a ) \(\dfrac{\left(x-3\right)^2}{3}-\dfrac{\left(2x-1\right)^2}{12}\le x\)

\(\Leftrightarrow4\left(x-3\right)^2-\left(2x-1\right)^2\le12x\)

\(\Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-4x+1\right)-12x\le0\)

\(\Leftrightarrow4x^2-24x+36-4x^2+4x-1-12x\le0\)

\(\Leftrightarrow-36x\le-35\)

\(\Leftrightarrow x\ge\dfrac{35}{36}\)

Vậy bất phương trình có nghiệm \(x\ge\dfrac{35}{36}\).

b ) \(2+\dfrac{3\left(x+1\right)}{3}< 3-\dfrac{x-1}{4}\)

\(\Leftrightarrow2+x+1< 3-\dfrac{x-1}{4}\)

\(\Leftrightarrow x+3< 3-\dfrac{x-1}{4}\)

\(\Leftrightarrow4\left(x+3\right)< 12-x+1\)

\(\Leftrightarrow4x+12+x< 13\)

\(\Leftrightarrow5x< 13-12\)

\(\Leftrightarrow5x< 1\)

\(\Leftrightarrow x< \dfrac{1}{5}\)

Vậy bất phương trình có nghiệm \(x< \dfrac{1}{5}\)

\(\dfrac{d}{dx}\left(f\left(x\right)\right)\equiv f'\left(x\right)\)

\(\dfrac{1}{sinx}dx=\dfrac{sinx}{sin^2x}dx=\dfrac{sinx}{1-cos^2x}dx=\dfrac{d\left(cosx\right)}{cos^2x-1}\)

a) \(\dfrac{\left(x-1\right)^2}{x-2}=\dfrac{\left(x-2\right)^2+2\left(x-2\right)+1}{x-2}=x-2+2+\dfrac{1}{x-2}\ge2+2\sqrt{\left(x-2\right).\dfrac{1}{x-2}}=4\)

GTNN là 4 khi x=3

Bài 2.

\(S_n=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)

\(\Rightarrow S_n=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)\(\Rightarrow S_n=\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)

Bài 1:

\(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{14}\)

\(=\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\)

\(=\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{-47}{60}\right):\dfrac{47}{24}\)

\(=\dfrac{7}{15}.3+\left(\dfrac{-47}{60}\right):\dfrac{47}{24}\)

\(=\dfrac{7}{5}+\left(\dfrac{-47}{60}\right):\dfrac{47}{24}\)

\(=\dfrac{7}{5}+\dfrac{-2}{5}\)

\(=\dfrac{5}{5}=1\)

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\\\dfrac{\dfrac{2}{3}}{x}+\dfrac{\dfrac{2}{3}}{y}+\dfrac{\dfrac{8}{9}}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\\\dfrac{\dfrac{2}{3}}{x}+\dfrac{\dfrac{14}{9}}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\left(1\right)\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\left(2\right)\end{matrix}\right.\)

Nhân cả hai vế (1) cho \(\dfrac{2}{3}\) ta có: \(\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{2}{3y}=\dfrac{5.2}{6.3}\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{2}{3y}=\dfrac{10}{18}\left(3\right)\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\left(4\right)\end{matrix}\right.\)

Lấy (4) trừ (3) ta có:

\(\dfrac{14}{9y}-\dfrac{2}{3y}=1-\dfrac{10}{18}\)\(\Leftrightarrow\dfrac{8}{9y}=\dfrac{4}{9}\)\(\Leftrightarrow y=2\Rightarrow x=\dfrac{1}{\dfrac{5}{6}-\dfrac{1}{2}}=3\)

Theo đề ra ta có hệ : 

 \(\left\{{}\begin{matrix}\dfrac{4}{a^2}=1\\\dfrac{1}{a^2}+\dfrac{\dfrac{3}{4}}{b^2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\\dfrac{\dfrac{3}{4}}{b^2}=\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)

Vậy (a,b) = (2,1) 

đúng

Bài 1:

\(a,E=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

\(b,E>0\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}>0\)

Mà: \(\sqrt{x}>0\\ \Rightarrow\sqrt{x}-1>0\\ \Leftrightarrow\sqrt{x}>1\\ \Leftrightarrow x>1\)

Bài 2:

\(a,G=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{1-\sqrt{x}}-\dfrac{2\sqrt{x}}{x-1}\right)\left(\sqrt{x}+1\right)\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\left(\sqrt{x}+1\right)\\ =\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\left(\sqrt{x}+1\right)\\ =\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\left(\sqrt{x}+1\right)\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\\ =\sqrt{x}-1\)

( Tiếp )

\(b,G=2\Leftrightarrow\sqrt{x}-1=2\\ \Leftrightarrow\sqrt{x}=3\\ \Leftrightarrow x=9\)

+,Ta có :A thuộc E => thay x=2 và y=0 vào E ta đc a^2=4 => a=2 (loại a=-2 vì a<0 )

+, Tương tự thay B vào E => 3b^2=3 =>b=1(loại b=-1 vì b <0)

=> vậy a =2 b =1 

học tốt ! :)))