\(\frac{5.4^{15}.9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)

= 1

đúng không ?

\(Q=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\)

\(\frac{1}{1\times10}+\frac{1}{2\times15}+\frac{1}{3\times20}+...+\frac{1}{98\times495}+\frac{1}{99\times500}\)

\(=\frac{1}{1\times2\times5}+\frac{1}{2\times3\times5}+\frac{1}{3\times4\times5}+...+\frac{1}{98\times99\times5}+\frac{1}{99\times100\times5}\)

\(=\frac{1}{5}\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\right)\)

\(=\frac{1}{5}\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(=\frac{1}{5}\times\left(1-\frac{1}{100}\right)=\frac{1}{5}\times\frac{99}{100}=\frac{99}{500}\)

\(\frac{1}{1\times10}+\frac{1}{2\times15}+\frac{1}{3\times20}+...+\frac{1}{98\times495}+\frac{1}{99\times500}\)

\(=\frac{1}{1\times2\times5}+\frac{1}{2\times3\times5}+\frac{1}{3\times4\times5}+...+\frac{1}{98\times90\times5}+\frac{1}{90\times100\times5}\)

\(=\frac{1}{5}\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\right)\)

\(=\frac{1}{5}\times\left(\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+...+\frac{99-98}{98\times99}+\frac{100-99}{99\times100}\right)\)

\(=\frac{1}{5}\times\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(=\frac{1}{5}\times\left(1-\frac{1}{100}\right)=\frac{99}{500}\)

cảm ơn bạn Xyz và bạn Đoàn Đức Hà nhìu (^-^) thank you !!!!!!!!!!!!!!!!!

\(4x^2-28x+49=\left(2x\right)^2-2\cdot2x\cdot7+7^2=\left(2x-7\right)^2\)

Khi x=4 thì \(4x^2-28x+49=\left(2x-7\right)^2=\left(2\cdot4-7\right)^2=1\)

Ta có : 

\(A=\frac{11+12+13+...+89}{21+22+23+...+99}\)

+) Xét dãy \(11;12;13;...;89\) : 

Số số hạng là : \(89-11+1=79\)

Tổng là : \(\frac{\left(89+11\right).79}{2}=3950\)\(\left(1\right)\)

+) Xét dãy \(21;22;23;...;99\) : 

Số số hạng là : \(99-21+1=79\)

Tổng là : \(\frac{\left(99+21\right).79}{2}=4740\)\(\left(2\right)\)

Từ (1) và (2) suy ra : \(A=\frac{11+12+13+...+89}{21+22+23+...+99}=\frac{3950}{4740}=\frac{5}{6}\)

Vậy \(A=\frac{5}{6}\)

Chúc bạn học tốt ~ 

Tính tử số:

Số số hạng:(89-11):1+1=79

Tổng:(89+11)x79:2=3950

Tính mẫu số:

Số số hạng:(99-21):1+1=79

Tổng:(99+21)x79:2=4740

\(\Rightarrow A=\frac{11+12+13+...+89}{21+22+23+...+99}=\frac{3950}{4740}=\frac{5}{6}\)

T*** mik nha!

\(n\in\left\{2;4;6;8;10;12\right\}\)

:))))

\(\text{Đ}k:n\in Z^+\)

để \(\dfrac{n}{14}\) có thể rút gonj và \(n< 14\\ \Rightarrow n;14⋮cho1s\text{ố }\\ ta.th\text{ấy}:14⋮1;2;7;14\\ \Rightarrow n\in\left\{2;7;14\right\}\)( do \(n\in Z^+\) ; \(\dfrac{1}{14}\) đã tối giản )

nhưng n <14 \(\Rightarrow n\ne14\)

\(+,n=2\Rightarrow\dfrac{n}{14}=\dfrac{2}{14}=\dfrac{1}{7}\\ +,n=7\Rightarrow\dfrac{n}{14}=\dfrac{7}{14}=\dfrac{1}{2}\)

vậy \(n\in\left\{2;7\right\}\)

a) ĐKXĐ: 

\(\left\{{}\begin{matrix}x^2-9\ne0\\x+3\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne-3\end{matrix}\right.\Leftrightarrow x\ne\pm3\) 

b) \(A=\dfrac{x+15}{x^2-9}-\dfrac{2}{x+3}\)

\(A=\dfrac{x+15}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{x+15-2x+6}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{21-x}{\left(x+3\right)\left(x-3\right)}\)

c) Thay x = - 1 vào A ta có: 

\(A=\dfrac{21-\left(-1\right)}{\left(-1+3\right)\left(-1-3\right)}=\dfrac{21+1}{2\cdot-4}=\dfrac{22}{-8}=-\dfrac{11}{4}\)