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HIHI, bài này thì bó tay lẫn cả chân
Vì mới học xong lớp 6 hoi.
Học tốt nha, nếu ko ai giải thì thử vào câu hỏi tương tự thử
Nha, học tốt !
#)Giải:
-Không sao mình biết cách làm mà, mình chỉ thử lòng ae thui !
Bài 2:
a) \(x^2-y^2+3x-3y=\left(x^2-y^2\right)+\left(3x-3y\right)\)
\(=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
b) \(5x-5y+x^2-2xy+y^2=\left(5x-5y\right)+\left(x^2-2xy+y^2\right)\)
\(=5\left(x-y\right)+\left(x-y\right)^2=\left(x-y\right)\left(x-y+5\right)\)
c) \(x^2-5x+4=x^2-x-4x+4=\left(x^2-x\right)-\left(4x-4\right)\)
\(=x\left(x-1\right)-4\left(x-1\right)=\left(x-1\right)\left(x-4\right)\)
a) \(4x^2-6x=2x\left(2x-3\right)\)
b) \(9x^4y^3+3x^2y^4=3x^2y^3\left(3x^2+y\right)\)
c) \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(5x+3\right)\left(x-y\right)\)
d) \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)
e) \(5\left(x+3y\right)-15x\left(x+3y\right)=\left(5-15x\right)\left(x+3y\right)\)
\(=5\left(1-3x\right)\left(x+3y\right)\)
f) \(2x^2\left(x+1\right)-4\left(x+1\right)=\left(2x^2-4\right)\left(x+1\right)\)
\(=\left(\sqrt{2}x-2\right)\left(\sqrt{2}x+2\right)\left(x+1\right)\)
- Toán lớp 8
Khóa học Toán lớp 8
Gửi câu trả lời của bạn
Bài 1
\(a,5x^2-10xy+5y^2\)
\(=5\cdot\left(x^2-2xy+y^2\right)\)
\(=5\cdot\left(x-y\right)^2\)
\(b,x^2-y^2+6y-9\)
\(=x^2-\left(y^2-6y+9\right)\)
\(=x^2-\left(y-3\right)^2\)
\(=\left(x-y+3\right)\cdot\left(x+y-3\right)\)
\(c,3x^4-75x^2y^2\)
\(=3x^2\cdot\left(x^2-25y^2\right)\)
\(=3x^2\cdot\left(x-5y\right)\cdot\left(x+5y\right)\)
\(d,x^4y+xy^4\)
\(=xy\left(x^3+y^3\right)\)
\(=xy\cdot\left(x+y\right)\cdot\left(x^2-xy+y^2\right)\)
a) ( 3 - x )( x2 + 2x - 7 ) + ( x - 3 )( x2 + x - 5 )
= ( 3 - x )( x2 + 2x - 7 ) - ( 3 - x )( x2 + x - 5 )
= ( 3 - x )( x2 + 2x - 7 - x2 - x + 5 )
= ( 3 - x )( x - 2 )
b) ( x - 5 )2 + 3( 5 - x )
= ( x - 5 )2 - 3( x - 5 )
= ( x - 5 )( x - 5 - 3 ) = ( x - 5 )( x - 8 )
c) 2x( x - 1 )2 - ( 1 - x )3
= 2x( 1 - x )2 - ( 1 - x )3
= ( 1 - x )2( 2x - 1 + x ) = ( 1 - x )2( 3x - 1 )
d) x2 + 8x + 16 = ( x + 4 )2
e) x2 - 4xy + 4y2 = ( x - 2y )2
g) 4x2 - 25y2 = ( 2x )2 - ( 5y )2 = ( 2x - 5y )( 2x + 5y )
h) 25( x + 1 )2 - 4( x - 3 )2
= 52( x + 1 )2 - 22( x - 3 )2
= ( 5x + 5 )2 - ( 2x - 6 )2
= ( 5x + 5 - 2x + 6 )( 5x + 5 + 2x - 6 )
= ( 3x + 11 )( 7x - 1 )
i) x3 + 27 = ( x + 3 )( x2 - 3x + 9 )
k) 8x3 - 125 = ( 2x )3 - 53 = ( 2x - 5 )( 4x2 + 10x + 25 )
l) x3 + 6x2 + 12x + 8 = ( x + 2 )3
m) -x3 + 9x2 - 27x + 27 = -( x3 - 9x2 + 27x - 27 ) = -( x - 3 )3
\(ab-a+b-a^2=\left(ab-a^2\right)+\left(b-a\right)=a.\left(b-a\right)+\left(b-a\right)=\left(b-a\right).\left(a+1\right)\\ \)
\(8-\left(x-1\right)^3=2^3-\left(x-1\right)^3=\left(2-x+1\right).\left(2^2+2.\left(x-1\right)+\left(x-1\right)^2\right)\)
\(=\left(3-x\right).\left(4+2x-2+x^2-2x+1\right)=\left(3-x\right).\left(3+x^2\right)\)
\(3x^2-8x-16=\left(3x^2+4x\right)-\left(12x+16\right)=x.\left(3x+4\right)-4.\left(3x+4\right)=\left(3x+4\right).\left(x-4\right)\)
\(ab-a+b-a^2\)
\(=\left(ab-a^2\right)+\left(b-a\right)\)
\(=a\left(b-a\right)+\left(b-a\right)\)
\(=\left(a+1\right)\left(b-a\right)\)
\(\text{a)}x^3-6x^2+12x-8\)
\(=x^3-2x^2-4x^2+8x+4x-8\)
\(=\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(4x-8\right)\)
\(=x^2\left(x-2\right)+4x\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)\left(x+2\right)^2\)
\(\text{b)}8x^2+12x^2y+6xy^2+y^3=\left(2x+y\right)^3\)
Bài 2:
\(\text{a) }x^7+1=\left(x^{\frac{7}{3}}\right)^3+1^3=\left(x^{\frac{7}{3}}+1\right)\left[\left(x^{\frac{7}{3}}\right)^2-x^{\frac{7}{3}}+1\right]=\left(x^{\frac{7}{3}}+1\right)\left(x^{\frac{14}{3}}-x^{\frac{7}{3}}+1\right)\)
\(\text{b) }x^{10}-1=\left(x^5\right)^2-1^2=\left(x^5-1\right)\left(x^5+1\right)\)
Bài 3:
\(\text{a) }69^2-31^2=\left(69-31\right)\left(69+31\right)=38.100=3800\)
\(\text{b) }1023^2-23^2=\left(1023-23\right)\left(1023+23\right)=1000.1046=1046000\)