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Bài 1 : \(\left(y+a\right)^3=y^3+3y^2a+3ya^2+a^3\)
Bài 2:
1. \(x^2-2x+1=\left(x-1\right)^2\)
2. \(x^2+2x+1=\left(x+1\right)^2\)
3. \(x^2-6x+9=\left(x-3\right)^2\)
4. \(x^2-10x+25=\left(x-5\right)^2\)
5. \(x^2+14x+49=\left(x+7\right)^2\)
6. \(x^2-22x+121=\left(x-11\right)^2\)
7. \(4x^2-4x+1=\left(2x-1\right)^2\)
8. \(x^2-4x+4=\left(x-2\right)^2\)
9. \(x^2-2xy+y^2=\left(x-y\right)^2\)
10. \(4x^2-4xy+y^2=\left(2x-y\right)^2\)
Bài 1 :
\(\left(y+a\right)^3=y^3+3y^2a+3ya^2+a^3\)
Bài 2 : mk lm tiếp phần còn lại thôi, mấy câu mk ko lm có ở bài trc rồi
\(x^2+14x+49=\left(x+7\right)^2\)
\(x^2-22x+121=\left(x-11\right)^2\)
\(4x^2-4x+1=\left(2x-1\right)^2\)
\(x^2-4x+4=\left(x-2\right)^2\)
\(x^2-2xy+y^2=\left(x-y\right)^2\)
\(4x^2-4xy+y^2=\left(2x-y\right)^2\)
1.
\(x^2-22x+12\) : biểu thức không phân tích được thành nhân tử nữa.
2.
\(9x^2+6x+1=(3x)^2+2.3x.1+1^2=(3x+1)^2\)
3.
\(x^2-10x+2\): không p. tích được thành nhân tử.
4.
\(x^3+1=x^3+1^3=(x+1)(x^2-x+1)\)
5.
\(8x^3-27y^3=(2x)^3-(3y)^3=(2x-3y)[(2x)^2+(2x)(3y)+(3y)^2]\)
\(=(2x-3y)(4x^2+6xy+9y^2)\)
6.
\((x+3y)^2-(3y+1)^2=[(x+3y)-(3y+1)][(x+3y)+(3y+1)]\)
\(=(x-1)(x+6y+1)\)
7.
\(4y^2-36x^2=(2y)^2-(6x)^2=(2y-6x)(2y+6x)=4(y-3x)(y+3x)\)
8.
\(27-(x+4)^3=3^3-(x+4)^3=[3-(x+4)][3^2+3(x+4)+(x+4)^2]\)
\(=-(x+1)(37+x^2+11x)\)
9.
\(25x^2-10xy+y^2=(5x)^2-2.5x.y+y^2=(5x-y)^2\)
10.
\(9x^6-12x^7+4x^8=x^6(9-12x+4x^2)=x^6[3^2-2.3.2x+(2x)^2]\)
\(=x^6(3-2x)^2\)
a) x2 - 2x - 4y2 - 4y
= (x2 - 4y2) - (2x + 4y)
= (x + 2y)(x - 2y) - 2(x + 2y)
= (x + 2y)(x - 2y - 2)
= (x + 2y)[x - 2(y + 1)]
b) x4 + 2x3 - 4x - 4
= (x4 - 4) + ( 2x3 - 4x)
= (x2 - 2)(x2 + 2) + 2x(x2 - 2)
= (x2 - 2)(x2 + 2 + 2x)
c) x3 + 2x2y - x -2y
= (x3 - x) + (2x2y - 2y)
= x(x2 - 1) + 2y(x2 - 1)
= (x + 2y)(x2 - 1)
1 ) x3 - 2x2 + x
= x( x2 - 2x + 1 )
= x ( x-1)2
2) 4x3 - 25x
= x ( 4x2 - 25)
= x( 2x-5) ( 2x +5)
11) \(x^2-y^2-4x+4\)
\(=\left(x^2-4x+4\right)-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-y-2\right)\left(x+y-2\right)\)
13) \(x^4+4=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a) \(4x^2-12x+9=\left(2x\right)^2-2.2x.3+3^2=\left(2x-3\right)^2\)
b) \(4x^2+4x+1=\left(2x\right)^2+2.2x.1+1^2=\left(2x+1\right)^2\)
c) \(1+12x+36x^2=1^2+2.6x.1+\left(6x\right)^2=\left(1+6x\right)^2\)
d) \(9x^2-24xy+16y^2=\left(3x\right)^2-2.3x.4y+\left(4y\right)^2=\left(3x-4y\right)^2\)
f) \(-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
g) \(-16a^4b^6-24a^5b^5-9a^6b^4=-\left(16a^4b^6+24a^5b^5+9a^6b^4\right)\)
\(=-\left[\left(4a^2b^3\right)^2+2.4a^2b^3.3a^3b^2+\left(3a^3b^2\right)^2\right]\)
\(=-\left(4a^2b^3+3a^3b^2\right)^2\)
h) \(25x^2-20xy+4y^2=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\) \(=\left(5x-2y\right)^2\)
i) \(25x^4-10x^2y+y^2=\left(5x^2\right)^2-2.5x^2.y+y^2=\left(5x^2-y\right)^2\)
\(a,3\left(x+4\right)-x^2-4x\)
\(=3\left(x+4\right)-\left(x^2+4x\right)\)
\(=3\left(x+4\right)-x\left(x+4\right)\)
\(=\left(3-x\right)\left(x+4\right)\)
\(a,3\left(x+4\right)-x^2-4x\)
\(=3\left(x+4\right)-\left(x^2+4x\right)\)
\(=3\left(x+4\right)-x\left(x+4\right)\)
\(=\left(3-x\right),\left(x+4\right)\)
1) x2 - 6x + 9 = (x2 - 3x) -(3x - 9) = x(x - 3) - 3(x - 3) = (x - 3)(x - 3)
2) 25 + 10x + x2 = x2 + 5x + 5x + 25 = x(x + 5) + 5(x + 5) = (x + 5)(x + 5)
5) x2 - 10x + 25 = x2 - 5x - 5x + 25 = x(x - 5) - 5(x - 5) = (x - 5)(x - 5)
6) x2 + 4xy + 4y2 = x2 + 2xy + 2xy + 4y2 = x(x + 2y) + 2y(x + 2y) = (x + 2y)(x + 2y)
7) (3x + 2)2 - 4 = (3x + 2)2 - 22 = (3x + 2 - 2)(3x + 2 + 2) = 3x(3x + 4)
8) 4x2 - 49 = (2x)2 - 72 = (2x - 7)(2x + 7)
9) \(\frac{9}{25}x^4-\frac{1}{4}=\left(\frac{2}{3}\right)^2.\left(x^2\right)^2-\left(\frac{1}{2}\right)^2=\left(\frac{2}{3}x^2\right)^2-\left(\frac{1}{2}\right)^2=\left(\frac{2}{3}x^2-\frac{1}{2}\right)\left(\frac{2}{3}x^2+\frac{1}{2}\right)\)
10) x32 - 1 = (x16)2 - 12 = (x16 - 1)(x16 + 1)
11 4x2 + 4x + 1 = 4x2 + 2x + 2x + 1 = 2x(2x + 1) + (2x + 1) = (2x + 1)(2x + 1)
12 x2 - 20x - 100 = x2 - 10x - 10x + 100 = x(x - 10) - 10(x - 10) = (x - 10)(x - 10)
13 y4 - 14y2 + 49 = y4 - 7y2 - 7y2 + 49 = y2(y2 - 7) - 7(y2 - 7) = (y2 - 7)2
Mk chỉ làm được đến đó thôi
Định làm hết nhưng bạn kia làm đúng rồi ! Còn 2 câu làm nốt !!!!
Bài giải
\(3,\text{ }\frac{1}{4}a^2+2ab+4b^2=\left(\frac{1}{2}a\right)^2+2ab+\left(2b\right)^2=\left(\frac{1}{2}a+2b\right)^2\)
\(4,\text{ }\frac{1}{9}-\frac{2}{3}y^4+y^8=\left(\frac{1}{3}\right)^2-2\cdot\frac{1}{3}y^4+\left(y^4\right)^2=\left(y^4+\frac{1}{3}\right)^2\)