a) 25x2 - 20xy + 4y
b) 1/36a2 - 1/4b2
c) 0.125 (a+2)3 - 1
d) x6 - 1
a)a2 – 4b2 b) x2 – y2 + 6y - 9
c) (2a + b)2 – a2 d) 16(x – 1)2 – 25(x + y)2
e)x2 + 10x + 25 f) 25x2 – 20xy + 4y2
g)9x4 + 24x2 + 16 h) x3 – 125
i)x6 – 1 k) x3 + 15x2 + 75x + 125
a) (a - 2b)x(a + 2b)
b) x2-(y-3)2
=> (x-y+3)(x+y-3)
c) (2a + b - a)(2a + b + a)
=> (a+b)(3a+b)
d) (4(x - 1))2 - (5(x + y))2
⇔ (4x - 4 - 5x - 5y)(4x - 4 + 5x + 5y)
⇔ -(x + 5y + 4)(9x + 5y + -4)
e) (x + 5)2
f) (5x - 2y)2
h) (x - 5)(x2 + 5x + 25)
k) (x + 5)3
Phân tích đa thức thành nhân tử:
a) 50x5-8x3
b) x4-5x2-4y2+10y
c) 36a2-b2+12a+1
d) x3+y3-xy2-x2y
e) 4x2+4x-3
f) 9x4+16x2-4
g) -6x2+5xy+4y2
h)(x2+4x)2+8(x2+4x)+15
i) 9x4+5x2+1
a: \(50x^5-8x^3\)
\(=2x^3\left(25x^2-4\right)\)
\(=2x^3\left(5x-2\right)\left(5x+2\right)\)
b: \(x^4-5x^2-4y^2+10y\)
\(=\left(x^2-2y\right)\left(x^2+2y\right)-5\left(x^2-2y\right)\)
\(=\left(x^2-2y\right)\left(x^2+2y-5\right)\)
c: \(36a^2+12a+1-b^2\)
\(=\left(6a+1\right)^2-b^2\)
\(=\left(6a+1-b\right)\left(6a+1+b\right)\)
d: \(x^3+y^3-xy^2-x^2y\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x+y\right)\cdot\left(x-y\right)^2\)
e: Ta có: \(4x^2+4x-3\)
\(=4x^2+6x-2x-3\)
\(=2x\left(2x+3\right)-\left(2x+3\right)\)
\(=\left(2x+3\right)\left(2x-1\right)\)
f: Ta có: \(9x^4+16x^2-4\)
\(=9x^4+18x^2-2x^2-4\)
\(=9x^2\left(x^2+2\right)-2\left(x^2+2\right)\)
\(=\left(x^2+2\right)\left(9x^2-2\right)\)
g: Ta có: \(-6x^2+5xy+4y^2\)
\(=-6x^2+8xy-3xy+4y^2\)
\(=-2x\left(3x-4y\right)-y\left(3x-4y\right)\)
\(=\left(3x-4y\right)\left(-2x-y\right)\)
h: Ta có: \(\left(x^2+4x\right)^2+8\left(x^2+4x\right)+15\)
\(=\left(x^2+4x\right)^2+3\left(x^2+4x\right)+5\left(x^2+4x\right)+15\)
\(=\left(x^2+4x+3\right)\cdot\left(x^2+4x+5\right)\)
\(=\left(x+1\right)\left(x+3\right)\left(x^2+4x+5\right)\)
Phân tích ra nhân tử(Phương pháp dùng hằng đẳng thức)
1) 9(a+b)2-4(a-2b)26
2) 9x2+12x+4
3) 4x4+20x2+25
4) 25x2-20xy+4y2
5) 9x4-12x2y+4y2
6) 4x4-16x2y3+16y6
7) 9x4-12x5+4x6
Làm giúp mình với, nhanh nha mình cần gấp
2) 9x2+ 12x+ 4
<=>(3x)2+ 2.3x.2+ 22 <=>(3x+ 2)2
3) 4x4+ 20x2+ 25
<=>(2x2)2+ 2.2x2.5+ 52 <=>(2x2+5)2
4) 25x2- 20xy+ 4y2
<=> (5x)2- 2.5x.2y+ (2y)2<=> (5x-2y)2
5) 9x4- 12x2y+ 4y2
<=> (3x2)2- 2.3x2.2.y+ (2y)2<=> (3x2- 2y)2
6) 4x4- 16x2y3+ 16y6
<=> (2x2)2- 2.2x2.4y3+ (4y3)2<=> (2x2- 4y3)2
7) 9x4- 12x5+ 4x6
<=> (3x2)2- 2.3x2.2x3+ (2x3)2<=> (3x2- 2x3)2
phân tích các đa thức sau thành nhân tử
a,36-4x2+20xy -25y2
b, x4+25x2+20x-4
helpppp.chiều đi học rùi
\(a,36-4x^2+20xy-25y^2\\ =36-\left(4x^2-20xy+25y^2\right)\\ =6^2-\left[\left(2x\right)^2-2.2x.5y+\left(5y\right)^2\right]\\ =6^2-\left(2x-5y\right)^2\\ =\left[6-\left(2x-5y\right)\right]\left[6+\left(2x-5y\right)\right]\\ =\left(6-2x+5y\right).\left(6+2x-5y\right)\)
a/
\(=6^2-\left[\left(2x\right)^2-2.2x.5y+\left(5y\right)^2\right]=\)
\(6^2-\left(2x-5y\right)^2=\left[6-\left(2x-5y\right)\right].\left[6+\left(2x-5y\right)\right]\)
Viết biểu thức 25 x 2 – 20 x y + 4 y 2 dưới dạng bình phương của một hiệu
A. 5 x – 2 y 2
B. 2 x – 5 y 2
C. 25 x – 4 y 2
D. 5 x + 2 y 2
Ta có 25 x 2 – 20 x y + 4 y 2 = ( 5 x ) 2 – 2 . 5 x . 2 y + ( 2 y ) 2 = ( 5 x – 2 y ) 2
Đáp án cần chọn là: A
Phan tích da thuc thanh nhan tu
0.125(a+2)^3-1
x^6-1
\(125\left(a+2\right)^3-1\)
\(=\left[5\left(a+2\right)\right]^3-1\)
\(=\left(5a+10\right)^3-1\)
\(=\left(5a+10-1\right)\left[\left(5a+10\right)^2-\left(5a+10\right)+1\right]\)
\(=\left(5a+10-1\right)\left[25a^2+100a+100-5a-10+1\right]\)
\(=\left(5a+9\right)\left[25a^2+95a+91\right]\)
b) \(x^6-1=\left(x^3\right)^2-1=\left(x^3-1\right)\left(x^3+1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
0,125(a+2)3-1
= (0,5)3(a+2)3-1
= (0,5)3(a+2)3-13
k bt làm nx ==
x6-1 = (x3)2-12=(x3-1)(x3+1)
\(0,125.\left(a+2\right)^3-1\)
\(=0,125.\left(a^3+3.a^2.2+3.a.2^2+2^3\right)-1\)
\(=0,125.\left(a^3+6a^2+12a+8\right)-1\)
\(=0,125a^3+0,75a^2+1,5a+1-1\)
\(=0,125a^3+0,75a^2+1,5a\)
\(x^6-1=\left(x^2\right)^3-1=\left(x^2\right)^3-1^3=\left(x^2-1\right)\left(x^4+x^2+1\right)\)
Tìm x:
a) (x+2)(x2-2x+4)=35
b) (25x2+5x+1)(5x-1)=-9
a) Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)=35\)
\(\Leftrightarrow x^3+8=35\)
\(\Leftrightarrow x^3=27\)
hay x=3
b) Ta có: \(\left(25x^2+5x+1\right)\left(5x-1\right)=-9\)
\(\Leftrightarrow125x^3-1=-9\)
\(\Leftrightarrow125x^3=-8\)
\(\Leftrightarrow x=-\dfrac{2}{5}\)
Phân tích thành tích:
a)25x2-9
b)64-(x-7)2
c)-100a2b2+1
d)x2-6x-8
a) \(=\left(5x\right)^2-3^2\) \(=\left(5x-3\right)\left(5x+3\right)\)
b) \(=8^2-\left(x-7\right)^2\) \(=\left(8-x+7\right)\left(8+x-7\right)\) \(=\left(15-x\right)\left(x+1\right)\)
c) \(=1-100a^2b^2\) \(=1-\left(10ab\right)^2\) \(=\left(1-10ab\right)\left(1+10ab\right)\)
d) Mình sửa đề chút nhé! Đề như trên thì không phân tích thành nhân tử được :)
\(x^2-6x+8\)
\(=x^2-2x-4x+8\) \(=x\left(x-2\right)-4\left(x-2\right)\) \(=\left(x-2\right)\left(x-4\right)\)
A. 1 :1:1:1
B. 3:1:2:3
C. 2:3:1:1
D. 6:5:2:9