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Đặt \(A=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(A=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+10=y\)
\(\Rightarrow\)\(A=y.\left(y+2\right)-24\)
\(A=y^2+2y+1-25\)
\(A=\left(y+1\right)^2-5^2\)
\(A=\left(y+1-5\right)\left(y+1+5\right)\)
\(A=\left(y-4\right)\left(y+6\right)\)
\(\Rightarrow A=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(A=\left[\left(x^2+x\right)+\left(6x+6\right)\right].\left(x^2+7x+16\right)\)
\(A=\left[x.\left(x+1\right)+6.\left(x+1\right)\right].\left(x^2+7x+16\right)\)
\(A=\left(x+1\right).\left(x+6\right).\left(x^2+7x+16\right)\)
Đặt \(B=\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(B=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
Đặt \(12x^2+11x-1=a\)
\(\Rightarrow B=a.\left(a+3\right)-4\)
\(B=a^2+3a-4\)
\(B=\left(a^2-a\right)+\left(4a-4\right)\)
\(B=a.\left(a-1\right)+4.\left(a-1\right)\)
\(B=\left(a-1\right)\left(a+4\right)\)
\(\Rightarrow B=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)
h) \(x^4+4=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\)
i) \(\left(1+x^2\right)^2-4x\left(1-x^2\right)=\left(1+x^2\right)^2+4x^3-4x=x^4+4x^3+2x^2-4x+1\)
Ta có :
\(x^2\left(x^4-1\right)\left(x^2+1\right)+1=x^2\left(x^2-1\right)\left(x^2+1\right)\left(x^2+2\right)+1\)
\(\Leftrightarrow x^2\left(x^2+1\right)\left(x^2-1\right)\left(x^2+2\right)+1=\left(x^4-x^2\right)\left(x^4+x^2-2\right)+1\)
Gọi \(x^4-x^2\) là t, ta có:
t(t-2)+1=\(t^2-2t+1=\left(t-1\right)^2=\left(x^4+x^2-1\right)^2\)
a) \(4x^3\left(x^2+x\right)-\left(x^2+x\right)=\left(x^2+x\right)\left(4x^3-1\right)\)
b)\(\left(1-2a+a^2\right)-\left(b^2-2bc+c^2\right)=\left(1-a\right)^2-\left(b-c\right)^2=\)\(\left(1-a+b-c\right)\left(1-a-b+c\right)\)
lm tiếp câu c
c) \(C=\left(x-7\right)\left(x-5\right)\left(x-4\right)\left(x-2\right)-72\)
\(=\left[\left(x-7\right)\left(x-2\right)\right]\left[\left(x-5\right)\left(x-4\right)\right]-72\)
\(=\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72\)
Đặt \(x^2-9x+17=a\) ta có:
\(C=\left(a-3\right)\left(a+3\right)-72\)
\(=a^2-9-72\)
\(=a^2-81=\left(a-9\right)\left(a+9\right)\)
Thay trở lại ta được: \(C=\left(x^2-9x++8\right)\left(x^2-9x+26\right)\)
phân tích đa thức thành nhân tử \(x^2\cdot\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)
\(x^2\cdot\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x+4\right)^2-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\)
a) 16x2 - ( x2 + 4 )2
= ( 4x )2 - ( x2 + 4 )2
= [ 4x - ( x2 + 4 ) ][ 4x + ( x2 + 4 ) ]
= ( -x2 + 4x - 4 )( x2 + 4x + 4 )
= [ -( x2 - 4x + 4 ) ]( x + 2 )2
= [ -( x - 2 )2 ]( x + 2 )2
b) ( x + y )3 + ( x - y )3
= [ ( x + y ) + ( x - y ) ][ ( x + y )2 - ( x + y )( x - y ) + ( x - y )2 ]
= ( x + y + x - y )[ x2 + 2xy + y2 - ( x2 - y2 ) + x2 - 2xy + y2 ]
= 2x( 2x2 + 2y2 - x2 + y2
= 2x( x2 + 3y2 )
\(P\left(x\right)=\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=\left[\left(4x+1\right)\left(3x+2\right)\right].\left[\left(12x-1\right)\left(x+1\right)\right]-4\)
\(=\left(12x^2+8x+3x+2\right).\left(12x^2+12x-x-1\right)-4\)
\(=\left(12x^2+11x+2\right).\left(12x^2+11x-1\right)-4\)
Đặt \(12x^2+11x=t\), ta có:
\(\left(t+2\right)\left(t-1\right)-4\)
\(=t^2-t+2t-2-4=t^2+t-6\)
\(=t^2-2t+3t-6\)
\(=t\left(t-2\right)+3\left(t-2\right)=\left(t-2\right)\left(t+3\right)\)
Thay \(t=12x^2+11x\), ta được:
\(P\left(x\right)=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)
Đs...
\(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)
\(=\left(x+4\right)^2\left(x^2-1\right)-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\)
\(=\left(x-1\right)\left(x+1\right)\left(x+4+1\right)\left(x+4-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+5\right)\left(x-3\right)\)
=.= hok tốt!!