Chú ý: \(a^2-1=\left(a-1\right)\left(a+1\right)\)

Áp dụng:

\(A=\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}...\frac{49.51}{50^2}=\frac{2.3.4^2.5^2...49^2.50.51}{3^2.4^2.5^2...50^2}=\frac{2.51}{3.50}=\frac{51}{75}\)

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Yêu cầu của bài là gì vậy. Tính A? hay Chứng minh A < 2 hoặc chứng minh A không phải là số nguyên

Chứng minh A < 2

\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)

\(< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)

\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=2-\frac{1}{50}< 2\)

Vậy A < 2

1² + 2² + 3² + .... + 50²

= 1(2 - 1) + 2.(3 - 1) + 3.(4 - 1) + ..... + 50(51 - 1)

= 1.2 - 1 + 2.3 - 2 + 3.4 - 3 + ..... + 50.51 - 50

= (1.2 + 2.3 + ... + 50.51) - (1 + 2 + ... + 50)

\(=\frac{50.51.52}{3}-\frac{50.51}{2}\)

\(=44200-1275\)

= 42925

\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)

\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=2-\frac{1}{50}< 2\)

\(\Rightarrow A< 2\)

2B= 2 + 2²+.......+2⁵⁰+2⁵¹

2B - B= (2+2²+.......+2⁵¹) - (1+2+......+2⁵⁰)

B=2⁵¹-1

ĐÚNG KO BẠN

a) Ta có: \(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)

\(\Leftrightarrow x\cdot\frac{2}{3}=\frac{1}{10}+\frac{1}{2}=\frac{6}{10}\)

hay \(x=\frac{6}{10}:\frac{2}{3}=\frac{6}{10}\cdot\frac{3}{2}=\frac{18}{20}=\frac{9}{10}\)

Vậy: \(x=\frac{9}{10}\)

b) Ta có: \(5\frac{4}{7}:x=13\)

\(\Leftrightarrow\frac{39}{7}:x=13\)

\(\Leftrightarrow x=\frac{39}{7}:13=\frac{39}{7}\cdot\frac{1}{13}=\frac{3}{7}\)

Vậy: \(x=\frac{3}{7}\)

c) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)

\(\Leftrightarrow\frac{14}{5}x-50=51\cdot\frac{2}{3}=34\)

\(\Leftrightarrow x\cdot\frac{14}{5}=84\)

\(\Leftrightarrow x=84:\frac{14}{5}=84\cdot\frac{5}{14}=\frac{420}{14}=30\)

Vậy: x=30

d) Ta có: \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)

\(\Leftrightarrow\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}=\frac{-1}{15}\)

hay \(x=\frac{1}{3}:\frac{-1}{15}=\frac{1}{3}\cdot\left(-15\right)=\frac{-15}{3}=-5\)

Vậy: x=-5

e) Ta có: \(8\frac{2}{3}:x-10=-8\)

\(\Leftrightarrow\frac{26}{3}:x=2\)

hay \(x=\frac{26}{3}:2=\frac{26}{3}\cdot\frac{1}{2}=\frac{26}{6}=\frac{13}{3}\)

Vậy: \(x=\frac{13}{3}\)

g) Ta có: \(x+30\%=-1.3\)

\(\Leftrightarrow x+\frac{3}{10}=\frac{-13}{10}\)

hay \(x=\frac{-13}{10}-\frac{3}{10}=\frac{-16}{10}=\frac{-8}{5}\)

Vậy: \(x=\frac{-8}{5}\)

i) Ta có: \(3\frac{1}{3}x+16\frac{3}{4}=-13.25\)

\(\Leftrightarrow x\cdot\frac{10}{3}+\frac{67}{4}=-\frac{53}{4}\)

\(\Leftrightarrow x\cdot\frac{10}{3}=\frac{-53}{4}-\frac{67}{4}=-30\)

\(\Leftrightarrow x=-30:\frac{10}{3}=-30\cdot\frac{3}{10}=\frac{-90}{10}=-9\)

Vậy: x=-9

k) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)

\(\Leftrightarrow x\cdot\frac{14}{5}-50=51\cdot\frac{2}{3}=34\)

\(\Leftrightarrow x\cdot\frac{14}{5}=34+50=84\)

hay \(x=84:\frac{14}{5}=84\cdot\frac{5}{14}=30\)

Vậy: x=30

m) Ta có: \(\left|2x-1\right|=\left(-4\right)^2\)

\(\Leftrightarrow\left|2x-1\right|=16\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=16\\2x-1=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=17\\2x=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{17}{2}\\x=\frac{-15}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{17}{2};\frac{-15}{2}\right\}\)

thank you nha!

a) \(47-\left[\left(45\cdot2^4-5^2\cdot12\right):14\right]\)

\(=47-\left[\left(45\cdot16-25\cdot12\right):14\right]\)

\(=47-\left[\left(720-300\right):14\right]\)

\(=47-\left[420:14\right]\)

\(=47-30=17\)

b) \(50-\left[\left(20-2^3\right):2+34\right]\)

\(=50-\left[\left(20-8\right):2+34\right]\)

\(=50-\left[12:2+34\right]\)

\(=50-\left[6+34\right]\)

\(=50-40=10\)

c) \(10^2-\left[60:\left(5^6:5^4-3\cdot5\right)\right]\)

\(=100-\left[60:\left(5^{6-4}-15\right)\right]\)

\(=100-\left[60:\left(5^2-15\right)\right]\)

\(=100-\left[60:\left(25-15\right)\right]\)

\(=100-\left[60:10\right]\)

\(=100-6=94\)

d) \(50-\left[\left(50-2^3\cdot5\right):2+3\right]\)

\(=50-\left[\left(50-8\cdot5\right):2+3\right]\)

\(=50-\left[\left(50-40\right):2+3\right]\)

\(=50-\left[10:2+3\right]\)

\(=50-\left[5+3\right]\)

\(=50-8=42\)

e) \(10-\left[\left(8^2-48\right)\cdot5+\left(2^3\cdot10+8\right)\right]:28\)

\(=10-\left[\left(64-48\right)\cdot5+\left(8\cdot10+8\right)\right]:28\)

\(=10-\left[16\cdot5+\left(80+8\right)\right]:28\)

\(=10-\left[80+88\right]:28\)

\(=10-168:28\)

\(=10-6=4\)

f) \(8697-\left[3^7:3^5+2\left(13-3\right)\right]\)

\(=8697-\left[3^{7-5}+2\cdot10\right]\)

\(=8697-\left[3^2+20\right]\)

\(=8697-\left[9+20\right]\)

\(=8697-29=8668\)

CHÚC BẠN HỌC TỐT!!!!!!!!!!!

Ta có : \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)\(=1+\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{50.50}\)

Vì \(\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3.3}< \frac{1}{2.3};..;\frac{1}{50.50}< \frac{1}{49.50}\)nên :

\(\Rightarrow\)  \(1+\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{50.50}\)\(< 1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)

Ta có : \(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

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

\(=1+\left(1-\frac{1}{50}\right)\)\(=1+\frac{49}{50}\)

Vì \(\frac{49}{50}< 1\)nên \(1+\frac{49}{50}< 2\)\(\Rightarrow\)\(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}< 2\)

\(\Rightarrow\)\(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)\(< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}< 2\)