Ta có: \(\frac{a}{3}=\frac{b}{4}=\frac{ab}{3b}=\frac{48}{3b}\)

\(\Rightarrow\frac{b}{4}=\frac{48}{3b}\Rightarrow b.3b=48.4\Rightarrow3b^2=192\)

\(\Rightarrow b^2=\frac{192}{3}=64\Rightarrow\orbr{\begin{cases}b=8\\b=-8\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}a=\frac{48}{8}=6\\a=\frac{48}{-8=-6}\end{cases}}\)

Vậy \(\left(a;b\right)=\left(6;8\right);\left(-6;-8\right)\)

Đặt \(\frac{a}{3}=\frac{b}{4}=k\)

\(\Rightarrow\hept{\begin{cases}a=3k\\b=4k\end{cases}}\)

Thay vào \(ab=48\)ta có :

\(3k.4k=48\)

\(\Leftrightarrow\)\(12.k^2=48\)

\(\Leftrightarrow\)\(k^2=4\)

\(\Leftrightarrow\)\(k^2=2^2\)hoặc \(\left(-2\right)^2\)

\(\Leftrightarrow\)\(k=\pm2\)

* Nếu \(k=2\) : \(\Rightarrow\hept{\begin{cases}a=3.2\\b=4.2\end{cases}}\)\(\Rightarrow\hept{\begin{cases}a=6\\b=8\end{cases}}\)

* Nếu \(k=-2\): \(\Rightarrow\hept{\begin{cases}a=3.\left(-2\right)\\b=4.\left(-2\right)\end{cases}}\)\(\Rightarrow\hept{\begin{cases}a=-6\\b=-8\end{cases}}\)

ta có: 

Lai cho toi đi

x= 0,625 hoặc x= 0,875                                             

ta có  kết quả như sau 

\(\left|x-\frac{1}{2}\right|\le\frac{1}{3}\)

\(\Leftrightarrow\frac{-1}{3}\le x-\frac{1}{2}\le\frac{1}{3}\)

\(\Leftrightarrow\frac{-1}{3}+\frac{1}{2}\le x\le\frac{1}{3}+\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{6}\le x\le\frac{5}{6}\)

6.

a) \(\left|x-\frac{1}{2}\right|\le\frac{1}{3}\)\(\Rightarrow\left|x-\frac{1}{2}\right|\le\left|\frac{1}{3}\right|\)

\(\Rightarrow\frac{-1}{3}\le x-\frac{1}{2}\le\frac{1}{3}\)

\(\Rightarrow\frac{1}{6}\le x\le\frac{5}{6}\)

b) \(\left|2x-\frac{1}{2}\right|>\left|-1,5\right|\)

\(\Rightarrow\orbr{\begin{cases}-1,5< 2x-\frac{1}{2}< 1,5\\2x-\frac{1}{2}< -1,5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\frac{-3}{2}< 2x-\frac{1}{2}< \frac{3}{2}\\2x-\frac{1}{2}< \frac{-3}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\frac{-3}{2}< 2x-\frac{1}{2}< \frac{3}{2}\\2x-\frac{1}{2}< \frac{-3}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-1< 2x< 2\\2x< -1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\frac{-1}{2}< x< 1\\x< \frac{-1}{2}\end{cases}}\)

7.

a) \(\left|-2x+1,5\right|=\frac{1}{4}\)

\(\Rightarrow\orbr{\begin{cases}-2x+\frac{3}{2}=\frac{1}{4}\\-2x+\frac{3}{2}=\frac{-1}{4}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-2x=\frac{7}{4}\\-2x=\frac{5}{4}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{-7}{8}\\x=\frac{-5}{8}\end{cases}}\)

b) \(\frac{3}{2}-\left|1\frac{1}{4}+3x\right|=\frac{1}{4}\)

\(\Rightarrow\left|\frac{5}{4}+3x\right|=\frac{3}{2}-\frac{1}{4}=\frac{5}{4}\)

\(\Rightarrow\orbr{\begin{cases}\frac{5}{4}+3x=\frac{5}{4}\\\frac{5}{4}+3x=\frac{-5}{4}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}3x=0\\3x=\frac{-5}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-5}{6}\end{cases}}\)

ta có :

\(B=\frac{x^2+15}{x^2+3}=1+\frac{12}{x^2+3}\le1+\frac{12}{3}=5\)

Vậy GTLN của B là 5

khi x=0

ta có A = 30/13

ta có :