1)
\(x^3+2x^2+x\\ =x^3+x^2+x^2+x\\ =x^2\left(x+1\right)+x\left(x+1\right)\\ =\left(x+1\right)\left(x^2+x\right)\\ =x\left(x+1\right)\left(x+1\right)\)
2)
\(x^4-4x^3+4x^2\\ =x^2\left(x^2-4x+4\right)\\ =x^2\left(x-2\right)^2\\ =x.x\left(x-2\right)\left(x-2\right)\)
3)
\(5x^3-10x^2+5x\\ =5x^3-5x^2-5x^2+5x\\ =5x^2\left(x-1\right)-5x\left(x-1\right)\\ =\left(x-1\right)\left(5x^2-5x\right)\\ =5x\left(x-1\right)\left(x-1\right)\)
4)
\(2x^3-12x^2+18x\\ =2x\left(x^2-6x+9\right)\\ =2x\left(x-3\right)^2\\ =2x\left(x-3\right)\left(x-3\right)\)
5)
\(8x^2y-8xy+2x\\ =2x\left(4xy-4y+1\right)\)
6)
\(5x^2y-35xy+60y\\ =5y\left(xy-7x+12\right)\)
1: =x(x^2+2x+1)
=x(x+1)^2
2: =x^2(x^2-4x+4)
=x^2(x-2)^2
3: =5x(x^2-2x+1)
=5x(x-1)^2
4: =2x(x^2-6x+9)
=2x(x-3)^2
5: =2x(4xy-4y+1)
6: \(=5y\left(xy-7y+12\right)\)
1) \(x^2-x\)
\(=x\left(x-1\right)\)
2) \(x^2-8x\)
\(=x\left(x-8\right)\)
3) \(x^2-12x\)
\(=x\left(x-12\right)\)
4) \(x^3-4x\)
\(=x\left(x^2-4\right)\)
5) \(3x^2-x\)
\(=x\left(3x-1\right)\)
6) \(2xy+y^2\)
\(=y\left(2x+y\right)\)
7) \(x^2-xy\)
\(=x\left(x-y\right)\)
8) \(x^3-x^2y\)
\(=x^2\left(x-y\right)\)
1: =x*x-x*1
=x(x-1)
2: =x*x-8*x
=x(x-8)
3: =x*x+12*x
=x(x+12)
4: =x(x^2-4)
=x(x-2)(x+2)
5: =x*3x-x*1
=x(3x-1)
6: =y*2x+y*y
=y(2x+y)
7: =x*x-x*y
=x(x-y)
8: =x^2*x-x^2*y
=x^2(x-y)
1. x² - 9 + 3x ( x - 3 ) 2. x⁴ + 5x² + 4x + 5 3. x⁴ + x² + 1 minh cam on
1: x^2-9+3x(x-3)
=(x-3)(x+3)+3x(x-3)
=(x-3)(x+3+3x)
=(4x+3)(x-3)
2: =x^4+5x^2+4x+5
=x^4+x^3+x^2-x^3-x^2-x+5x^2+5x+5
=(x^2+x+1)(x^2-x+5)
3: =x^4+2x^2+1-x^2
=(x^2+1)^2-x^2
=(x^2-x+1)(x^2+x+1)
1) \(x^2-9+3x\left(x-3\right)\)
\(=\left(x^2-9\right)+3x\left(x-3\right)\)
\(=\left(x+3\right)\left(x-3\right)+3x\left(x-3\right)\)
\(=\left(x-3\right)\left(x+3+3x\right)\)
\(=\left(x-3\right)\left(4x+3\right)\)
2) \(x^4+5x^2+4x+5\)
\(=x^4+x^3+x^2-x^3-x^2-x+5x^2+5x+5\)
\(=x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+5\left(x^2+x+1\right)\)
\(=\left(x^2-x+5\right)\left(x^2+x+1\right)\)
3) \(x^4+x^2+1\)
\(=x^4+x^3+x^2-x^3-x^2-x+x^2+x+1\)
\(=x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
a) Ta có:
\(\dfrac{1}{xy+x^2}=\dfrac{1}{x^2+xy}=\dfrac{1}{x\left(x+y\right)}\)
\(\dfrac{1}{x+y}=\dfrac{1\cdot x}{x\cdot\left(x+y\right)}=\dfrac{x}{x\left(x+y\right)}\)
b) Ta có:
\(\dfrac{1}{x^2+xy}=\dfrac{1}{x\left(x+y\right)}=\dfrac{1\cdot y}{xy\left(x+y\right)}=\dfrac{y}{xy\left(x+y\right)}\)
\(\dfrac{1}{xy+y^2}=\dfrac{1}{y\left(x+y\right)}=\dfrac{1\cdot x}{xy\left(x+y\right)}=\dfrac{x}{xy\left(x+y\right)}\)
c) Ta có:
\(\dfrac{2x}{x^2+4x+4}=\dfrac{2x}{x^2+2\cdot2\cdot x+2^2}=\dfrac{2x}{\left(x+2\right)^2}\)
\(\dfrac{x+1}{x+2}=\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x+2\right)}=\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)^2}\)
d: \(\dfrac{3x-24}{x+y}=\dfrac{x\left(3x-24\right)}{x\left(x+y\right)}=\dfrac{3x^2-24x}{x^2+xy}\)
\(\dfrac{xy+5}{x^2+xy}=\dfrac{xy+5}{x^2+xy}\)
e: \(\dfrac{2xy-3y^2}{x^2-3xy}=\dfrac{y\left(2x-3y\right)}{x\left(x-3y\right)}=\dfrac{3y\left(2x-3y\right)}{3x\left(x-3y\right)}\)
\(\dfrac{x}{3x-9y}=\dfrac{x}{3\left(x-3y\right)}=\dfrac{x^2}{3x\left(x-3y\right)}\)
f: \(\dfrac{1}{x-2}=\dfrac{x+1}{\left(x-2\right)\left(x+1\right)}\)
\(\dfrac{3}{x+1}=\dfrac{3\left(x-2\right)}{\left(x-2\right)\left(x+1\right)}=\dfrac{3x-6}{\left(x-2\right)\left(x+1\right)}\)
a) \(\dfrac{25x^3y^5z}{5x^2yz}\)
\(=\dfrac{5xy^4}{1}\)
\(=5xy^4\)
b) \(\dfrac{x^2-y^2}{x-y}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)}{x-y}\)
\(=x+y\)
c) \(\dfrac{x^3+1}{x+1}\)
\(=\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{x+1}\)
\(=x^2-x+1\)
d) \(\dfrac{15xy}{10y}\)
\(=\dfrac{3x}{2}\)
e) \(\dfrac{x^2+2x+1}{x^2-1}\)
\(=\dfrac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{x+1}{x-1}\)
f) \(\dfrac{x+1}{x^2+x}\)
\(=\dfrac{x+1}{x\left(x+1\right)}\)
\(=\dfrac{1}{x}\)
g) \(\dfrac{x^2+6x+9}{x^2-9}\)
\(=\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{x+3}{x-3}\)
h) \(\dfrac{x-4}{x^2-16}\)
\(=\dfrac{x-4}{\left(x-4\right)\left(x+4\right)}\)
\(=\dfrac{1}{x+4}\)
Bạn có thể phóng to lên không chứ ảnh mờ quá !
Cho a,b,c # 0 và a+b+c#0 thỏa mãn 1/a+1/b+1/c=1/a+b+c cmr 1/a^2017+1/b^2017+1/c^2017=1/a^2017+b^2017+c^2017
Lời giải:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$
$\Leftrightarrow (\frac{1}{a}+\frac{1}{b})+(\frac{1}{c}-\frac{1}{a+b+c})=0$
$\Leftrightarrow \frac{a+b}{ab}+\frac{a+b}{c(a+b+c)}=0$
$\Leftrightarrow (a+b)(\frac{1}{ab}+\frac{1}{c(a+b+c)})=0$
$\Leftrightarrow (a+b).\frac{ab+c(a+b+c)}{abc(a+b+c)}=0$
$\Leftrightarrow \frac{(a+b)(c+a)(c+b)}{abc(a+b+c)}=0$
$\Leftrightarrow (a+b)(c+a)(c+b)=0$
$\Leftrightarrow a+b=0$ hoặc $c+a=0$ hoặc $c+b=0$
Không mất tổng quát giả sử $a+b=0$
$\Leftrightarrow a=-b$.
Khi đó:
$\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{(-b)^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}$
$=\frac{-1}{b^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}$
$=\frac{1}{c^{2017}}=\frac{1}{(-b)^{2017}+b^{2017}+c^{2017}}$
$=\frac{1}{a^{2017}+b^{2017}+c^{2017}}$ (đpcm)
Lần sau bạn lưu ghi đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt nhất. Mọi người đọc đề của bạn dễ hiểu thì cũng sẽ dễ giúp hơn.
\(^{n^3}\)-3 chia hết cho n-3 tìm n
=>n^3-27+24 chia hết cho n-3
=>n-3 thuộc {1;-1;2;-2;3;-3;4;-4;6;-6;8;-8;12;-12;24;-24}
=>n thuộc {4;2;5;1;6;0;7;-1;9;-3;11;-5;15;-9;27;-21}
cho a b c d thỏa mãn a/a+b + b/b+c + c/c+d + d/d+a thuộc Z chứng minh rằng a+b+c+d là hợp số