\(\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2=0\)

Nhận xét:

\(\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2\ge0,\forall x\\\left(y+\dfrac{1}{2}\right)^2\ge0,\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2\ge0,\forall x,y\)

Dấu \("="\) xảy ra khi:

\(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+\dfrac{1}{2}=0\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{1}{2};y=-\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2};y=-\dfrac{1}{2}\)

Ta có: \(\dfrac{16}{6}=\dfrac{8}{3}=\dfrac{9}{3}-\dfrac{1}{3}=3-\dfrac{1}{3}\)

\(\dfrac{14}{5}=\dfrac{15}{5}-\dfrac{1}{5}=3-\dfrac{1}{5}\)

Vì \(\dfrac{1}{3}>\dfrac{1}{5}\) nên \(3-\dfrac{1}{3}< 3-\dfrac{1}{5}\) hay \(\dfrac{16}{6}< \dfrac{14}{5}\)

Đặt \(A=5^{100}-5^{99}+5^{98}-5^{97}+...+5^2-5\)

Ta có: \(5A=5^{101}-5^{100}+5^{99}-5^{98}+...+5^3-5^2\)

\(\Rightarrow5A+A=\left(5^{101}-5^{100}+5^{99}-5^{98}+...+5^3-5^2\right)+\left(5^{100}-5^{99}+5^{98}-5^{97}+...+5^2-5\right)\)

\(\Rightarrow6A=5^{101}-5\)

\(\Rightarrow A=\dfrac{5^{101}-5}{6}\)

Đề chưa được rõ bạn ơi!

Mỗi ngăn có số quyển là:
\(135:9=15\) (quyển)

Bốn ngăn trên có số quyển là:
\(15\times4=60\) (quyển)

Năm ngăn trên có số quyển là:

\(15\times5=75\) (quyển)

Đáp số:...

\(A=1+3+3^2+...+3^{2024}\)

\(\Rightarrow3A=3+3^2+3^3+...+3^{2025}\)

\(\Rightarrow3A-A=\left(3+3^2+3^3+...+3^{2025}\right)-\left(1+3+3^2+...+3^{2024}\right)\)

\(\Rightarrow2A=3^{2025}-1\)

\(\Rightarrow A=\dfrac{3^{2025}-1}{2}\)

Vậy \(A=\dfrac{3^{2025}-1}{2}\)

\(8^{11}\cdot9^4\cdot15^6:\left(4^{13}\cdot18^2\cdot3\cdot5^5\right)\)

\(=\left(2^3\right)^{11}\cdot\left(3^2\right)^4\cdot\left(3\cdot5\right)^6:\left[\left(2^2\right)^{13}\cdot\left(2\cdot3^2\right)^2\cdot3\cdot5^5\right]\)

\(=2^{33}\cdot3^8\cdot3^6\cdot5^6:\left(2^{26}\cdot2^2\cdot3^4\cdot3\cdot5^5\right)\)

\(=2^{33}\cdot3^{14}\cdot5^6:\left(2^{28}\cdot3^5\cdot5^5\right)\)

\(=\left(2^{33}:2^{28}\right)\cdot\left(3^{14}:3^5\right)\cdot\left(5^6:5^5\right)\)

\(=2^5\cdot3^9\cdot5\)

\(16^5:8^3\)

\(=\left(2^4\right)^5:\left(2^3\right)^3\)

\(=2^{4\cdot5}:2^{3\cdot3}\)

\(=2^{20}:2^9\)

\(=2^{20-9}\)

\(=2^{11}\)

\(=2048\)

\(27^3:3^4\)

\(=\left(3^3\right)^4:3^4\)

\(=3^{3\cdot4}:3^4\)

\(=3^{12}:3^4\)

\(=3^{12-4}\)

\(=3^8\)

Ta có:

+) \(\dfrac{103}{105}=\dfrac{105}{105}-\dfrac{2}{105}=1-\dfrac{2}{105}=1-\dfrac{6}{315}\)

+) \(\dfrac{205}{208}=\dfrac{208}{208}-\dfrac{3}{208}=1-\dfrac{3}{208}=1-\dfrac{6}{416}\)

Vì \(\dfrac{6}{315}>\dfrac{6}{416}\) nên \(1-\dfrac{6}{315}< 1-\dfrac{6}{416}\) hay \(\dfrac{103}{105}< \dfrac{205}{208}\)