phân tích đa thức sau thành nhân tử : (a+5)2+36+12(a+5) 3
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\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(a^5+a^3-a^2-1=a^3\left(a^2+1\right)-\left(a^2+1\right)=\left(a^3-1\right)\left(a^2+1\right)\)
\(a^5+a^3-a^2-1=a^3\left(a^2+1\right)-\left(a^2+1\right)=\left(a^3-1\right)\left(a^2+1\right)=\left(a-1\right)\left(a^2+a+1\right)\left(a^2+1\right)\)
\(a^5+a^3-a^2-1\)
\(=a^3\left(a^2+1\right)-\left(a^2+1\right)\)
\(=\left(a^2+1\right)\left(a^3-1\right)\)
a) xy+3x-7y-21
=x(y+3)-7(x+3)
=(x-7)(y+3)
b)2xy-15-6x-5y
=2x(y-3)-5(-3+y)
=(2x-5)(y-3)
c)2x^2y+2xy^2-2x-2y
=2x(xy-1)+2y(xy-1)
=(2x+2y)(xy-1)
x(x+3)-5x(x-5)-5(x+3)
=(x-5)(x+3)-5x(x-5)
=(x-5)(x+3-5x)
Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn
Đặt \(M=\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)
\(M=\left[\left(a+1\right)\left(a+7\right)\right]\left[\left(a+3\right)\left(a+5\right)\right]+15\)
\(M=\left(a^2+7a+a+7\right)\left(a^2+5a+3a+15\right)+15\)
\(M=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)
Đặt \(p=a^2+8a+11\)
\(\Rightarrow M=\left(p-4\right)\left(p+4\right)+15\)
\(\Rightarrow M=p^2-16+15\)
\(\Rightarrow M=p^2-1\)
\(\Rightarrow M=\left(p-1\right)\left(p+1\right)\)
Thay \(p=a^2+8a+11\)vào M, ta có :
\(M=\left(a^2+8a+11-1\right)\left(a^2+8a+11+1\right)\)
\(M=\left(a^2+8a+10\right)\left(a^2+8a+12\right)\)
a)\(x^2+7x+12\)
\(=x^2+x+6x+6\)
\(=x\left(x+1\right)+6\left(x+1\right)\)
\(=\left(x+1\right)\left(x+6\right)\)
A=( a +1)(a+3)(a+5)(a+7)+15
=(a+1)(a+7)(a+3)(a+5)+15
=(a2+8a+7)(a2+8a+15)+15
Đặt y=a2+8a+7 ta được :
y(y+8)+15=y2 + 8y +15
=y2 +3y+5y+15
=y(y+3) +5(y+3)
=(y+3)(y+5)
thay y=a2+8a+7 ta được
(a2+8a+7+3)(a2+8a+7+5)
=(a2+8a+10)(a2-2a-6a+12)
=(a2+8a+10)[a(a-2)-6(a-2)]
=(a2+8a+10)(a-2)(a-6)
=(a+5)2.(13a+65)+36