i giúp em vớiiiiii

\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)

\(=\dfrac{3}{\dfrac{2\left(2+1\right)}{2}}+\dfrac{3}{\dfrac{3\left(3+1\right)}{2}}+...+\dfrac{3}{\dfrac{2022\left(2022+1\right)}{2}}\)

\(=\dfrac{6}{2\left(2+1\right)}+\dfrac{6}{3\left(3+1\right)}+...+\dfrac{6}{2022\cdot2023}\)

\(=\dfrac{6}{2\cdot3}+\dfrac{6}{3\cdot4}+...+\dfrac{6}{2022\cdot2023}\)

\(=6\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2022\cdot2023}\right)\)

\(=6\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)

\(=6\cdot\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)=6\cdot\dfrac{2021}{4046}=\dfrac{12126}{4046}< 3\)

mà \(3< \dfrac{10}{3}\)

nên \(M< \dfrac{10}{3}\)

mọi người ơi giúp em vs ạ , e đang rất cần 

\(1+2+...+n=\dfrac{\left(\dfrac{n-1}{1}+1\right).\left(n+1\right)}{2}=\dfrac{n\left(n+1\right)}{2}\)

\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)

\(=3\left(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+...+2022}\right)\)

\(=3\left(\dfrac{1}{\dfrac{2.\left(2+1\right)}{2}}+\dfrac{1}{\dfrac{3.\left(3+1\right)}{2}}+...+\dfrac{1}{\dfrac{2022.\left(2022+1\right)}{2}}\right)\)

\(=3\left(\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{2022.2023}\right)\)

\(=3.2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)\)

\(=6.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)

\(=6.\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)\)

\(=6.\dfrac{2021}{4046}=3.\dfrac{2021}{2023}=\dfrac{6063}{2023}=\dfrac{18189}{6069}\)

\(\dfrac{10}{3}=\dfrac{20230}{6069}>\dfrac{18189}{6069}=M\)

\(7^4=7^3\cdot7=7^3+7^3\cdot6>7^3+1\)

\(S=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}+...+\frac{1}{\sqrt{2025}-\sqrt{2024}}\)

Ta nhận xét thấy mỗi số hạng trong S đều dương. Từ đó ta đặt

\(A=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}+...+\frac{1}{\sqrt{2024}-\sqrt{2023}}\left(A>0\right)\)

\(\Rightarrow S=A+\frac{1}{\sqrt{2025}-\sqrt{2024}}=A+\frac{\sqrt{2025}+\sqrt{2024}}{\left(\sqrt{2025}-\sqrt{2024}\right)\left(\sqrt{2025}+\sqrt{2024}\right)}\)

\(=A+\sqrt{2025}+\sqrt{2024}>\sqrt{2025}=45\)

Vậy \(S>45\)

PS: Phan Thanh Tịnh xem lại bài giải nhé bạn

Ta có : 1 = (n + 1) - n =\(\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}\right)^2\)

\(=\left(\sqrt{n+1}\right)^2-\sqrt{n+1}.\sqrt{n}+\sqrt{n+1}.\sqrt{n}+\left(\sqrt{n}\right)^2\)

\(=\sqrt{n+1}.\left(\sqrt{n+1}-\sqrt{n}\right)+\sqrt{n}.\left(\sqrt{n+1}-\sqrt{n}\right)\)

\(=\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n-1}+\sqrt{n}\right)\)\

\(\Rightarrow\frac{1}{\sqrt{n+1}-\sqrt{n}}=\sqrt{n+1}+\sqrt{n}\)

Áp dụng vào bài toán,ta có :

\(S=\sqrt{1}+\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2025}-\sqrt{2024}=\sqrt{2025}\)= 45

Vậy S = 45

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Bằng 5^57/7,71 cách giải 12:0,1+7/^1-729=5^57/7,71

5^57/7,71-3:3x2+2:4=5^57/7,71

Chúc bạn học giỏi nhe :)))) 👍👍👍👍👍👍👍👍👍 

Vậy thì D sẽ > 1/64 nha

Chu pa pi pồ nhà nhố.....!!!!