A = (1 + 31 + 32 + 33) + (34 + 35 +36 + 37) + ...+ (324 + 325 + 326 + 327) + (328 + 229 + 330)

A = (1 + 31 + 32 + 33) + 34.(1 + 31 + 32 + 33) + ...+ 324.(1 + 31 + 32 + 33) + (328 + 229 + 330)

A = 40 + 34.40 + ....+ 324.40 + (328 + 229 + 330)

A = 40.(1 + 34 + ...+ 324) + (328 + 229 + 330)

\(\dfrac{x}{y}=\dfrac{4}{5}\\ =>\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{2x+y}{2.4+5}=\dfrac{26}{13}=2\\ =>x=8;y=10\)

\(\dfrac{x}{3}=\dfrac{y}{5}=>\dfrac{x}{21}=\dfrac{y}{35};\dfrac{x}{7}=\dfrac{z}{4}=>\dfrac{x}{21}=\dfrac{z}{12}\\ =>\dfrac{x}{21}=\dfrac{y}{35}=\dfrac{z}{12}=\dfrac{3x-2y+z}{3.21-2.35+12}=\dfrac{5}{5}=1\\ =>x=21;y=35;z=12\)

a) \(A=\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

\(2A=\dfrac{2}{4^2}+\dfrac{2}{6^2}+...+\dfrac{2}{100^2}\)

\(< \dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\)

\(< \dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\dfrac{1}{2}-\dfrac{1}{100}< \dfrac{1}{2}\)

Suy ra \(A< \dfrac{1}{4}\)

Do đó \(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2^2}+\dfrac{1}{4}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\). 

b) \(\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+..+\dfrac{2499}{2500}\)

\(=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+1-\dfrac{1}{4^2}+...+1-\dfrac{1}{50^2}\)

\(=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)\)

\(>49-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)\)

\(=49-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

\(=49-\left(1-\dfrac{1}{50}\right)=48+\dfrac{1}{50}>48\)

c) \(\dfrac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}=\dfrac{1+5\left(1+5+...+5^8\right)}{1+5+5^2+...+5^8}>5\)

\(\dfrac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}=\dfrac{1+3\left(1+3+...+3^8\right)}{1+3+3^2+...+3^8}=\dfrac{1}{1+3+3^2+...+3^8}+3< 4\)

Do đó ta có đpcm. 

\(x\) và \(y\) là đại lượng tỉ lệ thuận nên \(y=kx\left(k\ne0\right)\)

\(10=y_1+y_2=kx_1+kx_2=k\left(x_1+x_2\right)=5\)

suy ra \(k=2\).

Vậy \(y=2x\).

Đặt : `a/3=b/4=k`

`=>a=3k;b=4k`

Ta có :

`ab=48=>3k.4k=48`

`=>12k^2=48`

`=>k^2=4`

`=>k=2;k=-2`

Vậy `(a;b)=(6;8);(-6;-8)`

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