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Đặt: \(x^2-6x+1=a;x^2+1=b\)
Khi đó đa thức này có dạng:
\(2a^2+5ab+2b^2=2a^2+4ab+ab+2b^2\)
\(=2a\left(a+2b\right)+b\left(a+2b\right)=\left(a+2b\right)\left(2a+b\right)\)
Thay lại a và b thì được:
\(\left(a+2b\right)\left(2a+b\right)=\left(x^2-6x+1+2x^2+2\right)\left(2x^2-12x+2+x^2+1\right)\)
\(=\left(3x^2-6x+3\right)\left(3x^2-12x+3\right)\)
\(=9\left(x-1\right)^2\left(x^2-4x+1\right)\)
Vậy ...
c) Đặt \(A=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
Đặt \(x^2+3x+1,5=a\)
\(\Rightarrow A=\left(a-0,5\right)\left(a+0,5\right)-6\)
\(\Rightarrow A=a^2-0,25-6\)
\(\Rightarrow A=a^2-\frac{25}{4}\)
\(\Rightarrow A=\left(a-\frac{5}{2}\right)\left(a+\frac{5}{2}\right)\)
Thay \(a=x^2+3x+0,5\)vào A ta có :
\(A=\left(x^2+3x+0,5-\frac{5}{2}\right)\left(x^2+3x+0,5+\frac{5}{2}\right)\)
\(A=\left(x^2+3x-2\right)\left(x^2+3x+3\right)\)
c, Đặt \(x^2+3x+2=a\)
Ta có : \(\left(a-1\right)a-6=a^2-a-6=\left(a^2-3a\right)+\left(2a-6\right)\)
\(=a\left(a-3\right)+2\left(a-3\right)\)
\(=\left(a+2\right)\left(a-3\right)\)
\(=\left(x^2+3x+4\right)\left(x^2+3x-1\right)\)
Câu d làm tương tự .
Gợi ý : (x+3)(x+5) = x2 + 8x + 15
đặt bằng a rồi giải tiếp
a, 4y(x-1)-(1-x)
=(x-1)(4y+1)
b,3x(z+2)+5(-x-2)
=3x(z+2)-5(x+2)
=(z+2)(3x-5)
\(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=\)\(\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)
\(=\)\(\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\)\(\left(3x-2\right)\left(3x-6\right)\)
\(=\)\(3\left(x-2\right)\left(3x-2\right)\)
Chúc bạn học tốt ~
Ta có: \(3x^2\left(y-x\right)+6x^2\left(x-y\right)^2\)
\(=3x^2\left(y-x\right)+6x^2\left(y-x\right)^2\)
\(=3x^2\left(y-x\right)\left[1-2\left(y-x\right)\right]\)
\(=3x^2\left(y-x\right)\left(2x-2y+1\right)\)
3x2( y - x ) + 6x2( x - y )2
= 3x2( y - x ) + 6x2( y - x )2
= 3x2( y - x )[ 1 + 2( y - x ) ]
= 3x2( y - x )( 2y - 2x + 1 )
\(P\left(x\right)=\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=\left[\left(4x+1\right)\left(3x+2\right)\right].\left[\left(12x-1\right)\left(x+1\right)\right]-4\)
\(=\left(12x^2+8x+3x+2\right).\left(12x^2+12x-x-1\right)-4\)
\(=\left(12x^2+11x+2\right).\left(12x^2+11x-1\right)-4\)
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...
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 \(12x^2+11x-1=a\)
\(\Rightarrow B=a.\left(a+3\right)-4\)
\(B=a^2+3a-4\)
\(B=\left(a^2-a\right)+\left(4a-4\right)\)
\(B=a.\left(a-1\right)+4.\left(a-1\right)\)
\(B=\left(a-1\right)\left(a+4\right)\)
\(\Rightarrow B=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)
Ta có (6x+5)2(3x+2)(x+1)-35
= (36x2+60x+25)(3x2+5x+2)-35 (1)
Đặt a=3x2+5x+2
=> 12a+1= 12(3x2+5x+2)+1 =36x2+60x+25
Thay a=3x2+5x+2 vào (1) ta được
(12a+1).a-35=12a2+a-35
= 12a2-20a+21a-35
= 4a(3a-5)+7(3a-5)
= (3a-5)(4a+7) (2)
Thay 3x2+5x+2=a vào (2) ta được
(9x2+15x+6-5)(12x2+20x+8+7)
= (9x2+15x+1)(12x2+20x+15)
Ta có: \(\left(6x+5\right)^2\left(3x+2\right)\left(x+1\right)-35\)
\(=\left(36x^2+60x+25\right)\left(3x^2+5x+2\right)-35\)(1)
Đặt \(3x^2+5x+2=y\)
\(\left(1\right)=\left(12y+1\right)y-35\)
\(=12y^2+y-35\)
\(=\left(3y-5\right)\left(4y+7\right)\)
\(=\left(9x^2+15x+1\right)\left(12x^2+20x+15\right)\)