\(\Leftrightarrow x-\dfrac{1}{4}=\dfrac{20}{27}\)

hay x=20/27+1/4=107/108

\(\dfrac{x-1}{4}=\dfrac{5}{3}.\dfrac{4}{9}\Leftrightarrow\dfrac{x-1}{4}=\dfrac{20}{27}\Rightarrow27x-27=80\)

\(\Leftrightarrow27x=107\Leftrightarrow x=\dfrac{107}{27}\)

=>x-1/4=20/27

hay x=107/108

A = 5 + \(\frac{15}{4}\)|3x+7| + 3

Vì |3x+7| lớn hơn hoặc bằng 0                  Với mọi x

=>|3x+7| + 3  lớn hơn hoặc bằng 0 + 3          Với mọi x

=> \(\frac{15}{4}\)|3x+7| + 3 lớn hơn hoặc bằng 3     Với mọi x

=>5 + \(\frac{15}{4}\)|3x+7| + 3 lớn hơn hoặc bằng 5 + 3       Với mọi x

hay C lớn hơn hoặc bằng 8

Dấu = xảy ra <=> |3x+7| = 0

                    <=> 3x + 7 = 0

                    <=> 3x       = 0 + 7

                    <=> 3x       = 7

                    <=>  x        =  7 : 3

                    <=>  x        = \(\frac{7}{3}\)

Vậy biểu thức A đạt GTLN bằng 8 tại x =\(\frac{7}{3}\)

xong rùi đó 

thank zì^^

\(x-\dfrac{1}{4}=\dfrac{5}{6}\times\dfrac{4}{9}\)

<=>\(x-\dfrac{1}{4}=\dfrac{10}{27}\)

<=>\(x=\dfrac{10}{27}+\dfrac{1}{4}=\dfrac{67}{108}\)

\(x-\dfrac{1}{4}=\dfrac{5}{6}\times\dfrac{4}{9}\\ \Rightarrow x-\dfrac{1}{4}=\dfrac{10}{27}\\ \Rightarrow x=\dfrac{10}{27}+\dfrac{1}{4}\\ \Rightarrow x=\dfrac{67}{108}\)

a) \(3-2x>4\)

\(\Leftrightarrow-2x>1\)

\(\Leftrightarrow x< \frac{-1}{2}\)

b) \(\frac{2}{3-x}-\frac{9}{3+x}=\frac{1}{2}\)ĐKXĐ : \(x\pm3\)

\(\Leftrightarrow\frac{-4\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}-\frac{18\left(x-3\right)}{2\left(x-3\right)\left(x+3\right)}=\frac{\left(x-3\right)\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow-4x-13-18x+54=x^2-9\)

\(\Leftrightarrow x^2+22x-50=0\)

\(\Leftrightarrow x^2+2\cdot x\cdot11+11^2-171=0\)

\(\Leftrightarrow\left(x+11\right)^2=\left(\pm\sqrt{171}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{171}-11\\x=-\sqrt{171}-11\end{cases}}\)( thỏa )

Vậy....

\(a,\)\(3-2x>4\)

\(\Rightarrow-2x>1\)

\(\Rightarrow x< \frac{-1}{2}\)

Cho \(x=\frac{1}{4}\Rightarrow2.f\left(\frac{1}{\frac{1}{4}}\right)=\left(\frac{1}{4}\right)^2\)

\(\Rightarrow2f\left(4\right)=\frac{1}{16}\Rightarrow f\left(4\right)=\frac{1}{32}\)

7/48 - (1/2 x 2 + 1/6 x 4 + 1/8 x 5 + 1/12 x 7 + 1/14 x 8) : x = 0

7/48 - (1 + 2/3 + 5/8 + 7/12 + 4/7) : x = 0 (đã rút gọn)

7/48 - (336/336 + 224/336 + 210/336 + 196/336 + 192/336) : x = 0 (quy đồng)

7/48 - 193/56 : x  = 0

193/56 : x = 0 + 7/48

193/56 : x = 7/48

              x = 193/56 : 7/48

              x = 1158/49

\(\dfrac{2}{3}\times\left(x+\dfrac{4}{5}\right)=\dfrac{-1}{3}\\ x+\dfrac{4}{5}=\dfrac{-1}{3}:\dfrac{2}{3}\\ x+\dfrac{4}{5}=\dfrac{-1}{3}\times\dfrac{3}{2}\\ x+\dfrac{4}{5}=\dfrac{-1}{2}\\ x=\dfrac{-1}{2}-\dfrac{4}{5}\\ x=\dfrac{-5}{10}-\dfrac{8}{10}\\ x=\dfrac{-13}{10}\)

\(\dfrac{2}{3}.\left(x+\dfrac{4}{5}\right)=-\dfrac{1}{3}\)

    \(\left(x+\dfrac{4}{5}\right)=\left(-\dfrac{1}{3}:\dfrac{2}{3}\right)\)

    \(\left(x+\dfrac{4}{5}\right)=\left(-\dfrac{1}{3}.\dfrac{3}{2}\right)=-\dfrac{1}{2}\)

      \(x\)            \(=\left(-\dfrac{1}{2}\right)-\dfrac{4}{5}\)

      \(x\)            \(=\left(-\dfrac{1}{2}\right)+\left(-\dfrac{4}{5}\right)\)

      \(x\)            \(=-\dfrac{13}{10}\)