110,5:34=3,25

905,2:7,3=124

483:35=13,8

91,08:3,6=25,3

3:6,25=0,48

\(=6\cdot\dfrac{1}{27}-\dfrac{4}{9}=\dfrac{2}{9}-\dfrac{4}{9}=-\dfrac{2}{9}\)

\(A=\dfrac{\sqrt{3x-9}}{x}=\sqrt{\dfrac{3x-9}{x^2}}=\sqrt{-\dfrac{9}{x^2}+\dfrac{3}{x}}\)\(=\sqrt{\dfrac{1}{4}-\left(\dfrac{9}{x^2}+2.\dfrac{3}{x}.\dfrac{1}{2}+\dfrac{1}{4}\right)}\)

\(=\sqrt{\dfrac{1}{4}-\left(\dfrac{3}{x}-\dfrac{1}{2}\right)^2}\le\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(\dfrac{3}{x}-\dfrac{1}{2}=0\Leftrightarrow x=6\) (tm)

Vậy \(A_{max}=\dfrac{1}{2}\)

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

\(=\sqrt{\dfrac{1}{4}-\left(\dfrac{2}{x}-\dfrac{1}{2}\right)^2}\le\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(\dfrac{2}{x}-\dfrac{1}{2}=0\Leftrightarrow x=4\) (tm)

Vậy \(B_{max}=\dfrac{1}{2}\)

\(C=\dfrac{\sqrt{x-4}}{3x}=\sqrt{\dfrac{x-4}{9x^2}}=\sqrt{-\dfrac{4}{9x^2}+\dfrac{1}{9x}}=\sqrt{\dfrac{1}{144}-\left(\dfrac{4}{9x^2}-2.\dfrac{2}{3x}.\dfrac{1}{12}+\dfrac{1}{144}\right)}\)

\(=\sqrt{\dfrac{1}{144}-\left(\dfrac{2}{3x}-\dfrac{1}{12}\right)^2}\le\sqrt{\dfrac{1}{144}}=\dfrac{1}{12}\)

Dấu "=" xảy ra khi \(x=8\) (tm)

Vậy \(C_{max}=\dfrac{1}{12}\)

a) Ta có: \(\dfrac{\sqrt{a-4}}{a}+\dfrac{\sqrt{b-4}}{b}+\dfrac{\sqrt{c-4}}{c}=\dfrac{3}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{a-4}=a\\2\sqrt{b-4}=b\\2\sqrt{c-4}=c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{a-4}=\dfrac{1}{2}a\\\sqrt{b-4}=\dfrac{b}{2}\\\sqrt{c-4}=\dfrac{c}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-4=\dfrac{1}{4}a^2\\b-4=\dfrac{1}{4}b^2\\c-4=\dfrac{1}{4}c^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{4}a^2-a+4=0\\\dfrac{1}{4}b^2-b+4=0\\\dfrac{1}{4}c^2-c+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2-4a+16=0\\b^2-4b+16=0\\c^2-4c+16=0\end{matrix}\right.\Leftrightarrow a,b,c\in\varnothing\)

\(\dfrac{3}{5}+\dfrac{1}{1}=\dfrac{3}{5}+\dfrac{5}{5}=\dfrac{8}{5}\)

8/5

\(\dfrac{3}{5}+\dfrac{1}{1}=\dfrac{3}{5}+1=\dfrac{8}{5}\)

( 2 + 3 + 4 - 5 ) x 2 = 4 x 2 

                               = 8

CHÚC EM HỌC TỐT \(\left(\cdot o\cdot\right)\)

( 2 + 3 + 4 - 5 ) x 2 = 8 nhé em 

                                 chúc em học tốt

Ta lấy 2+3=5 ! Xong lấy 5 + 4 = 9 ! Rồi 9 - 5 = 4 ! Cuối cùng là ta lấy 4 x 2 = 8 nhé ! 
HT
!!!!!
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\(x-\dfrac{1}{3}x=x\left(1-\dfrac{1}{3}\right)=\dfrac{2}{3}x\)

\(=x\left(1-\dfrac{1}{3}\right)=\dfrac{2}{3}x\)

\(x-\dfrac{x}{3}=\dfrac{3x-x}{3}=\dfrac{2x}{3}\)

Ta có: \(\left|2x\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=x+3\left(x\ge0\right)\\-2x=x+3\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3\\-2x-x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

Vậy: S={3;-1}

8,75

5/10

bằng 8,75