`#040911`

a,

\(\dfrac{1}{2}\cdot\left(x-4\right)-\dfrac{1}{4}\cdot\left(x-\dfrac{4}{3}\right)=2\cdot\left(x-\dfrac{1}{2}\right)\)

\(\Rightarrow\dfrac{1}{2}x-2-\dfrac{1}{4}x+\dfrac{1}{3}=2x-1\\\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{4}x-2x\right)=2-\dfrac{1}{3}-1\\ \Rightarrow-\dfrac{7}{4}x=\dfrac{2}{3}\\ \Rightarrow x=\dfrac{2}{3}\div\left(-\dfrac{7}{4}\right)\\ \Rightarrow x=-\dfrac{8}{21}\)

Vậy, \(x=-\dfrac{8}{21}\)

b,

\(\dfrac{3}{4}-\left(x-\dfrac{1}{2}\right)^2=-\dfrac{11}{2}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{3}{4}-\left(-\dfrac{11}{2}\right)\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\left(\pm\dfrac{5}{2}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{5}{2}\\x-\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}+\dfrac{1}{2}\\x=-\dfrac{5}{2}+\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy, \(x\in\left\{-2;3\right\}\)

c,

\(\dfrac{3}{16}+1\dfrac{1}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\\ \Rightarrow\dfrac{17}{16}\cdot\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\div\dfrac{17}{16}\\ \Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{17}\)

Bạn xem lại đề có sai kh nhỉ?

c) \(\dfrac{3}{16}+\dfrac{1}{\dfrac{1}{16}}\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}\)

\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{3}{4}-\dfrac{3}{16}\)

\(\Rightarrow16\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}\)

\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{16}:16\)

\(\Rightarrow\left(x-\dfrac{2}{3}\right)^2=\dfrac{9}{256}=\left(\dfrac{3}{16}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{3}=\dfrac{3}{16}\\x-\dfrac{2}{3}=-\dfrac{3}{16}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{16}+\dfrac{2}{3}\\x=-\dfrac{3}{16}+\dfrac{2}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{41}{48}\\x=\dfrac{23}{48}\end{matrix}\right.\)

(x - 3)⁴ =16

=> 2⁴ = 16

=> x- 3 = 2

=> x = 5

4^(x-3) =16

=> 4² = 16

=> x-3 = 2 

=> x = 5

(x - 3)⁴ = 16

(x - 3)⁴ = 2⁴ = (-2)⁴

=> \(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

4^x-3 = 16

4^x-3 = 4²

=> x - 3 = 2

=> x = 2 + 3

=> x = 5

a, \(\left(\frac{4}{9}.\frac{3}{7}\right).\frac{7}{4}=\frac{4}{9}\left(\frac{3}{7}.\frac{7}{4}\right)=\frac{4}{9}.\frac{3}{4}=\frac{3}{9}=\frac{1}{3}\)

b, \(\left(\frac{6}{5}.\frac{4}{5}\right).\frac{25}{16}=\frac{6}{5}\left(\frac{4}{5}.\frac{25}{16}\right)=\frac{6}{5}.\frac{5}{4}=\frac{6}{4}=\frac{3}{2}\)

c, \(\left(\frac{7}{8}.\frac{16}{9}\right).\frac{3}{14}=\frac{7}{8}\left(\frac{16}{9}.\frac{3}{14}\right)=\frac{7}{8}.\frac{8}{21}=\frac{7}{21}=\frac{1}{3}\)

a, $\left(\frac{4}{9}.\frac{3}{7}\right).\frac{7}{4}=\frac{4}{9}\left(\frac{3}{7}.\frac{7}{4}\right)=\frac{4}{9}.\frac{3}{4}=\frac{3}{9}=\frac{1}{3}$

b, $\left(\frac{6}{5}.\frac{4}{5}\right).\frac{25}{16}=\frac{6}{5}\left(\frac{4}{5}.\frac{25}{16}\right)=\frac{6}{5}.\frac{5}{4}=\frac{6}{4}=\frac{3}{2}$

c, $\left(\frac{7}{8}.\frac{16}{9}\right).\frac{3}{14}=\frac{7}{8}\left(\frac{16}{9}.\frac{3}{14}\right)=\frac{7}{8}.\frac{8}{21}=\frac{7}{21}=\frac{1}{3}$

( 4/9 x 3/7 ) x 7/4 = 4/9 x ( 3/7 x 7/4 ) =4/9 x 21/28 = 84/252

( 6/5 x 4/5 ) x 25/16 = ( 25/16 x 4/5 ) x 6/5 = 100/80 x 6/5 = 600/400

bài2 \(x\times\dfrac{15}{16}-x\times\dfrac{4}{16}=2\) 

     \(x\times\dfrac{11}{16}=2\) 

     \(x=2:\dfrac{11}{16}\) 

    \(x=\dfrac{32}{11}\)

Bài 1 : 

 \(\dfrac{x}{16}\times\left(2017-1\right)=2\)

          \(\dfrac{x}{16}\times2016=2\)

                      \(\dfrac{x}{16}=\dfrac{2}{2016}\)

                         \(x=\dfrac{2}{2016}\times16\)

                         \(x=\dfrac{1}{63}\)

1- (5\(\dfrac{4}{9}\) +x+7\(\dfrac{7}{18}\)) : 15\(\dfrac{3}{4}\) = 0 

1- (\(\dfrac{49}{9}+x+\dfrac{133}{18}\)) : \(\dfrac{63}{4}=0\) 

 (\(\dfrac{49}{9}+\dfrac{133}{18}\)+\(x\) ) : \(\dfrac{63}{4}\) = 1 - 0 

       (\(\dfrac{77}{6}\) + \(x\) ) : \(\dfrac{63}{4}\) = 1

         \(\dfrac{77}{6}+x\)         = 1 x \(\dfrac{63}{4}\) 

          \(\dfrac{77}{6}\) + \(x\)          = \(\dfrac{63}{4}\)

                  \(x\)            = \(\dfrac{63}{4}\) - \(\dfrac{77}{6}\)

                  \(x=\) \(\dfrac{35}{12}\)

Tại x = 16 => x +1 = 17

Thay vào A ta được:

A = x⁴ - (x+1)x³ + (x+1)x² - (x+1)x + 20

A= x⁴ -(x⁴ + x³)  + (x³ + x²)  -(x² + x) +20

A= x⁴ - x⁴ - x³ + x³ + x² - x² -x + 20

A= - x+20

Mà  x = 16

=> A= -16 + 20 = 4

Vậy A= 4 khi x =16