\(\frac{1+2+3+4+...+8+9}{11+12+...+18+19}\)

\(\Rightarrow\frac{\left(1+9\right).9:2}{\left(11+9\right).9:2}=\frac{1+9}{11+19}=\frac{10}{30}=\frac{1}{3}\)

a, 8

b, 43046721( hình như đề sai phải bạn )

c, 43046721

d, ......

Dùng latex

\(\dfrac{9^{14}\cdot25^5\cdot8^7}{18^2\cdot625^3\cdot24^3}\)

\(=\dfrac{3^{42}\cdot5^{10}\cdot2^{21}}{2^2\cdot3^4\cdot5^{12}\cdot2^9\cdot3^3}=\dfrac{3^{42}\cdot5^{10}\cdot2^{21}}{2^{11}\cdot3^7\cdot5^{12}}\)

\(=\dfrac{3^{35}}{5^2}\cdot2^{10}\)

a) \(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y\cdot2x=4xy\)

b) \(\left(a+b\right)^3+\left(a-b\right)^3-2a^3\)

\(=a^3+3a^2b+3ab^2+b^3+a^2-3a^2b+3ab^2-b^3-2a^3\)

\(=6ab^2\)

c) \(9^8\cdot2^8-\left(18^4-1\right)\left(18^4+1\right)\)

\(=18^8-\left(18^8-1\right)=1\)

a) \(\left(x+y\right)^2-\left(x-y\right)^2=x^2+2xy+y^2-\left(x^2-2xy+y^2\right)\)

\(=x^2+2xy+y^2-x^2+2xy-y^2\)

\(=\left(x^2-x^2\right)+\left(y^2-y^2\right)+\left(2xy+2xy\right)\)

\(=4xy\)

a) (x+y)²-(x-y)²

=(x+y+x-y)(x+y-x+y)

=2x.2y=4xy

b) (a+b)³+(a-b)³-2a³

=(a+b+a-b)(a²+2ab+b²-a²+b²+a²-2ab+b²)-2a³

=2a.(a²+3b²)-2a³

=2a³+6ab²-2a³

=6ab²

\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)

\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)

\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)

\(=6x^2y\)

\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)

\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)

\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)

1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy

2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3

=6x^2y

3: =(x+y-x+y)^2=(2y)^2=4y^2

4: =(2x+3-2x-5)^2=(-2)^2=4

5: =18^8-18^8+1=1

a) Ta có: \(P=\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)

\(=\dfrac{x+3+\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)

\(=\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

b) Ta có: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

\(=5+\sqrt{2}-4-\sqrt{2}\)

=1

Thay x=1 vào P, ta được:

\(P=\dfrac{1+1}{1+3}=\dfrac{2}{4}=\dfrac{1}{2}\)