Tính nhanh:
(x^3-x^2-5x-3):(x+1)
K
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Những câu hỏi liên quan
H
3
22 tháng 7 2017
1) \(2x^2+5x-3=0\)
\(\Leftrightarrow2x^2+6x-x-3=0\)
\(\Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{2}\end{cases}}}\)
22 tháng 7 2017
\(2x^2+5x-3=0\)
\(\Leftrightarrow2x^2+2x+3x-3=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)
NT
1
14 tháng 9 2015
Tính tổng: 1x2 + 2x3 + 3x4 + 4x5 +.............+ 99x100
Gọi biểu thức trên là A, ta có :
A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
3A= 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
3A = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
3A = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
3A = 99x100x101
A = 99x100x101 : 3
A = 333300
18 tháng 8 2019
a,5x^2 - 10xy + 5y^2 - 20z^2
=5(x^2 -2xy +y^2-4z^2 )
=5[(x-y)^2-(2z)^2 ]
=5 .(x-y-2z)(x-y+2z)
b,.= (5x^2+5xy)-(x+y)
=5x(x+y)-(x+y)
=(x+y)(5x-1)
d,x2 - 4x + 3 = x2 - x - 3x + 3
= x(x - 1) - 3(x - 1) = (x -1)(x - 3)
e,x2 - x - 6 = x2 +2x - 3x - 6
= x(x + 2) - 3(x + 2)
= (x + 2)(x - 3)
f,x2 - x - 6 = x2 +2x - 3x - 6
= x(x + 2) - 3(x + 2)
= (x + 2)(x - 3)
g,2x^2(3x - 5)
= 2x^2 x 3x - 2x^2 x 5
= 6x^3 - 10x^2
18 tháng 8 2019
\(\text{1) }\)
\(\text{a) }5x^2-10xy+5y^2-20z^2\)
\(=5\left(x^2-2xy+y^2-4z^2\right)\)
\(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)
\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=5\left(x-y+2z\right)\left(x-y-2z\right)\)
\(\text{b) }5x^2+5xy-x-y\)
\(=\left(5x^2-x\right)+\left(5xy-y\right)\)
\(=x\left(5x-1\right)+y\left(5x-1\right)\)
\(=\left(5x-1\right)\left(x+y\right)\)
\(\text{c) }2\left(x+4\right)-x^2+16\)
\(=2\left(x+4\right)-\left(x^2-16\right)\)
\(=2\left(x+4\right)-\left(x+4\right)\left(x-4\right)\)
\(=\left(x+4\right)\left(2-x+4\right)\)
\(=\left(x+4\right)\left(6-x\right)\)
\(\text{d) }x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=\left(x^2+3x\right)+\left(x+3\right)\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+3\right)\left(x+1\right)\)
\(\text{e) }x^2+5x-6\)
\(=x^2+6x-x-6\)
\(=\left(x^2+6x\right)-\left(x+6\right)\)
\(=x\left(x+6\right)-\left(x+6\right)\)
\(=\left(x+6\right)\left(x-1\right)\)
T
10 tháng 5 2018
\(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{5.6}\\ \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\\ \)
\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{6}\)
\(\frac{1}{2}+\frac{1}{2}-\frac{1}{6}\)
\(1-\frac{1}{6}\\ \frac{5}{6}\)
k nha bn
Ta có \(x^3-x^2-5x-3=x^3+x^2-2x^2-2x-3x-3\)
\(=x^2.\left(x+1\right)-2x.\left(x+1\right)-3.\left(x+1\right)\)
\(=\left(x+1\right).\left(x^2-2x-3\right)\)
Do đó \(\left(x^3-x^2-5x-3\right):\left(x+1\right)=\left(x+1\right).\left(x^2-2x-3\right):\left(x+1\right)\)
\(=x^2-2x-3\)