Ta có : \(x\left(x-y\right)-y\left(x-y\right)=\frac{3}{10}-\frac{-3}{50}\)

\(\left(x-y\right)^2=\frac{9}{25}\)

\(\Rightarrow\orbr{\begin{cases}x-y=\frac{3}{5}\\x-y=\frac{-3}{5}\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=\frac{3}{10}:\frac{3}{5}=\frac{1}{2}\\x=\frac{3}{10}:\frac{-3}{5}=\frac{-1}{2}\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}y=\frac{-3}{50}:\frac{3}{5}=\frac{-1}{10}\\y=\frac{-3}{50}:\frac{-3}{5}=\frac{1}{10}\end{cases}}\)

Vậy \(x=\frac{1}{2};y=\frac{-1}{10}\)hoặc  \(x=\frac{-1}{2};y=\frac{1}{10}\)

A = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

A < \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

A < \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

A < 1 - \(\frac{1.}{100}\)

A < \(\frac{99}{100}< \frac{199}{100}\)

=> A < \(\frac{199}{100}\)

b,

S = \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{99}{10^2}\)

S = \(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{9.11}{10.10}\)

S = \(\frac{1.3.2.4.3.5.4.6.5.7...9.11}{2.2.3.3.4.4...10.10}\)

S = \(\frac{1.2.3^2.4^2.5^2...9^2.10.11}{2^2.3^3.4^2...10^2}\)

S = \(\frac{1.11}{2.10}\)

S = \(\frac{11}{20}\)

Ta có: \(\frac{1}{2}A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{100}{2^{101}}\)

\(A-\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}-\frac{100}{2^{101}}\)

Ta có: \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{100}}=1-\frac{1}{2^{100}}< 1\)

\(\Rightarrow\frac{1}{2}A< 1-\frac{100}{2^{101}}\)

\(\Rightarrow A< 2-\frac{200}{2^{101}}< 2\)

Vậy A<2

Ta có 

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

..............

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

=> S < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

S < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(S< 1-\dfrac{1}{100}< 1\)(do 1/100 >0)

ĐPcm

Giải:

\(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\) 

Ta có:

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\) 

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\) 

\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\) 

\(...\) 

\(\dfrac{1}{99^2}=\dfrac{1}{99.99}< \dfrac{1}{98.99}\) 

\(\dfrac{1}{100^2}=\dfrac{1}{100.100}< \dfrac{1}{99.100}\) 

\(\Rightarrow S< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\) 

\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\) 

\(\Rightarrow S< \dfrac{1}{1}-\dfrac{1}{100}< 1\) 

\(\Rightarrow S< 1\) 

Vậy S < 1.

Ta có 

\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{99}{2^{99}}+\frac{100}{2^{100}}\)

\(2A=1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{99}{2^{98}}+\frac{100}{2^{99}}\)

Suy ra \(A=2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\frac{100}{2^{100}}\)

Đặt \(n=\frac{1}{2}\) thì \(A=1+n+n^2+...+n^{99}-\frac{100}{2^{100}}\)

Xét \(B=1+n+n^2+...+n^{99}\Leftrightarrow B.n=n+n^2+n^3+...+n^{100}\)

\(\Leftrightarrow B.n=\left(1+n+n^2+...+n^{99}\right)+\left(n^{100}-1\right)\)

\(\Leftrightarrow B.n=B+n^{100}-1\Leftrightarrow B\left(n-1\right)=n^{100}-1\Leftrightarrow B=\frac{n^{100}-1}{n-1}\)

Suy ra \(A=\frac{\frac{1}{2^{100}}-1}{\frac{1}{2}-1}-\frac{100}{2^{100}}=2\left(1-\frac{1}{2^{100}}\right)-\frac{100}{2^{100}}=-\frac{102}{2^{100}}+2< 2\)

Vậy A < 2

Lời giải:

$A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}$

$3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}$

$\Rightarrow 3A-A=1-\frac{1}{3^{100}}$

$\Rightarrow 2A=1-\frac{1}{3^{100}}<1$

$\Rightarrow A< \frac{1}{2}$

$\Rightarrow A< B$

Ta có :

\(\frac{a}{b}=\frac{c}{d}=\frac{2c}{2d}=\frac{a+c}{b+d}=\frac{a+2c}{b+2d}\)(Áp dụng tính chất dãy tỉ số bằng nhau)

\(\Leftrightarrow\left(a+2c\right)\left(b+d\right)=\left(a+c\right)\left(b+2d\right)\)(Nhân chéo)

(Đpcm)

Từ \(\left(a+2c\right)\left(b+d\right)=\left(a+c\right)\left(b+2d\right)\)

\(\Rightarrow\frac{a+2c}{b+2d}=\frac{a+c}{b+d}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}\Rightarrow\frac{a+2c}{b+2d}=\frac{a+c}{b+d}\Leftrightarrow\frac{bk+2dk}{b+2d}=\frac{bk+dk}{b+d}}\)

Xét VT \(\frac{bk+2dk}{b+2d}=\frac{k\left(b+2d\right)}{b+2d}=k\left(1\right)\)

Xét VP \(\frac{bk+dk}{b+d}=\frac{k\left(b+d\right)}{b+d}=k\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left(a+2c\right)\left(b+d\right)=\left(a+c\right)\left(b+2d\right)\left(đpcm\right)\)

Chúc bạn học tốt !!!