Phân tích thành nhân tử:
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+12\right)-165x^2\)
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H9
HT.Phong (9A5)
CTVHS
23 tháng 7 2023
\(S=x^6-8\)
\(S=\left(x^2\right)^3-2^3\)
\(S=\left(x^2-2\right)\left(x^4+2x^2+4\right)\)
⇒ Chọn C
MH
17 tháng 10 2023
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)-24\)
\(=\left(x^2+7x+11\right)^2-25\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
T
17 tháng 10 2023
\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)
Đặt \(y=x^2+7x+10\), khi đó biểu thức trở thành:
\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(x^2+7x+10+1-5)(x^2+7x+10+1+5)\\=(x^2+7x+6)(x^2+7x+16)\\=(x^2+x+6x+6)(x^2+7x+16)\\=[x(x+1)+6(x+1)](x^2+7x+16)\\=(x+1)(x+6)(x^2+7x+16)\\Toru\)
AH
Akai Haruma
Giáo viên
29 tháng 1 2023
Bạn nên tách bài ra để đăng. Không nên đăng 1 loạt như thế này.
29 tháng 1 2023
1: \(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)
=(x^2+x)^2+3(x^2+x)-10
=(x^2+x+5)(x^2+x-2)
=(x^2+x+5)(x+2)(x-1)
2: \(=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\)
\(=\left(x^2+5ax\right)^2+10a^2\left(x^2+5ax\right)+25a^2\)
\(=\left(x^2+5ax+5a^2\right)^2\)
3: \(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
5: \(M=\left(n+1\right)\left(n^2+2n\right)+360\)
=n(n+1)(n+2)+360 chia hết cho 6
6A
7D
PC
1
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 1 2019
\(\left(x+2\right)\left(x+12\right)\left(x+4\right)\left(x+6\right)=165x^2\)
\(\Leftrightarrow\left(x^2+14x+24\right)\left(x^2+10x+24\right)=165x^2\)
Nhận thấy \(x=0\) không phải nghiệm, pt đã cho tương đương:
\(\left(x+14+\dfrac{24}{x}\right)\left(x+10+\dfrac{24}{x}\right)=165\)
Đặt \(x+\dfrac{24}{x}+10=t\):
\(t\left(t+4\right)-165=0\Leftrightarrow t^2+4t-165=0\Rightarrow\left[{}\begin{matrix}t=11\\t=-15\end{matrix}\right.\)
TH1: \(x+\dfrac{24}{x}+10=11\Leftrightarrow x^2-x+24=0\) (vô nghiệm)
TH2: \(x+\dfrac{24}{x}+10=-15\Leftrightarrow x^2+25x+24=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-24\end{matrix}\right.\)
2 tháng 8 2016
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)
\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10+16+8\right)+16\)
\(=\left(x^2+10x+16\right)^2+2.\left(x^2+10x+16\right).4+4^2\)
\(=\left(x^2+10x+16+4\right)^2\)
\(=\left(x^2+10+20\right)^2\)
2 tháng 8 2016
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+16\)
\(=\left(x^2+8x+2x+16\right)
\left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\left(1\right)\)
\(\text{Đặt }x^2+10x+\frac{16+24}{2}=t\)
\(\text{hay }x^2+10x+20=t\)
\(\left(1\right)\Rightarrow\left(t-4\right)\left(t+4\right)+16\)
\(=t^2-4^2+16\)
\(=t^2-16+16\)
\(=t^2\)
\(=\left(x^2+10x+20\right)^2\)
24 tháng 9 2019
\(\left(x-2\right)\left(x-4\right)\left(x-6\right)\left(x-8\right)+16\)
\(=\left[\left(x-2\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x-6\right)\right]+16\)
\(=\left(x^2-10x+16\right)\left(x^2-10x+24\right)+16\)(1)
Đặt \(x^2-10x+20=t\)thay vào (1) ta được :
\(\left(t-4\right)\left(t+4\right)+16\)
\(=t^2-16+16\)
\(=t^2\)Thay \(t=x^2-10x+20\)ta được :
\(\left(x^2-10x+20\right)^2\)
\(=\left(x^2-2.5.x+25-25+20\right)^2\)
\(=\left[\left(x-5\right)^2-5\right]^2\)
\(=\left(x-5-\sqrt{5}\right)^2\left(x-5+\sqrt{5}\right)^2\)
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+12\right)-165x^2\)
\(=\left[\left(x+2\right)\left(x+12\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]-165x^2\)
\(=\left(x^2+14x+24\right)\left(x^2+10x+24\right)-165x^2\)
\(=\left(x^2+12x+24+2x\right)\left(x^2+12x+24-2x\right)-165x^2\)
\(=\left(x^2+12x+24\right)^2-4x^2-165x^2\)
\(=\left(x^2+12x+24\right)^2-169x^2\)
\(=\left(x^2+12x+24-13x\right)\left(x^2+12x+24+13x\right)\)
\(=\left(x^2-x+24\right)\left(x^2+25x+24\right)\)
\(=\left(x^2-x+24\right)\left(x^2+x+24x+24\right)\)
\(=\left(x^2-x+24\right)\left[x\left(x+1\right)+24\left(x+1\right)\right]\)
\(=\left(x^2-x+24\right)\left(x+1\right)\left(x+24\right)\)