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a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)(1)
Đặt \(x^2+5x+4=t\)
\(\Rightarrow\left(1\right)=t\left(t+2\right)-15=t^2+2t+1-16\)
\(=\left(t+1\right)^2-4^2=\left(t+5\right)\left(t-3\right)\)
\(=\left(x^2+5x+9\right)\left(x^2+5x+1\right)\)
b) \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(2x+5+x-9\right)\left(2x+5-x+9\right)\)
\(=\left(3x-4\right)\left(x+14\right)\)
1/(x+2)(x+3)(x+4)(x+5)-24
=(x+2)(x+5)(x+3)(x+4)
=(x+2)(x-2+7)(x+3)(x-3+7)
=[(x+2)(x-2)+7x+14][(x+3)(x-3)+7x+21]
=(x2-4+7x+14)(x2-9+7x+21)
=(x2+10+7x)(x2+12+7x)
2/(x2+x)2+4(x2+x)-12
=(x2+x)2+4(x2+x)+22-16
=(x2+x+2)2-42
=(x2+x+2+4)(x2+x+2-4)
=(x2+x+6)(x2+x-2)
3/(x2+x+1)(x2+x+2)-12
=(x2+x+1)(x2+x+-1+3)-12
=(x2+x+1)(x2+x+-1)+3(x2+x+1)-12
=(x2+x)-1+3(x2+x)+3-12
=(x2+x)(x2+x+3)-10
làm đến đây thì mk bí, bạn giúp suy nghĩ nốt nha
4/nó là nhân tử sẵn rồi mà
\(3/\)
\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
\(=\left(x^2+x+1\right)\left(x^2+x+1+1\right)-12\)
\(=\left(x^2+x+1\right)^2+x^2+x+1-12\)
\(=\left(x^2+x+1\right)^2+4\left(x^2+x+1\right)-3\left(x^2+x+1\right)-12\)
\(=\left(x^2+x+1\right)\left(x^2+x+1+4\right)-3\left(x^2+x+1+4\right)\)
\(=\left(x^2+x+1-3\right)\left(x^2+x+1+4\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)
Đặt \(x^2+5x+4=t\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=t.\left(t+2\right)-15\)
\(=t^2+2t+1-16\)
\(=\left(t+1\right)^2-4^2\)
\(=\left(t-3\right)\left(t+5\right)\)
\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)
Ta có :
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-15\)
\(=\left[x\left(x+4\right)+1\left(x+4\right)\right]\left[x\left(x+3\right)+2\left(x+3\right)\right]-15\)
\(=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)
\(=\left(x^2+5x+4\right)\left[\left(x^2+5x+4\right)+2\right]-15\)(1)
Đặt \(x^2+5x+4=y\)thì (1) trở thành :
\(y\left(y+2\right)-15\)
\(=y^2+2y-15\)
\(=y^2+5y-3y-15\)
\(=\left(y^2+5y\right)-\left(3y+15\right)\)
\(=y\left(y+5\right)-3\left(y+5\right)\)
\(=\left(y-3\right)\left(y+5\right)\)(2)
Thay \(y=x^2+5x+4\)thì (2) trở thành:
\(\left(x^2+5x+4-3\right)\left(x^2+5x+4+5\right)\)
\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)
\(x^8+x^7+1\)
\(=x^8+x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-xx+1\)
\(=\left(x^8-x^6+x^5-x^3+x^2\right)\)
\(+\left(x^7-x^5+x^4-x^2+x\right)\)
\(+\left(x^6-x^4+x^3-x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
b)\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)-24\)4
\(=\left[\left(x-1\right)\left(x-4\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]-24\)
\(=\left(x^2-4x-x+4\right)\left(x^2-3x-2x+6\right)-24\)
\(=\left(x^2-5x+4\right)\left(x^2-5x+4+2\right)-24\)
\(\)Đặt \(x^2-5x+4\)là a,ta có
\(=a\left(a+2\right)-24\)
\(=a^2+2a-24\)
\(=a^2+6a-4a-24\)
\(=a\left(a+6\right)-4\left(a+6\right)\)
\(=\left(a+6\right)\left(a-4\right)\)
Hay \(\left(x^2-5x+4+6\right)\left(x^2-5x+4-4\right)\)
\(=\left(x^2-5x+10\right)\left(x^2-5\right)\)
Câu hỏi của Huỳnh Bảo Nguyên - Toán lớp 8 - Học toán với OnlineMath
Mk làm òi nhé !
a) ( x + 1 )( x + 2 )( x + 3 )( x + 4 ) - 15
= [ ( x + 1 )( x + 4 ) ][ ( x + 2 )( x + 3 ) ] - 15
= ( x2 + 5x + 4 )( x2 + 5x + 6 ) - 15 (*)
Đặt t = x2 + 5x + 4
(*) trở thành
t( t + 2 ) - 15
= t2 + 2t - 15
= t2 - 3t + 5t - 15
= t( t - 3 ) + 5( t - 3 )
= ( t - 3 )( t + 5 )
= ( x2 + 5x + 4 - 3 )( x2 + 5x + 4 + 5 )
= ( x2 + 5x + 1 )( x2 + 5x + 9 )
b) ( x + 2 )( x + 3 )2( x + 4 ) - 12
= [ ( x + 2 )( x + 4 ) ]( x + 3 )2 - 12
= ( x2 + 6x + 8 )( x2 + 6x + 9 ) - 12 (*)
Đặt t = x2 + 6x + 8
(*) trở thành
t( t + 1 ) - 12
= t2 + t - 12
= t2 - 3t + 4t - 12
= t( t - 3 ) + 4( t - 3 )
= ( t - 3 )( t + 4 )
= ( x2 + 6x + 8 - 3 )( x2 + 6x + 8 + 4 )
= ( x2 + 6x + 5 )( x2 + 6x + 12 )
= ( x2 + x + 5x + 5 )( x2 + 6x + 12 )
= [ x( x + 1 ) + 5( x + 1 ) ]( x2 + 6x + 12 )
= ( x + 1 )( x + 5 )( x2 + 6x + 12 )
a, Gọi\(A=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)
Đặt\(y=x^2+5x+4\)
\(\Rightarrow A=y\left(y+2\right)-15\)
\(=y^2+2y-15\)
\(=\left(x-3\right)\left(x+5\right)\)
Hay\(A=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)
Vậy...
b,Gọi\(B=\left(x+2\right)\left(x+3\right)^2\left(x+4\right)-12\)
\(=\left(x^2+6x+8\right)\left(x^2+6x+9\right)-12\)
Đặt\(z=x^2+6x+8\)
\(\Rightarrow B=z\left(z+1\right)-12\)
\(=z^2+z-12\)
\(=\left(z-3\right)\left(z+4\right)\)
Hay\(B=\left(x^2+6x+5\right)\left(x^2+6x+12\right)\)
Vậy...
Linz