phân tích đa thức thành nhân tử: x(x^2-4)-3x+6
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7 tháng 10 2021
a: \(x^4+x^2+2x+6\)
\(=x^4-2x^3+3x^2+2x^3-4x^2+6x+2x^2-4x+6\)
\(=\left(x^2-2x+3\right)\left(x^2+2x+2\right)\)
LT
0
KL
1
DD
Đoàn Đức Hà
Giáo viên
7 tháng 7 2021
\(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)-3x^2\)
\(=4\left[\left(x+5\right)\left(x+12\right)\right]\left[\left(x+6\right)\left(x+10\right)\right]-3x^2\)
\(=4\left(x^2+17x+60\right)\left(x^2+16x+60\right)-3x^2\)
\(=\left(2x^2+34x+120\right)\left(2x^2+32x+60\right)-3x^2\)
\(=\left(2x^2+33x+120\right)^2-x^2-3x^2\)
\(=\left(2x^2+33x+120-2x\right)\left(2x^2+33x+120+2x\right)\)
\(=\left(2x+15\right)\left(x+8\right)\left(2x^2+35x+120\right)\)
TN
1
ML
10 tháng 7 2015
\(4\left(x+5\right)\left(x+12\right)\left(x+6\right)\left(x+10\right)-3x^2\)
\(=2\left(x^2+60+17x\right).2\left(x^2+60+16x\right)-3x^2\)
\(=\left(2x^2+120+33x+x\right)\left(2x^2+120+33x-x\right)-3x^2\)
\(=\left(2x^2+120+33x\right)^2-x^2-3x^2\)
\(=\left(2x^2+120+33x\right)^2-4x^2\)
\(=\left(2x^2+120+33x+2x\right)\left(2x^2+120+33x-2x\right)\)
\(=\left(2x^2+35x+120\right)\left(2x^2+31x+120\right)\)
\(=\left(2x^2+35x+120\right)\left(x+8\right)\left(2x+15\right)\)
6 tháng 11 2021
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
22 tháng 11 2023
a: \(2y\left(x+2\right)-3x-6\)
\(=2y\left(x+2\right)-3\left(x+2\right)\)
\(=\left(x+2\right)\left(2y-3\right)\)
b: \(3\left(x+4\right)-x^2-4x\)
\(=3\left(x+4\right)-\left(x^2+4x\right)\)
\(=3\left(x+4\right)-x\left(x+4\right)\)
\(=\left(x+4\right)\left(3-x\right)\)
c: \(2\left(x+5\right)-x^2-4x\)
\(=2x+10-x^2-4x\)
\(=-x^2-2x+10\)
\(=-x^2-2x-1+11\)
\(=11-\left(x^2+2x+1\right)\)
\(=11-\left(x+1\right)^2\)
\(=\left(\sqrt{11}-x-1\right)\left(\sqrt{11}+x+1\right)\)
d: \(x^2+6x-3x-18\)
\(=\left(x^2+6x\right)-\left(3x+18\right)\)
\(=x\left(x+6\right)-3\left(x+6\right)\)
\(=\left(x+6\right)\left(x-3\right)\)
PL
2
T
25 tháng 8 2023
\(3x\cdot\left(x-y\right)^2-6\cdot\left(y-x\right)\)
\(=3x\left(x-y\right)^2+6\left(x-y\right)\)
\(=\left(x-y\right)\left[3x\left(x-y\right)+6\right]\)
\(=\left(x-y\right)\left(3x^2-3xy+6\right)\)
TQ
0
HN
2
NN
31 tháng 10 2020
Đặt \(x^2+3x+1=t\)
\(\Rightarrow\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6=t.\left(t+1\right)-6\)
\(=t^2+t-6=\left(t^2-2t\right)+\left(3t-6\right)\)
\(=t\left(t-2\right)+3\left(t-2\right)=\left(t-2\right)\left(t+3\right)\)
\(=\left(x^2+3x+1-2\right)\left(x^2+3x+1+3\right)\)
\(=\left(x^2+3x-1\right)\left(x^2+3x+4\right)\)
G
31 tháng 10 2020
\(A=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
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 :
\(A=\left(x^2+3x+1+6\right)\left(x^2+3x+1-1\right)\)
\(=\left(x^2+3x+7\right)\left(x^2+3x\right)\)
\(x.\left(x^2-4\right)-3x+6\)
\(=x.\left(x+2\right).\left(x-2\right)-3.\left(x-2\right)\)
\(=\left(x^2+2x\right).\left(x-2\right)-3.\left(x-2\right)\)
\(=\left(x-2\right).\left(x^2+2x-3\right)\)
\(=\left(x-2\right).\left(x^2-x+3x-3\right)\)
\(=\left(x-2\right).[x.\left(x-1\right)+3.\left(x-1\right)]\)
\(=\left(x-2\right).\left(x-1\right).\left(x+3\right)\)