\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right).n}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{\left(n-1\right)}-\dfrac{1}{n}=1-\dfrac{1}{n}\le1-\dfrac{1}{2}=\dfrac{1}{2}< \dfrac{2}{3}\)

\(\Rightarrow\)đpcm

a) \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

- ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

\(P\Leftrightarrow\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}-\dfrac{3x-3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)\(\Leftrightarrow\dfrac{2x-6+x+3\sqrt{x}-3x-3}{x-9}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(\Leftrightarrow\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(\Leftrightarrow\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x-3}}{\sqrt{x}+1}\)

\(\Leftrightarrow\dfrac{3}{\sqrt{x}+1}\)

b)\(P< \dfrac{-1}{3}\Leftrightarrow\dfrac{3}{\sqrt{x}+1}< -\dfrac{1}{3}\)

\(\Leftrightarrow-\dfrac{1}{3}-\dfrac{3}{\sqrt{x}+1}>0\)

\(\Leftrightarrow-\left(\dfrac{1}{3}+\dfrac{3}{\sqrt{x}+1}\right)>0\) (Sai)

\(\Rightarrow\)Không có giá trị nào của \(x\) để \(P< -\dfrac{1}{3}\)

c) \(P=\dfrac{3}{\sqrt{x}+1}\)Chỉ xđ đc GTLN là 3 mình nghĩ vậy!
Thông cảm!

Than đá, dầu mỏ, vàng, kim cương, sắt,… tập trung chủ yếu ở

A. phần phía Tây.

B. phần phía Đông.

C. phần phía Nam.

D. phần phía Bắc.

Q= \(\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\\ \Leftrightarrow\left(\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\\ \left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\right)\\ \Leftrightarrow\dfrac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\left(\dfrac{\left(a-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\dfrac{\left(a-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\\ \Leftrightarrow\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\left(\dfrac{a-1-a+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\\ \)

\(\Leftrightarrow\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{1}\\ \Leftrightarrow\dfrac{\sqrt{a}-2}{\sqrt{a}}\)

Xong rồi đó bạn!

Nếu muốn bạn có thể trục căn thức: ĐA:\(\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{a}\)

\(\sqrt{7x}-\sqrt{2x-3}=0\)

\(\Leftrightarrow\sqrt{7x}=\sqrt{2x-3}\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3\ge0\\7x=2x-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\5x=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x=\dfrac{-3}{5}\end{matrix}\right.\)

(Vô lí )

Vậy pt vô nghiệm.

\(\sqrt{2x-1}=3\Leftrightarrow2x-1=9\Leftrightarrow2x=10\Leftrightarrow x=5\)

Có thể thử lại bằng MT

Có thể giúp mình không ạ!

a) \(\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{b^3}{\left(1+a\right)\left(1+c\right)}+\dfrac{c^3}{\left(1+a\right)\left(1+b\right)}\) biết abc=1

b) \(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}\ge a+b+c\)

c) \(\dfrac{ab}{a^5+ab+b^5}+\dfrac{bc}{b^5+bc+c^5}+\dfrac{ac}{a^5+ac+c^5}\) biết abc=1

Xin cảm ơn các bạn trước ạ!