Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left(x+8\right)^2-2\left(x+8\right)\left(x-2\right)+\left(x-2\right)^2\)
\(=\left[\left(x+8\right)-\left(x-2\right)\right]^2\)
\(=\left(x+8-x+2\right)^2\)
\(=10^2\)
\(=2^2.5^2\)
b)\(x^3-4x^2-12x+27=\left(x^3+27\right)-\left(4x^2+12x\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)-4x\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-3x+9-4x\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
c)\(x^3+6x^2+11x+6=x^3+x^2+5x^2+5x+6x+6\)
\(=x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+5x+6\right)\)
\(=\left(x+1\right)\left(x^2+2x+3x+6\right)\)
\(=\left(x+1\right)\left[x\left(x+2\right)+3\left(x+2\right)\right]\)
\(=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
d)\(x^3+6x^2-13x-42=x^3-3x^2+9x^2-27x+14x-42\)
\(=x^2\left(x-3\right)+9x\left(x-3\right)+14\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+9x+14\right)\)
\(=\left(x-3\right)\left(x^2+2x+7x+14\right)\)
\(=\left(x-3\right)\left[x\left(x+2\right)+7\left(x+2\right)\right]\)
\(=\left(x-3\right)\left(x+2\right)\left(x+7\right)\)
a)18x2-12x
=3x(6x-4)
b)3x2-11x+6
=x(3x-11+6)
=x(3x-5)
c)x3+6x2+11x+6
=x2(x+23
\(18x^2-12x\)
\(=6x\left(3x-2\right)\)
\(3x^2-11x+6\)
\(=3x^2-9x-2x+6\)
\(=3x\left(x-3\right)-2\left(x-3\right)\)
\(=\left(x-3\right)\left(3x-2\right)\)
a x2 - 7x - 6 = x2 -6x-x-6
= x(x-6)+(x-6)
= (x+1)(x+6)
b x3 + 4x2 - 7x -10
=x3 - 2x2 + 6x2 -12x+5x-10
= x2(x-2) + 6x(x-2) + 5(x-2)
= (x2 +6x+5)(x-2)
=(x2 +x+5x+5)(x-2)
=[x(x+1)+5(x+1)](x-2)
=(x+5)(x+1)(x-2)
c x3 +x2-2
=x3-x2+2x2-2x+2x-2
=x2(x-1)+2x(x-1)+2(x-1)
=(x2+2x+2)(x-1)
d x2 +5x+6
=x2 +2x+3x+6
=x(x+2)+3(x+2)
=(x+3)(x-2)
Con 1 y e nhung luoi lam
mk chỉnh lại đề
\(x^2-7x+6\)
\(=x^2-x-6x+6\)
\(=x\left(x-1\right)-6\left(x-1\right)\)
\(=\left(x-1\right)\left(x-6\right)\)
Sorry nhé, mk mới lớp 6 thui.
Nên ko bit làm, mong đừng giận.
Nhưng xin tk nhé, đang thiếu sp trầm trọng.
Ai tk mk mk sẽ tk lại !
chị vào câu hỏi tương tự
kham khảo nha
chúc chị
thành công trong
học tập
a) \(x^2-x-2=x^2+x-2x-2=x\left(x+1\right)-2\left(x+1\right)\)
\(=\left(x+1\right)\left(x-2\right)\)
a) \(x^2-x-2=x^2-2x+x-2=x\left(x-2\right)+\left(x-2\right)=\left(x-2\right)\left(x+1\right)\)
b) \(x^3-19x-30==x^3+2x^2-2x^2-4x-15x-30=x^2\left(x+2\right)-2x\left(x+2\right)-15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+2x-15\right)=\left(x+2\right)\left(x-3\right)\left(x+5\right)\)
c) \(x^3-6x^2+11x-6=x^3-x^2-5x^2+5x+6x-6=x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\)
\(x^3+6x^2+11x+6=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
a) x3 - 6x2 + 11x - 6
= ( x3 - 2x2 ) - ( 4x2 - 8x ) + ( 3x - 6 )
= x2( x - 2 ) - 4x( x - 2 ) + 3( x - 2 )
= ( x - 2 )( x2 - 4x + 3 )
= ( x - 2 )( x2 - x - 3x + 3 )
= ( x - 2 )[ x( x - 1 ) - 3( x - 1 ) ]
= ( x - 2 )( x - 1 )( x - 3 )
b) x3 - 6x2 - 9x + 14
= ( x3 - x2 ) - ( 5x2 - 5x ) - ( 14x - 14 )
= x2( x - 1 ) - 5x( x - 1 ) - 14( x - 1 )
= ( x - 1 )( x2 - 5x - 14 )
= ( x - 1 )( x2 + 2x - 7x - 14 )
= ( x - 1 )[ x( x + 2 ) - 7( x + 2 ) ]
= ( x - 1 )( x + 2 )( x - 7 )
c) x3 + 6x2 + 11x + 6
= ( x3 + 2x2 ) + ( 4x2 + 8x ) + ( 3x + 6 )
= x2( x + 2 ) + 4x( x + 2 ) + 3( x + 2 )
= ( x + 2 )( x2 + 4x + 3 )
= ( x + 2 )( x2 + x + 3x + 3 )
= ( x + 2 )[ x( x + 1 ) + 3( x + 1 ) ]
= ( x + 2 )( x + 1 )( x + 3 )
e) x6 - 9x3 + 8
Đặt t = x3
bthuc <=> t2 - 9t + 8
= t2 - t - 8t + 8
= t( t - 1 ) - 8( t - 1 )
= ( t - 1 )( t - 8 )
= ( x3 - 1 )( x3 - 8 )
= ( x - 1 )( x2 + x + 1 )( x - 2 )( x2 + 2x + 4 )