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1/
a, \(4x^2+36xy+81y^2=\left(2x+9y\right)^2\)
b, \(12y+\frac{9}{100}y^2+400=\left(\frac{3}{10}y+20\right)^2\)
2/
\(2bc+b^2+c^2-a^2=\left(b+c\right)^2-a^2=\left(a+b+c\right)\left(b+c-a\right)=2p\left(b+c-a\right)\) (1)
Ta có: a+b+c=2p => b+c=2p-a (2)
Thay (2) và (1) ta có:
\(2p\left(2p-a-a\right)=2p\left(2p-2a\right)=4p\left(p-a\right)\) (đpcm)
3/
Gọi 2 số tự nhiên chẵn là 2k và 2k+2 (k thuộc N)
Theo bài ra ta có: \(\left(2k+2\right)^2-\left(2k\right)^2=36\)
=> \(\left(2k+2-2k\right)\left(2k+2+2k\right)=36\)
=>\(2\left(4k+2\right)=36\)
=>\(8k+4=36\)
=>\(8k=32\)
=> k = 4
=> \(2k=8;2k+2=10\)
Vậy...
8x3+30x2+150x+125
=(2x)3+3. (2x)2.5+ 3.2x.52+53
=(2x+5)3
8x3 + 30x2 + 150x + 125 = (2x)3 + 3. (2x)2.5 + 3. 2x. 52 + 53 = (2x + 5)3
\(4a^2-4a+1\)
\(=\left(2a\right)^2-2.2a.1+1^2\)
\(=\left(2a-1\right)^2\)
ms lên lớp 8 à hk tốt nha
a) y 2 – 4 xy 2 + 4 x 2 y 2 . b) x 2 − 32 5 xy + 256 25 y 2 .
c) 1 9 a 2 b 4 − c 6 . d) a 4 − 8 9 a + 16 81 .
a) \(x^4-30x+31x-30\)
= \(x^4+x-30x^2+30x-30\)
=\(x\left(x^3+1\right)-30\left(x^2-x+1\right)\)
= \(x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)\)
= \(\left(x^2-x+1\right)\left(x^2+x-30\right)\)
=
b: \(=\left(x^4-x^3+x^2-x+1\right)\left(x^4+x^3+x^2+x+1\right)\)
a: \(=x^4-5x^3+5x^3-25x^2-5x^2+25x+6x-30\)
\(=\left(x-5\right)\left(x^3+5x^2-5x+6\right)\)
\(=\left(x-5\right)\left(x^3+6x^2-x^2-6x+x+6\right)\)
\(=\left(x-5\right)\left(x+6\right)\left(x^2-x+1\right)\)
\(a,\)
với \(a=100\)
\(=>9x^2+30x+25=\left(3x\right)^2+2.3.5x+5^2=\left(3x_{ }+5\right)^2\)
\(b,\)
với \(a=\dfrac{1}{25}\)
\(25x^2-2x+\dfrac{1}{25}=\left(5x\right)^2-2.5.x.\dfrac{1}{5}+\left(\dfrac{1}{5}\right)^2=\left(5x-\dfrac{1}{5}\right)^2\)
\(c,\)
với \(a=6\)
\(=>x^2+2.3.x+3^2=\left(x+3\right)^2\)
\(d.\)
với \(a=\dfrac{4}{3}\)
\(=>\left(2x\right)^2-2.2.\dfrac{1}{3}x+\left(\dfrac{1}{3}\right)^2=\left(2x-\dfrac{1}{3}\right)^2\)
\(a.x^2+x+\frac{1}{4}=x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2\)
\(=\left(x+\frac{1}{4}\right)^2\)
b) \(x^2+12xy+36xy^2=x^2+2.x.y+y^2\)
\(a.x^2+x+\frac{1}{4}=x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2\)
\(=\left(x+\frac{1}{4}\right)^2\)
b) \(x^2+12xy+36xy^2=x^2+2.x.y+y^2\)
\(a.x^2+x+\frac{1}{4}=x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2\)
\(=\left(x+\frac{1}{4}\right)^2\)
b) \(x^2+12xy+36xy^2=x^2+2.x.y+y^2\)
\(a.x^2+x+\frac{1}{4}=x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2\)
\(=\left(x+\frac{1}{4}\right)^2\)
b) \(x^2+12xy+36xy^2=x^2+2.x.y+y^2\)
\(=\left(x+y\right)^2\) c) \(4x^2-12xy+9y^2=\left(2x\right)^2-2.2x.3y+\left(3y\right)^2\) \(=\left(2x-3y\right)^2\)d) \(x^2-2x+4=x^2-2.x.4+4^2\)
\(=\left(x-4\right)^2\)
e) \(25x^2+4y^2-20xy=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\)
\(=\left(5x-2y\right)^2\)
^...^ ^_^ Bài làm có gì ko hiểu bạn cứ hỏi nhé ^_^
mạng của mk bị lỗi bạn xem cái phần cuối cùng nhé xl bạn nhiều vì mạng của mk bị lỗi
a) \(8a^2xy-18b^2xy=2xy\left(4a^2-9b^2\right)=2xy\left(2a-3b\right)\left(2a+3b\right)\)
b) \(32a^2b^2-4=4\left(8a^2b^2-1\right)\)
c) \(x^2-49z^2-4xy+4y^2=\left(x^2-4xy+4y^2\right)-49z^2\)
\(=\left(x-2y\right)^2-\left(7z\right)^2=\left(x-2y+7z\right)\left(x-2y-7z\right)\)
d) \(3x^2+6x+3-3y^2=3\left(x^2+2x+1-y^2\right)=3.\left[\left(x+1\right)^2-y^2\right]\)
\(=3\left(x-y+1\right)\left(x+y+1\right)\)
e) \(12x^2y-12y^3+36xy+27y=3y\left(4x^2-4y^2+12x+9\right)\)
\(=3y\left[\left(4x^2+12x+9\right)-4y^2\right]=3y\left[\left(2x+3\right)^2-\left(2y\right)^2\right]\)
\(=3y\left(2x-2y+3\right)\left(2x+2y+3\right)\)
a) 8a2xy - 18b2xy
= 2xy( 4a2 - 9b2 )
= 2xy( [ ( 2a )2 - ( 3b )2 ]
= 2xy( 2a - 3b )( 2a + 3b )
b) 32a2b2 - 4
= 4( 8a2b2 - 1 )
c) x2 - 49z2 - 4xy + 4y2
= ( x2 - 4xy + 4y2 ) - 49z2
= ( x - 2y )2 - ( 7z )2
= ( x - 2y - 7z )( x - 2y + 7z )
d) 3x2 + 6x + 3 - 3y2
= 3( x2 + 2x + 1 - y2 )
= 3[ ( x2 + 2x + 1 ) - y2 ]
= 3[ ( x + 1 )2 - y2 ]
= 3( x - y + 1 )( x + y + 1 )
e) 12x2y - 12y3 + 36xy + 27y
= 3y( 4x2 - 4y2 + 12x + 9 )
= 3y[ ( 4x2 + 12x + 9 ) - 4y2 ]
= 3y[ ( 2x + 3 )2 - ( 2y )2 ]
= 3y( 2x - 2y + 3 )( 2x + 2y + 3 )