a-1+b\right)\)
1) Tìm x, y, z
a) 9x2 +y2 + 2z2 – 18x +4z – 6y +20 = 0
b) 5x2 +5y2 +8xy+2y – 2x+2 = 0
c) 5x2 +2y2 + 4xy – 2x + 4y +5 = 0
d) x2 + 4y2 + z2 =2x + 12y – 4z – 14
e) x2 +y2 – 6x + 4y +2= 0
2) Phân tích đa thức thành nhân tử
a) 3xy2 – 3x3 – 6xy +3x
b) 3x2 + 11x + 6
c) –x3 – 4xy2 + 4x2y +16x
d) xz – x2 – yz +2xy – y2
e) 4x2 – y2 – 6x + 3y
f) X4 – x3 – 10x2 + 2x +4
g) (x3 – x2 + x)(121 – 25y2 – 10y) – (x3 – x2 + x) – (121 – 25y2 – 10y) +1
h) X4 – 14x3 + 71x2 – 154x + 120
Giúp mik vs cần gấp!!!
\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11
e: Ta có: \(x^2-6x+y^2+4y+2=0\)
\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Dấu '=' xảy ra khi x=3 và y=-2
1) Tìm x, y, z
a) 9x2 +y2 + 2z2 – 18x +4z – 6y +20 = 0
b) 5x2 +5y2 +8xy+2y – 2x+2 = 0
c) 5x2 +2y2 + 4xy – 2x + 4y +5 = 0
d) x2 + 4y2 + z2 =2x + 12y – 4z – 14
e) x2 +y2 – 6x + 4y +2= 0
2) Phân tích đa thức thành nhân tử
a) 3xy2 – 3x3 – 6xy +3x
b) 3x2 + 11x + 6
c) –x3 – 4xy2 + 4x2y +16x
d) xz – x2 – yz +2xy – y2
e) 4x2 – y2 – 6x + 3y
f) X4 – x3 – 10x2 + 2x +4
g) (x3 – x2 + x)(121 – 25y2 – 10y) – (x3 – x2 + x) – (121 – 25y2 – 10y) +1
h) X4 – 14x3 + 71x2 – 154x + 120
Giúp mik với mik đang cần rất gấp ạ!!!
Bài 1: thu gọn biểu thức
b) (x-y)(x2-2x+y)
c) (x2-y)(x+y2)-(x-y)(x2+xy+y2)
d) 3x(2xy-z)-5y(x2-2)+3xz
b: (x-y)(x^2-2x+y)
\(=x^3-2x^2+xy-x^2y+2xy-y^2\)
\(=x^3-2x^2-x^2y+3xy-y^2\)
c: \(\left(x^2-y\right)\left(x+y^2\right)-\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3+x^2y^2-xy-y^3-\left(x^3-y^3\right)\)
\(=x^2y^2-xy\)
d: \(3x\left(2xy-z\right)-5y\left(x^2-2\right)+3xz\)
\(=6x^2y-3xz-5x^2y+10y+3xz\)
\(=x^2y+10y\)
Bài 1: Rút gọn các biểu thức:
a. (2x - 1)2 - 2 (2x - 3)2 + 4
b. (3x + 2)2 + 2 (2 + 3x) (1 - 2y) + (2y - 1)2
c. (x2 + 2xy)2 + 2 (x2 + 2xy) y2 + y4
d. (x - 1)3 + 3x (x - 1)2 + 3x2 (x -1) + x3
e. (2x + 3y) (4x2 - 6xy + 9y2)
f. (x - y) (x2 + xy + y2) - (x + y) (x2 - xy + y2)
g. (x2 - 2y) (x4 + 2x2y + 4y2) - x3 (x – y) (x2 + xy + y2) + 8y3
a: Ta có: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
b: \(\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(1-2y\right)^2\)
\(=\left(3x+2+1-2y\right)^2\)
\(=\left(3x-2y+3\right)^2\)
1. Rút gọn biểu thức:
a. (2x-3)(4x2+6x+9)-2x(4x2-1)
b.(x+y)2+2(x+y)(x-y)+(x-y)2
2.Phân tích đa thức sau thành nhân tử:
a. 2x2y+4xy+2y c. x2-8x+7
b.9x2+6xy-4z2+y2 d. x3+4x2+x-6
1b.=2((x+y)+(x+y)(x-y)+(x-y))=2(x2-y2+x+y+x-y)=2(x2-y2+2x)=2x2-2y2+4x
2a.=4xy+4xy+2y=8xy+2y=2y(4x+1)
b.=(3x)2+2.3x.y+y2-(2z)2=(3x+y)2-(2z)2=(3x+y-2z)(3x+y+2z)
c.=x2-x-7x+7=x(x-1)-7(x-1)=(x-1)(x-7)
\(\left(x+y\right)^2+2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)^2\)
\(=\left(2x\right)^2\)
\(=4x^2\)
hk tốt
^^