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Tìm số dư trong phép chia (2023\(\left(2023^{2024}+2024^{2025}+2025^{2026}\right)^{10}\)chia cho 111
a. Xét phân số trung gian là \(\dfrac{72}{78}\) , ta thấy:
\(\dfrac{72}{73}>\dfrac{72}{78}\)
\(\dfrac{58}{78}< \dfrac{72}{78}\)
\(\Rightarrow\dfrac{72}{73}>\dfrac{58}{78}\)
b. Xét phân số trung gian là \(\dfrac{n}{n+2}\) , ta thấy:
\(\dfrac{n}{n+3}< \dfrac{n}{n+2}\)
\(\dfrac{n}{n+2}< \dfrac{n+1}{n+2}\)
\(\Rightarrow\dfrac{n}{n+3}< \dfrac{n+1}{n+2}\)
c. Ta có: \(\dfrac{10^{11}-1}{10^{12}-1}< 1\) (vì tử < mẫu)
\(\Rightarrow\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{\left(10^{11}-1\right)+11}{\left(10^{12}-1\right)+11}=\dfrac{10^{11}+10}{10^{12}+10}=\dfrac{10^{10}+1}{10^{11}+1}\)
Vậy \(\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{10^{10}+1}{10^{11}+1}\)
d. Xét phân số trung gian là \(\dfrac{1}{4}\) , ta thấy:
\(\dfrac{12}{47}>\dfrac{12}{48}=\dfrac{1}{4}\)
\(\dfrac{19}{77}< \dfrac{19}{76}=\dfrac{1}{4}\)
\(\Rightarrow\dfrac{12}{47}>\dfrac{19}{77}\)
\(\dfrac{2^{19}+27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(=\dfrac{2^{19}+\left(3^3\right)^3+5.3.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.4\right)^{10}}\)
\(=\dfrac{2^{19}.3^9+3.5.2^{18}.3^8}{3^9.2^9.2^{10}+3^{10}.4^{10}}\)
\(=\dfrac{2^{19}.3^9+5.2^{18}.3^9}{3^9.2^{19}+3^{10}.\left(2^2\right)^{10}}\)
\(=\dfrac{2^{18}.3^9.\left(2.5\right)}{3^9.2^{19}+3^{10}.2^{20}}\)
\(=\dfrac{2^{18}.3^9.7}{2^{19}.3^9.\left(1+3.2\right)}\)
\(=\dfrac{7}{2\left(1+6\right)}\)
\(=\dfrac{7}{2.7}\)
\(=\dfrac{1}{2}\)
a) \(5^{20}và2550^{10}\)
\(5^{20}=\left(5^2\right)^{10}=25^{10}< 2550^{10}\)
=> \(5^{20}< 2550^{10}\)
b) \(999^{10}và999999^5\)
\(999^{10}=\left(999^2\right)^5=1998^5< 999999^5\)
=> \(999^{10}< 999999^5\)
c) \(\left(\dfrac{-1^{300}}{5}\right)và\left(\dfrac{-1^{500}}{3}\right)\)
\(\left(\dfrac{-1^{300}}{5}\right)=\dfrac{-1}{5}\)
\(\left(\dfrac{-1^{500}}{3}\right)=\dfrac{-1}{3}\)
\(\dfrac{-1}{5}=\dfrac{-3}{15}\)
\(\dfrac{-1}{3}=\dfrac{-5}{15}\)
=> \(\dfrac{-3}{15}>\dfrac{-5}{15}\)
=> \(\left(\dfrac{-1^{300}}{5}\right)>\left(\dfrac{-1^{500}}{3}\right)\)
a, \(A=\dfrac{10^{15}+1}{10^6+1}>1\);\(B=\dfrac{10^6+1}{10^{17}+1}< 1\)
⇒\(A>B\)
b, \(D=\dfrac{2^{2007}+3}{2^{2006}-1}=\dfrac{2^{2008}+6}{2^{2007}-2}\)
Ta có : \(\dfrac{2^{2008}-3}{2^{2007}-1}< \dfrac{2^{2008}-3}{2^{2007}-2}< \dfrac{2^{2008}+6}{2^{2007}-2}\)
⇒ \(C< D\)
c, \(M=\dfrac{3}{8^3}+\dfrac{7}{8^4}=\dfrac{3}{8^3}+\dfrac{3}{8^4}+\dfrac{4}{8^4}\)
\(N=\dfrac{7}{8^3}+\dfrac{3}{8^4}=\dfrac{3}{8^3}+\dfrac{4}{8^3}+\dfrac{3}{8^4}\)
Vì \(\dfrac{4}{8^4}< \dfrac{4}{8^3}\)
⇒ \(M< N\)
\(A=\dfrac{25^{10}+1}{25^{10}-1}=\dfrac{25^{10}-1+2}{25^{10}-1}=\dfrac{25^{10}-1}{25^{10}-1}+\dfrac{2}{25^{10}-1}=1+\dfrac{2}{25^{10}-1}\left(1\right)\)
\(B=\dfrac{25^{10}-1}{25^{10}-3}=\dfrac{25^{10}-3+2}{25^{10}-3}=\dfrac{25^{10}-3}{25^{10}-3}+\dfrac{2}{25^{10}-3}=1+\dfrac{2}{25^{10}-3}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow A< B\)
Ta có : \(\dfrac{25^{10}+1}{25^{10}-1}=\dfrac{25^{10}-1+2}{25^{10}-1}=\dfrac{25^{10}-1}{25^{10}-1}+\dfrac{2}{25^{10}-1}\)\(=1+\dfrac{2}{25^{10}-1}\)
Ta có : \(\dfrac{25^{10}-1}{25^{10}-3}=\dfrac{25^{10}-3+2}{25^{10}-3}=\dfrac{25^{10}-3}{25^{10}-3}+\dfrac{2}{25^{10}-3}=1+\dfrac{2}{25^{10}-3}\)
Vì \(25^{10}-1>25^{10}-3\)
\(\Rightarrow\dfrac{2}{25^{10}-1}< \dfrac{2}{25^{10}-3}\)
\(\Rightarrow1+\dfrac{2}{25^{10}-1}< 1+\dfrac{2}{25^{10}-3}\)
\(\Rightarrow A< B\)
Vậy \(A< B\)
Đây nhé bé
Câu1
Vì \(\mid x \mid \geq 0 \Rightarrow \mid x \mid + 1 \geq 1\).
Do đó \(\left(\right. \mid x \mid + 1 \left.\right)^{10} \geq 1^{10} = 1\).
Suy ra:
\(A = \left(\right. \mid x \mid + 1 \left.\right)^{10} + 2023 \geq 1 + 2023 = 2024.\)
Dấu “=” chỉ xảy ra khi \(\mid x \mid = 0 \Leftrightarrow x = 0\).
\(\Rightarrow\) Giá trị nhỏ nhất của \(A\) là \(\boxed{2024}\), đạt tại \(x = 0\).
Câu 2 ( câu này kiến thức nâng cao nhé em nên là khi em đọc lời giải sẽ có khó hiểu nhé )
Đặt \(n = 2022\). Khi đó:
\(A = \frac{n^{2022} + 1}{n^{2023} + 1} , B = \frac{n^{2021} + 1}{n^{2022} + 1} .\)
Xét tổng quát với \(a_{k} = \frac{n^{k} + 1}{n^{k + 1} + 1} , \left(\right. n > 1 \left.\right)\).
Ta gọi k là luỹ thừa của cơ số
\(a_{k} > a_{k - 1} \textrm{ }\textrm{ } \Longleftrightarrow \textrm{ }\textrm{ } \left(\right. n^{k} + 1 \left.\right)^{2} > \left(\right. n^{k + 1} + 1 \left.\right) \left(\right. n^{k - 1} + 1 \left.\right) .\)
Xét hiệu:
\(\left(\right.n^{k}+1\left.\right)^2-\left(\right.n^{k+1}+1\left.\right)\left(\right.n^{k-1}+1\left.\right)=-n^{k-1}\left(\right.n-1\left.\right)^2<0\)
Vậy \(a_{k} < a_{k - 1}\), tức dãy \(\left(\right. a_{k} \left.\right)\) giảm dần theo \(k\)
Do đó:
\(A = a_{2022} < a_{2021} = B .\)
\(\Rightarrow B>A\)
Câu3
Ta đổi : \(27 = 3^{3}\), \(9 = 3^{2}\), \(125 = 5^{3}\).
\(\frac{5^{16} \cdot \left(\right. 3^{3} \left.\right)^{7}}{\left(\right. 5^{3} \left.\right)^{5} \cdot \left(\right. 3^{2} \left.\right)^{11}} = \frac{5^{16} \cdot 3^{21}}{5^{15} \cdot 3^{22}} = 5^{16 - 15} \cdot 3^{21 - 22} = \frac{5}{3} .\)
Vậy kết quả bằng \(\frac{5}{3}\).
Câu 3:
\(\frac{5^{16}\cdot27^7}{125^5\cdot9^{11}}\)
\(=\frac{5^{16}\cdot\left(3^3\right)^7}{\left(5^3\right)^5\cdot\left(3^2\right)^{11}}=\frac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}\)
\(=\frac53\)
Câu 2:
\(2022A=\frac{2022^{2023}+2022}{2022^{2023}+1}=1+\frac{2021}{2022^{2023}+1}\)
\(2022B=\frac{2022^{2022}+2022}{2022^{2022}+1}=1+\frac{2021}{2022^{2022}+1}\)
Ta có: \(2022^{2023}+1>2022^{2022}+1\)
=>\(\frac{2021}{2022^{2023}+1}<\frac{2021}{2022^{2022}+1}\)
=>\(\frac{2021}{2022^{2023}+1}+1<\frac{2021}{2022^{2022}+1}+1\)
=>2022A<2022B
=>A<B
Câu 1:
\(\left|x\right|\ge0\forall x\)
=>\(\left|x\right|+1\ge1\forall x\)
=>\(\left(\left|x\right|+1\right)^{10}\ge1^{10}=1\forall x\)
=>\(\left(\left|x\right|+1\right)^{10}+2023\ge1+2023=2024\forall x\)
Dấu '=' xảy ra khi x=0
1) So sánh các lũy thừa
a.
4444\(^{3333}\) và 3333\(^{4444}\)
4444\(^{3333}\) =(4\(^3\)\()\) \(^{111}\)
3333\(^{4444}\) =\((\)3\(^4\)\()\) \(^{111}\)
\(\rightarrow\) (4\(^3\)\()\) \(^{111}\) =64\(^{111}\) ; \((\)3\(^4\)\()\) \(^{111}\) =81\(^{111}\)
\(\rightarrow\)64\(^{111}\) <81\(^{111}\)
\(\Rightarrow\) 4444\(^{3333}\) < 3333\(^{4444}\)
Lười làm quá,ý còn lại bn làm tương tự,có ý lấy số chung để so sánh,có ý lấy số mũ để so sánh,có ý như trên.
Ta có :
\(\dfrac{10^{2023}}{10^{2024}}=\dfrac{10^{2022}}{10^{2023}}\)
mà \(\dfrac{10^{2023}}{10^{2024}}>\dfrac{10^{2023}-3}{10^{2024}-3}\)
\(\dfrac{10^{2022}}{10^{2023}}< \dfrac{10^{2022}+1}{10^{2023}+1}\)
\(\Rightarrow\dfrac{10^{2023}-3}{10^{2024}-3}< \dfrac{10^{2022}+1}{10^{2023}+1}\)