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\(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)-3x^2\)
\(=4\left[\left(x+5\right)\left(x+12\right)\right]\left[\left(x+6\right)\left(x+10\right)\right]-3x^2\)
\(=4\left(x^2+17x+60\right)\left(x^2+16x+60\right)-3x^2\)
\(=\left(2x^2+34x+120\right)\left(2x^2+32x+60\right)-3x^2\)
\(=\left(2x^2+33x+120\right)^2-x^2-3x^2\)
\(=\left(2x^2+33x+120-2x\right)\left(2x^2+33x+120+2x\right)\)
\(=\left(2x+15\right)\left(x+8\right)\left(2x^2+35x+120\right)\)
\(4\left(x+5\right)\left(x+12\right)\left(x+6\right)\left(x+10\right)-3x^2\)
\(=2\left(x^2+60+17x\right).2\left(x^2+60+16x\right)-3x^2\)
\(=\left(2x^2+120+33x+x\right)\left(2x^2+120+33x-x\right)-3x^2\)
\(=\left(2x^2+120+33x\right)^2-x^2-3x^2\)
\(=\left(2x^2+120+33x\right)^2-4x^2\)
\(=\left(2x^2+120+33x+2x\right)\left(2x^2+120+33x-2x\right)\)
\(=\left(2x^2+35x+120\right)\left(2x^2+31x+120\right)\)
\(=\left(2x^2+35x+120\right)\left(x+8\right)\left(2x+15\right)\)
a: \(x^4+x^2+2x+6\)
\(=x^4-2x^3+3x^2+2x^3-4x^2+6x+2x^2-4x+6\)
\(=\left(x^2-2x+3\right)\left(x^2+2x+2\right)\)
Đặt \(x^2+3x+1=t\)
\(\Rightarrow\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6=t.\left(t+1\right)-6\)
\(=t^2+t-6=\left(t^2-2t\right)+\left(3t-6\right)\)
\(=t\left(t-2\right)+3\left(t-2\right)=\left(t-2\right)\left(t+3\right)\)
\(=\left(x^2+3x+1-2\right)\left(x^2+3x+1+3\right)\)
\(=\left(x^2+3x-1\right)\left(x^2+3x+4\right)\)
\(A=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
Đặt \(x^2+3x+1=a\)ta có :
\(a\left(a+1\right)-6\)
\(=a^2+a-6\)
\(=a^2+6a-a-6\)
\(=\left(a^2+6a\right)-\left(a+6\right)\)
\(=a\left(a+6\right)-\left(a+6\right)\)
\(=\left(a+6\right)\left(a-1\right)\)
Thay \(a=x^2+3x+1\)vào A ta có :
\(A=\left(x^2+3x+1+6\right)\left(x^2+3x+1-1\right)\)
\(=\left(x^2+3x+7\right)\left(x^2+3x\right)\)
4( x+5) ( x+6) (x+10) ( x+12) -3x2
=4(x+5)(x+12)(x+6)(x+10)-3x2
=4.(x2+17x+60)(x2+16x+60)-3x2
Đặt t=x2+16x+60 ta được:
4.(t+x).t-3x2
=4t2+4tx-3x2
=4t2-2tx+6tx-3x2
=2t.(2t-x)+3x.(2t-x)
=(2t-x)(2t+3x)
thay t=x2+16x+60 ta được:
[2.(x2+16x+ 60)-x][2.(x2+16x+60)+3x]
=(2x2+32x+120-x)(2x2+32x+120+3x)
=(2x2+31x+120)(2x2+35x+120)
=(2x2+16x+15x+120)(2x2+35x+120)
=[2x.(x+8)+15.(x+8)](2x2+35x+120)
=(x+8)(2x+15)(2x2+35x+120)
4( x+5) ( x+6) (x+10) ( x+12) -3x 2
=4(x+5)(x+12)(x+6)(x+10)-3x 2
=4.(x 2+17x+60)(x 2+16x+60)-3x 2
Đặt t=x 2+16x+60 ta được: 4.(t+x).t-3x 2
=4t 2+4tx-3x 2
=4t 2 -2tx+6tx-3x 2
=2t.(2t-x)+3x.(2t-x)
=(2t-x)(2t+3x)
thay t=x 2+16x+60 ta được: [2.(x 2+16x+ 60)-x][2.(x 2+16x+60)+3x]
=(2x 2+32x+120-x)(2x 2+32x+120+3x)
=(2x 2+31x+120)(2x 2+35x+120)
=(2x 2+16x+15x+120)(2x 2+35x+120)
=[2x.(x+8)+15.(x+8)](2x 2+35x+120)
=(x+8)(2x+15)(2x 2+35x+120)
a, 3x(x^2-4)=0
+3x=0=>x=0
+x^2-4=0
=>x^2=4
=>x=+-2
c,x(x+2)-3x-6=0
x(x+2)-3(x+2)=0
(x+2)(x-3)=0
TH1 :x+2=0
x=-2
TH2 : x-3=0
x=3
câu b bạn chờ mình chúc nha
nhớ k cho mình
\(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
Đặt \(\left(x^2+3x+1\right)=a\), ta được:
\(a\left(a+1\right)-6\)\(=a^2+a-6\)\(=\left(a^2+3a\right)-\left(2a+6\right)\)\(=a\left(a+3\right)-2\left(a+3\right)\)
\(=\left(a+3\right)\left(a-2\right)\)
Thay \(a=\left(x^2+3x+1\right)\), ta được:
\(=\left(x^2+3x+1+3\right)\left(x^2+3x+1-2\right)\)
\(=\left(x^2+3x+4\right)\left(x^2+3x-1\right)\)
\(x.\left(x^2-4\right)-3x+6\)
\(=x.\left(x+2\right).\left(x-2\right)-3.\left(x-2\right)\)
\(=\left(x^2+2x\right).\left(x-2\right)-3.\left(x-2\right)\)
\(=\left(x-2\right).\left(x^2+2x-3\right)\)
\(=\left(x-2\right).\left(x^2-x+3x-3\right)\)
\(=\left(x-2\right).[x.\left(x-1\right)+3.\left(x-1\right)]\)
\(=\left(x-2\right).\left(x-1\right).\left(x+3\right)\)