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a.\(9.\left(x+5\right)^2-\left(x-7\right)^2\)
\(=3^2.\left(x+5\right)^2-\left(x-7\right)^2\)
\(=\left(x+8\right)^2-\left(x-7\right)^2\)
\(=\left(x+8-x+7\right)\left(x+8+x-7\right)\)
\(=15\left(2x+1\right)\)
\(=30x+15\)
b.\(49\left(y-4\right)^2-9\left(y-2\right)^2\)
\(=7^2.\left(y-4\right)^2-3^2\left(y-2\right)^2\)
\(=\left(y+3\right)^2-\left(y+1\right)^2\)
\(=\left(y+3-y-1\right)\left(y+3+y+1\right)\)
\(=2\left(2y+4\right)\)
\(=4y+8\)
a./ \(=\left[3\left(x+5\right)\right]^2-\left(x-7\right)^2=\left(3x+15-x+7\right)\left(3x+15+x-7\right)=\left(2x+22\right)\left(4x+8\right).\)
\(=8\left(x+11\right)\left(x+2\right).\)
b./ \(=\left[7\left(y-4\right)\right]^2-\left[3\left(y-2\right)\right]^2=\left(7y-28+3y-6\right)\left(7y-28-3y+6\right)=\left(10y-34\right)\left(4y-22\right)\)
\(=4\left(5y-17\right)\left(2y-11\right)\)
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\(49\left(y-4\right)^2-9\left(y+2\right)^2=7^2.\left(y-4\right)^2-3^2.\left(y+2\right)^2\)
\(=\left(7y-28\right)^2-\left(3y+6\right)^2\)
\(=\left(7y-28+3y+6\right).\left(7y-28-3y-6\right)\)
\(=4.\left(5y-11\right).\left(2y-17\right)\)
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49( y - 4 )2 - 9 ( y + 2 )2
= 72 ( y - 4 )2 - 32 ( ( y + 2 )2
= [ 7 ( y - 4 ) ]2 - [ 3 ( y + 2 ) ]2
= ( 7y - 4 )2 - ( 3y + 2 )2
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a)27x2.(y-1)9x3.(1-y)
=27x2.(y-1)+9x3.(y-1)
=9x2(y-1)[3+x]
b)8x3 + 1/27
=(2x)3 + (\(\frac{1}{3}\))3
= (2x+\(\frac{1}{3}\))(\(\left(2x\right)^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2\)
= (2x+\(\frac{1}{3}\))(\(\left(2x\right)^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2\)
a) 27x2 ( y - 1) - 9x3 ( 1 - y)
=27x2 (y-1) + 9x3 ( y - 1 )
= (27x2 + 9x3) ( y -1 )
=9x2 ( x + 3) ( y - 1)
b)8x3+1/27
\(=\left(2x\right)^3+\left(\frac{1}{3}\right)^3\)
\(=\left(\frac{2x}{9}+\frac{1}{27}\right)\left(36x^2-6x+1\right)\)
c)49 ( y - 4 )2 - 9 ( y + 2)2
= [7(y - 4)]2 - [3(y + 2)]2
= (7y - 28 + 3y + 6)(7y - 28 - 3y - 6)
= (10y - 22)(4y - 34)
= 4(5y - 11)(2y - 34)
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a, \(\left(x+1\right)^2-25=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
b, \(\left(xy+4\right)^2-4\left(x+y\right)^2=\left(xy+4\right)^2-\left(2x+2y\right)^2=\left(xy+4-2x-2y\right)\left(xy+4+2x+2y\right)\)
c, xem lại đề nhé
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a)49(y-4)2-9(y+2)2
=[7(y-4)]2-[3(y+2)]2
=(7y-28+3y+6)(7y-28-3y-6)
=(10y-22)(4y-34)
=4(5y-11)(2y-34)
b)8x3+1/27
=(2x)3+(1/3)3
=(2x+1/3)[(2x)2-2x.1/3+(1/3)2]
=(2x+1/3)(4x2-2/3x+1/9)
c)125-x6
=53-(x2)3
=(5-x2)(52+5.x2+x4)
=(5-x)(25+5x2+x4)
49(y-4)2-9(y+2)2 = [7(y-4)]2-[3(y+2)]2=[7(y-4)-3(y+2)].[7(y-4)+3(y+2)] = (7y-28-3y-6)(7y-28+3y+6) = (4y-34)(10y-22)=4(2y-17)(5y-11)
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a) \(x^3+3.2x^2y+3.2^2.x.y^2+\left(2y\right)^3=\left(x+2y\right)^3\)
b) áp dụng HDT : \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(\Rightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)=3x\left(x+2\right)\)
c) cũng áp dụng hdt :\(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\left[3\left(x+5\right)\right]^2-\left(x-7\right)^2=\left[3\left(x+5\right)-x+7\right]\left[3\left(x+5\right)+x-7\right]\)\(=\left(3x+15-x+7\right)\left(2x+15+x-7\right)=\left(2x+22\right)\left(3x+8\right)=2\left(x+11\right)\left(3x+8\right)\)
d) áp dụng típ \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\left[5\left(x-y\right)\right]^2-\left[4\left(x+y\right)\right]^2=\left[5\left(x-y\right)-4\left(x+y\right)\right]\left[5\left(x-y\right)+4\left(x+y\right)\right]\)
\(=\left(5x-5y-4x-4y\right)\left(5x-5y+4x+4y\right)=\left(x-9y\right)\left(9x-y\right)\)
e)Áp dụng típ Hdt như trên
\(\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2=\left[7\left(y-4\right)-3\left(y+2\right)\right]\left[7\left(y-4\right)+3\left(y+2\right)\right]\)
\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)=\left(4y-34\right)\left(11y-22\right)\)
\(=2\left(2y-17\right).11\left(y-2\right)=22\left(2y-17\right)\left(y-2\right)\)
Bạn 1 cái t i c k nha thật sự rất cảm ơn
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1. \(125x^3+y^6=\left(5x\right)^3+\left(y^2\right)^3\)
\(=\left(5x+y^2\right)\left[\left(5x\right)^2-5x.y^2+\left(y^2\right)^2\right]\)
\(=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)
2. \(4x\left(x-2y\right)+8y\left(2y-x\right)\)
\(=4x\left(x-2y\right)-8y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(4x-8y\right)\)
3. \(25\left(x-y\right)^2-16\left(x+y\right)^2\)
\(=\left[5\left(x-y\right)\right]^2-\left[4\left(x+y\right)\right]^2\)
\(=\left[5\left(x-y\right)-4\left(x+y\right)\right]\left[5\left(x-y\right)+4\left(x+y\right)\right]\)
\(=\left(5x-5y-4x-4y\right)\left(5x-5y+4x+4y\right)\)
\(=\left(x-9y\right)\left(9x-y\right)\)
4. \(x^4-x^3-x^2+1\)
\(=x^3\left(x-1\right)-\left(x^2-1\right)\)
\(=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^3-x-1\right)\)
5. \(a^3x-ab+b-x\)
\(=a^3x-x-ab+b\)
\(=x\left(a^3-1\right)-b\left(a-1\right)\)
\(=x\left(a-1\right)\left(a^2+a+1\right)-b\left(a-1\right)\)
\(=\left(a-1\right)\left[x\left(a^2+a+1\right)-b\right]\)
6. \(x^3-64=x^3-4^3\)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
7. \(0,125\left(a+1\right)^3-1\)
\(=\left[0,5\left(a+1\right)\right]^3-1^3\)
\(=\left[0,5\left(a+1\right)-1\right]\left\{\left[0,5\left(a+1\right)\right]^2+\left[0,5\left(a+1\right).1\right]+1^2\right\}\)
\(=\left[0,5\left(a+1-2\right)\right]\left[0,25a^2+0,5a+0,25+0,5a+0,5+1\right]\)
\(=\left[0,5\left(a-1\right)\right]\left(0,25a^2+a+1,75\right)\)
8. \(9\left(x+5\right)^2-\left(x-7\right)^2\)
\(=\left[3\left(x+5\right)\right]^2-\left(x-7\right)^2\)
\(=\left(3x+15-x+7\right)\left(3x+15+x-7\right)\)
\(=\left(2x+22\right)\left(4x+8\right)\)
9. \(49\left(y-4\right)^2-9\left(y+2\right)^2\)
\(=\left[7\left(y-4\right)\right]^2-\left[3\left(y+2\right)\right]^2\)
\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)\)
\(=\left(4y-34\right)\left(10y-22\right)\)
10. \(x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(xy-1\right)\)
11. \(x^3+3x^2+3x+1-27z^3\)
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
12. \(x^2-y^2-x+y=\left(x-y\right)\left(x+y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-1\right)\)
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a)
\(-25x^6-y^8+10x^3y^4=-(25x^6+y^8-10x^3y^4)\)
\(=-[(5x^3)^2-2.5x^3y^4+(y^4)^2]\)
\(=-(5x^3-y^4)^2\)
b) \(49(y-4)^2-9(y+2)^2=7^2(y-4)^2-3^2(y+2)^2\)
\(=(7y-28)^2-(3y+6)^2=(7y-28-3y-6)(7y-28+3y+6)\)
\(=(4y-34)(10y-22)=4(2y-17)(5y-11)\)
c)
\(x^2+2xy+y^2-xz-yz=(x^2+2xy+y^2)-(xz+yz)\)
\(=(x+y)^2-z(x+y)=(x+y)(x+y-z)\)
d)
\(1+(a+b+c)+(ab+bc+ac)+abc\)
\(=(1+a+b+ab)+(bc+ac+abc+c)\)
\(=(1+a+b+ab)+c(b+a+ab+1)\)
\(=(a+b+ab+1)(c+1)\)
\(=[(a+ab)+(b+1)](c+1)\)
\(=[a(b+1)+(b+1)](c+1)\)
\(=(a+1)(b+1)(c+1)\)