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DN
1
2 tháng 8 2017
\(x^3-x^2-21x+45\)
\(=\left(x^3-3x^2\right)+\left(2x^2-6x\right)+\left(-15x+45\right)\)
\(=x^2\left(x-3\right)+2x\left(x-3\right)-15\left(x-3\right)\)
\(=\left(x^2+2x-15\right)\left(x-3\right)\)
\(=\left[\left(x^2-3x\right)+\left(5x-15\right)\right]\left(x-3\right)\)
\(=\left[x\left(x-3\right)+5\left(x-3\right)\right]\left(x-3\right)\)
\(=\left(x+5\right)\left(x-3\right)^2\)
23 tháng 10 2021
Tham khảo:https://hoc247.net/hoi-dap/toan-8/phan-tich-da-thuc-x-7-x-2-1-thanh-nhan-tu-faq417522.html
23 tháng 10 2021
\(=x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2+x^2-x^2+x-x+1\\ =\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
G
10 tháng 8 2023
\(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)
\(=4a^2b^2-2ab\left(a^2+b^2-c^2\right)+2ab\left(a^2+b^2-c^2\right)-\left(a^2+b^2-c^2\right)^2\)
\(=2ab\left[2ab-\left(a^2+b^2-c^2\right)\right]+\left(a^2+b^2-c^2\right)\left[2ab-\left(a^2+b^2-c^2\right)\right]\)
\(=\left(2ab+a^2+b^2-c^2\right)\left(2ab-a^2-b^2+c^2\right)\)
\(=\left(a^2+ab+ab+b^2-c^2\right)\left[c^2-\left(a^2-ab-ab+b^2\right)\right]\)
\(=\left[a\left(a+b\right)+b\left(a+b\right)-c^2\right]\left[c^2-\left(a\left(a-b\right)-b\left(a-b\right)\right)\right]\)
\(=\left[\left(a+b\right)^2-c^2\right]\left[c^2-\left(a-b\right)^2\right]\)
\(=\left[\left(a+b\right)^2-c\left(a+b\right)+c\left(a+b\right)-c^2\right]\left[c^2+c\left(a-b\right)-c\left(a-b\right)-\left(a-b\right)^2\right]\)
\(=\left[\left(a+b\right)\left(a+b-c\right)+c\left(a+b-c\right)\right]\left[c\left(c+a-b\right)-\left(a-b\right)\left(c+a-b\right)\right]\)
\(=\left(a+b+c\right)\left(a+b-c\right)\left(c+a-b\right)\left(c-a+b\right)\)
28 tháng 12 2022
x^3-2x-4
=x^3-2x-8+4 (Ta thấy - 8 + 4 là bằng -4 nên ta thêm vào thì cũng giống nhau)
=(x^3-8)-(2x-4) (Nhóm hạng tử)
=(x-2)(x^2+2x+4)-2(x-2) \([\)(Hằng đẳng thức 6) và ta thấy -2 là nhân tử chung\(]\)
=(x-2)(x^2+2x+4-x+2) (Rút gọn)
=(x-2)(x^2+x+6)
17 tháng 11 2021
Đặt \(x^2+x+1=t\)
\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=t\left(t+1\right)-12=t^2+t-12=\left(t^2+t+\dfrac{1}{4}\right)-\dfrac{49}{4}=\left(t+\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2=\left(t+\dfrac{1}{2}-\dfrac{7}{2}\right)\left(t+\dfrac{1}{2}+\dfrac{7}{2}\right)=\left(t-3\right)\left(t+4\right)=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
17 tháng 11 2021
\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
= \(\left(x^2+x+1\right)\left[\left(x^2+x+1\right)+1\right]-12\)
= \(\left(x^2+x+1\right)^2\left(x^2+x+1\right)-12\)
= \(\left(x^2+x+1\right)\left(x^2+x+1\right)-3\left(x^2+x+1\right)+4\left(x^2+x+1\right)-4.3\)
= \(\left(x^2+x+1\right)\left(x^2+x-2\right)+4\left(x^2+x-2\right)\)
= \(\left(x^2+x+5\right)\left(x^2+x-2\right)\)
6 tháng 7 2017
Ta có \(x^4+4=\left(x^2\right)^2+2^2=\left(x^2+2\right)^2-2.x^2.2=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\)
23 tháng 11 2021
\(ax-2x-a^2+2a=x\left(a-2\right)-a\left(a-2\right)=\left(a-2\right)\left(x-a\right)\)
Bài 6:
a: \(A=n^2\left(n-1\right)+2n\left(1-n\right)\)
\(=n^2\left(n-1\right)-2n\left(n-1\right)\)
\(=\left(n-1\right)\left(n^2-2n\right)=n\left(n-1\right)\left(n-2\right)\)
Vì n;n-1;n-2 là ba số nguyên liên tiếp
nên n(n-1)(n-2)⋮3!
=>n(n-1)(n-2)⋮6
=>A⋮6
b: \(M=\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=\left(12x^2+12x-x-1\right)\left(12x^2+8x+3x+2\right)-4\)
\(=\left(12x^2+11x-1\right)\left(12x^2+11x+2\right)-4\)
\(=\left(12x^2+11x\right)^2+2\left(12x^2+11x\right)-\left(12x^2+11x\right)-2-4\)
\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-6\)
\(=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)
Bài 4:
a: \(A=x\left(x-y\right)^2-y\left(x-y\right)^2+xy^2-x^2y\)
\(=\left(x-y\right)^2\cdot\left(x-y\right)+xy\left(y-x\right)\)
\(=\left(x-y\right)^3-xy\left(x-y\right)\)
Khi x-y=5 và xy=4 thì \(A=5^3-4\cdot5=125-20=105\)
b: \(B=65^2-35^2+83^2-17^2\)
\(=\left(65-35\right)\left(65+35\right)+\left(83-17\right)\left(83+17\right)\)
\(=100\cdot30+100\cdot66=100\cdot96=9600\)
Bài 3:
a: \(4x\cdot\left(x+3\right)-x-3=0\)
=>4x(x+3)-(x+3)=0
=>(x+3)(4x-1)=0
=>\(\left[\begin{array}{l}x+3=0\\ 4x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-3\\ x=\frac14\end{array}\right.\)
b: \(x^2+4x=0\)
=>x(x+4)=0
=>\(\left[\begin{array}{l}x=0\\ x+4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-4\end{array}\right.\)
c: \(9x^2-\left(2x-1\right)^2=0\)
=>\(\left(3x\right)^2-\left(2x-1\right)^2=0\)
=>(3x-2x+1)(3x+2x-1)=0
=>(x+1)(5x-1)=0
=>\(\left[\begin{array}{l}x+1=0\\ 5x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-1\\ x=\frac15\end{array}\right.\)
d: \(\left(x^3-1\right)-\left(x-1\right)\left(x^2-5\right)=0\)
=>\(\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-5\right)=0\)
=>\(\left(x-1\right)\left(x^2+x+1-x^2+5\right)=0\)
=>(x-1)(x+6)=0
=>\(\left[\begin{array}{l}x-1=0\\ x+6=0\end{array}\right.=>\left[\begin{array}{l}x=1\\ x=-6\end{array}\right.\)