\(\left(2x+5\right)^4=3^4.5^4\\ \left(2x+5\right)^4=15^4\)

\(\Rightarrow2x+5=15\)

\(\Rightarrow2x=15-5=10\)

\(\Rightarrow x=10:2=5\)

\(B=16.\left(13+87\right)+100.\left(84+2\right)\\ B=16.100+100.86\\ B=100.\left(16+86\right)=100.102\\ B=10200\)

                                                 𝓖𝓲𝓪̉𝓲:

𝓖𝓸̣𝓲 𝓼𝓸̂́ 𝓽𝓾̛̣ 𝓷𝓱𝓲𝓮̂𝓷 𝓷𝓱𝓸̉ 𝓷𝓱𝓪̂́𝓽 𝓬𝓸́ 𝓷𝓪̆𝓶 𝓬𝓱𝓾̛̃ 𝓼𝓸̂́ 𝓬𝓪̂̀𝓷 𝓽𝓲̀𝓶 𝓵𝓪̀ 𝔁.

𝓣𝓪 𝓬𝓸́: \(x:8\)  𝓭𝓾̛ 7 

            \(x:9\)  𝓭𝓾̛ 7 

            \(x:10\)  𝓭𝓾̛ 7

            \(x:11\)  𝓭𝓾̛ 7

            \(x:12\)  𝓭𝓾̛ 7

\(\Rightarrow x-7⋮8,9,10,11,12\Rightarrow x-7\in BCNN\left(8;9;10;11\right)\)

𝓣𝓪 𝓽𝓲̀𝓶 𝓑𝓒𝓝𝓝(8;9;10;11;12)

\(8=2^3;9=3^2;10=2.5;11=11;12=2^2.3\)

\(\Rightarrow BCNN\left(8;9;10;11;12\right)=2^3.3^2.5.11=3960\)

\(\Rightarrow BCNN\left(8;9;10;11;12\right)\in BC\left(3960\right)=\left\{0;3960;7920;11880;...\right\}\)

𝓜𝓪̀ 𝔁 𝓼𝓸̂́ 𝓽𝓾̛̣ 𝓷𝓱𝓲𝓮̂𝓷 𝓷𝓱𝓸̉ 𝓷𝓱𝓪̂́𝓽 𝓬𝓸́ 𝓷𝓪̆𝓶 𝓬𝓱𝓾̛̃ 𝓼𝓸̂́ 𝓷𝓮̂𝓷 𝔁 𝓵𝓪̀ 11880

𝓥𝓪̣̂𝔂 𝓼𝓸̂́ 𝓬𝓪̂̀𝓷 𝓽𝓲̀𝓶 𝓵𝓪̀ 11880

đpcm chứ, nghĩa là điều phải chứng minh.

\(\dfrac{2}{x}=\dfrac{x}{8}\) nên \(2\times8=x\times x\) 

\(\Rightarrow16=x\times x\)

\(\Rightarrow4\times4=x\times x\)

\(\Rightarrow x=4\)

 Thay vào \(\dfrac{x}{3}=\dfrac{4}{y}\) \(\Rightarrow\dfrac{4}{3}=\dfrac{4}{y}\)

\(\Rightarrow4\times y=3\times4\)

\(\Rightarrow4\times y=12\)

\(\Rightarrow12\div4=3=y\)

Vậy x=4; y=3

Đề phải như này chứ 123123.139-139139.123

123123.139-139139.123

= 123.1001.139-139.1001.123

= 0

𝓥𝓲̀ \(\left(x-13+y\right)^2\ge0;\left(x-6-y\right)^2\ge0\)

\(\Rightarrow\left(x-13+y\right)^2+\left(x-6-y\right)^2\ge0\)

𝓓𝓪̂́𝓾 𝓫𝓪̆̀𝓷𝓰 𝔁𝓪̉𝔂 𝓻𝓪 𝓴𝓱𝓲 \(\left(x-13+y\right)^2=0;\left(x-6-y\right)^2=0\) 

\(\Rightarrow\left(x-13+y\right)^2=0\)                             \(\Rightarrow\left(x-6-y\right)^2=0\)

\(x-13+y=0\)                                      \(x-6-y=0\)

\(x+y=13\)                                            \(x-y=6\)

\(\Rightarrow\)𝔁 𝓵𝓪̀ 1 𝓼𝓸̂́  𝓵𝓸̛́𝓷 𝓱𝓸̛𝓷 𝔂 𝓫𝓸̛̉𝓲 𝓿𝓲̀ 𝓴𝓱𝓲 𝔁-𝔂 𝓴𝓮̂́𝓽 𝓺𝓾𝓪̉ 𝓵𝓪̀ 1 𝓼𝓸̂́ 𝓷𝓰𝓾𝔂𝓮̂𝓷 𝓭𝓾̛𝓸̛𝓷𝓰

\(\Rightarrow x=\left(13+6\right)\div2=9,5\)                                  

\(\Rightarrow y=13-9,5=3,5\) 

𝓥𝓪̣̂𝔂 𝔁=9,5 𝓿𝓪̀ 𝔂=3,5                          

\(A=\left(1+1+...+1\right)-\left(\dfrac{1}{9}+\dfrac{2}{10}+...+\dfrac{92}{100}\right)\)𝓒𝓸́ 92 𝓼𝓸̂́ 1

\(A=\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+...+\left(1-\dfrac{92}{100}\right)\)

\(A=\dfrac{8}{9}+\dfrac{8}{10}+...+\dfrac{8}{100}\)

\(A=8.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)\)

\(B=\dfrac{1}{45}+\dfrac{1}{50}+...+\dfrac{1}{500}\)

\(B=\dfrac{1}{5}.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{8.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)}{\dfrac{1}{5}.\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{100}\right)}\\ \Rightarrow\dfrac{A}{B}=\dfrac{8}{\dfrac{1}{5}}=40\)

𝓥𝓪̣̂𝔂 𝓽𝓲̉ 𝓼𝓸̂́ 𝓬𝓾̉𝓪 𝓐 𝓿𝓪̀ 𝓑 𝓵𝓪̀ 40

\(8,A=\left(\dfrac{3}{1.8}+\dfrac{3}{8.15}+...+\dfrac{3}{106.113}\right)-\left(\dfrac{25}{50.55}+\dfrac{25}{55.60}+...+\dfrac{25}{95.100}\right)\\ A=\dfrac{3}{7}.\left(\dfrac{7}{1.8}+\dfrac{7}{8.15}+...+\dfrac{7}{106.113}\right)-5.\left(\dfrac{5}{50.55}+\dfrac{5}{55.60}+...+\dfrac{5}{95.100}\right)\\ A=\dfrac{3}{7}.\left(1-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{15}+...+\dfrac{1}{106}-\dfrac{1}{113}\right)-5.\left(\dfrac{1}{50}-\dfrac{1}{55}+\dfrac{1}{55}-\dfrac{1}{60}+...+\dfrac{1}{95}-\dfrac{1}{100}\right)\\ A=\dfrac{3}{7}.\left(1-\dfrac{1}{113}\right)-5.\left(\dfrac{1}{50}-\dfrac{1}{100}\right)\)

\(A=\dfrac{3}{7}.\dfrac{112}{113}-5.\dfrac{1}{100}=\dfrac{48}{113}-\dfrac{1}{20}\\ A=\dfrac{847}{2260}\)

\(7,M=\left(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0,25+\dfrac{1}{5}}{1\dfrac{1}{6}-0,875+0,7}\right)\div\dfrac{2023}{2024}\\ M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\right)\div\dfrac{2023}{2024}\\ M=\left(\dfrac{2\times\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\times\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}\right)}\right)\div\dfrac{2023}{2024}\)

\(M=\left(\dfrac{2}{7}-\dfrac{1}{\dfrac{7}{2}}\right)\div\dfrac{2023}{2024}=\left(\dfrac{2}{7}-\dfrac{2}{7}\right)\div\dfrac{2023}{2024}\\ M=0\div\dfrac{2023}{2024}=0\)

𝓥𝓪̣̂𝔂 𝓜=0