Phan Thị Dung
Giới thiệu về bản thân
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
a) Tính tổng các chữ số của A ta thấy:
1+2+3 chia hết cho 3
4+5+6 chia hết cho 3
...
97+98+99 chia hết cho 3
100 + 101 = 201 chia hết cho 3
A có tổng các chữ số chia hết cho 3 nên A chia hết cho 3 ⇒ A là hợp số.
b) Vẫn tính tổng của A, nhưng theo cách:
1+2+3+...+9 chia hết cho 9
11+12+13+...+19 chia hết cho 9
...
91+92+93+...+99 chia hết cho 9
10+20+30+...+90 chia hết cho 9
100+101 không chia hết cho 9
Nên A không chia hết cho 9.
Do A chia hết cho 3 nên A viết được dưới dạng: A = 3B. Và B không chia hết cho 3 vì A không chia hết cho 9.
⇒ A không phải là 1 số chính phương.
\(A=\dfrac{10^{2004}+1}{10^{2003}+1}>\dfrac{10^{2004}+1+9}{10^{2003}+1+9}=\dfrac{10^{2004}+10}{10^{2003}+10}.\\ =\dfrac{10\left(10^{2003}+1\right)}{10\left(10^{2002}+1\right)}=\dfrac{10^{2003}+1}{10^{2002}+1}=B.\\ \Rightarrow A>B.\)
1. Giải:
Do \(5x+13B\in\left(2x+1\right)\Rightarrow5x+13⋮2x+1.\)
\(\Rightarrow2\left(5x+13\right)⋮2x+1\Rightarrow10x+26⋮2x+1.\)
\(\Rightarrow5\left(2x+1\right)+21⋮2x+1.\)
Do 5(2x+1)⋮2x+1⇒ Ta cần 21⋮2x+1.
⇒ 2x+1 ϵ B(21)=\(\left\{1;3;7;21\right\}.\)
Ta có bảng:
2x+1 | 1 | 3 | 7 | 21 |
x | 0 | 1 | 3 | 10 |
TM | TM | TM | TM |
Vậy xϵ\(\left\{0;1;3;10\right\}.\)
2. Giải:
Do (2x-18).(3x+12)=0.
⇒ 2x-18=0 hoặc 3x+12=0.
⇒ 2x =18 3x =-12.
⇒ x =9 x =-4.
Vậy xϵ\(\left\{-4;9\right\}.\)
3. S= 1-2-3+4+5-6-7+8+...+2021-2022-2023+2024+2025.
S= (1-2-3+4)+(5-6-7+8)+...+(2021-2022-2023+2024)+2025 Có 506 cặp.
S= 0 + 0 + ... + 0 + 2025.
⇒S= 2025.