a, Vì \(\left|\frac{2}{5}-x\right|\ge0\Rightarrow\frac{5}{2}\left|\frac{2}{5}-x\right|\ge0\Rightarrow-\frac{5}{2}\left|\frac{2}{5}-x\right|\Rightarrow D=3-\frac{5}{2}\left|\frac{2}{5}-x\right|\le3\)

Dấu "=" xảy ra khi \(\frac{5}{2}\left|\frac{2}{5}-x\right|=0\Rightarrow x=\frac{2}{5}\)

Vậy GTLN của D = 3 khi x = 2/5

b, Vì \(\left|\frac{5}{3}-x\right|\ge0\Rightarrow P=-\left|\frac{5}{3}-x\right|\le0\)

Dấu "=' xảy ra khi x = 5/3

VẬy GTLN của P = 0 khi x = 5/3

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

           Vì \(-\left|x+\frac{3}{2}\right|\)\(\le\)0

        Suy ra:\(\frac{1}{4}-\left|x+\frac{3}{2}\right|\le\frac{1}{4}\)

      Dấu = xảy ra khi \(x+\frac{3}{2}=0\)

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

Vậy Max A=\(\frac{1}{4}\) khi \(x=-\frac{3}{2}\)

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

        Vì \(-\left|x-\frac{4}{3}\right|\le0;-\left|y+\frac{1}{2}\right|\le0\)

               Suy ra:\(\frac{5}{3}-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\le\frac{5}{3}\)

     Dấu = xảy ra khi \(x-\frac{4}{3}=0;x=\frac{4}{3}\)

                                 \(y+\frac{1}{2}=0;y=-\frac{1}{2}\)

Vậy Max B=\(\frac{5}{3}\) khi \(x=\frac{4}{3};y=-\frac{1}{2}\)

a/ Ta có ; \(\left|x+\frac{3}{2}\right|\ge0\Rightarrow-\left|x+\frac{3}{2}\right|\le0\Rightarrow\frac{1}{4}-\left|x+\frac{3}{2}\right|\le\frac{1}{4}\)

Vậy BT đạt giá trị lớn nhất bằng 1/4 khi x = -3/2

b/ \(\begin{cases}\left|x-\frac{4}{3}\right|\ge0\\\left|y+\frac{1}{2}\right|\ge0\end{cases}\) \(\Rightarrow\begin{cases}-\left|x-\frac{4}{3}\right|\le0\\-\left|y+\frac{1}{2}\right|\le0\end{cases}\) 

\(\Rightarrow-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\le0\)

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

Vậy BT đạt giá trị lớn nhất bằng 5/3 khi x = 4/3 , y = -1/2

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

Ta thấy: \(\left|x+\frac{3}{2}\right|\ge0\)

\(\Rightarrow-\left|x+\frac{3}{2}\right|\le0\)

\(\Rightarrow\frac{1}{4}-\left|x+\frac{3}{2}\right|\le\frac{1}{4}-0=\frac{1}{4}\)

\(\Rightarrow A\le\frac{1}{4}\)

Dấu = khi \(x=-\frac{3}{2}\)

Vậy MaxA=\(\frac{1}{4}\Leftrightarrow x=-\frac{3}{2}\)

b)Đặt \(B=\frac{5}{3}-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\)

Ta thấy: \(\begin{cases}\left|x-\frac{4}{3}\right|\\\left|y+\frac{1}{2}\right|\end{cases}\ge0\)

\(\Rightarrow\begin{cases}-\left|x-\frac{4}{3}\right|\\-\left|y+\frac{1}{2}\right|\end{cases}\)\(\le0\)

\(\Rightarrow-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\le0\)

\(\Rightarrow\frac{5}{3}-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\le\frac{5}{3}-0=\frac{5}{3}\)

\(\Rightarrow B\le\frac{5}{3}\)

Dấu = khi \(\begin{cases}x=\frac{4}{3}\\y=-\frac{1}{2}\end{cases}\)

Vậy MaxB=\(\frac{5}{3}\Leftrightarrow\)\(\begin{cases}x=\frac{4}{3}\\y=-\frac{1}{2}\end{cases}\)

a) Ta có: \(\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2\ge0\)(với mọi x,y)

=>\(C=\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\ge-10\)

Dấu "=" xảy ra khi x=-2;y=1/5

Vậy GTNN của C là -10 tại x=-2;y=1/5

b)Ta có: \(\left(2x-3\right)^2\ge0\Rightarrow\left(2x-3\right)^2+5\ge0\Rightarrow D=\frac{4}{\left(2x-3\right)^2+5}\le\frac{4}{5}\)

Dấu "=" xảy ra khi: x=3/2

Vậy GTLN của D là : 4/5 tại x=3/2

b)B có GTLN <=> (2x-3)²+5 có GTNN

Vì (2x-3)² > 0 với mọi x

=>(2x-3)²+5 > 5 với mọi x

=>GTNN của (2x-3)²+5 là 5

=>D = \(\frac{4}{\left(2x-3\right)^2+5}\) < \(\frac{4}{5}\)

=>GTLN của D là 4/5

Dấu "=" xảy ra <=> (2x-3)²=0<=>x=3/2

Vậy..............

sau 3 phút có kết quả tuy bạn http://olm.vn/hoi-dap/question/772291.html

A) \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}\)

\(=\frac{\left(x+10\right)-\left(x+3\right)}{\left(x+3\right)\left(x+10\right)}+\frac{\left(x+21\right)-\left(x+10\right)}{\left(x+10\right)\left(x+21\right)}+\frac{\left(x+34\right)-\left(x+21\right)}{\left(x+21\right)\left(x+34\right)}\)

\(=\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}\)

\(=\frac{1}{x+3}-\frac{1}{x+34}\)

\(=\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}\)\(=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Rightarrow\left(x+34\right)-\left(x+3\right)=x\)

\(\Rightarrow x=31\)

Vậy, x = 31 

Bạn áp dụng: \(\frac{k}{x\cdot\left(x+k\right)}=\frac{1}{x}-\frac{1}{x+k}\) với    \(x,k\inℝ;x\ne0;x\ne-k\)

Chứng minh: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)

B) \(\frac{\left(x-4\right)-\left(x-7\right)}{\left(x-7\right)\left(x-4\right)}+\frac{\left(x-7\right)-\left(x-13\right)}{\left(x-13\right)\left(x-7\right)}+\frac{\left(x-13\right)-\left(x-28\right)}{\left(x-28\right)\left(x-13\right)}\)

\(=\frac{1}{x-7}-\frac{1}{x-4}+\frac{1}{x-13}-\frac{1}{x-7}+\frac{1}{x-28}-\frac{1}{x-13}\)

\(=\frac{1}{x-28}-\frac{1}{x-4}=-\frac{5}{2}+\frac{1}{x-28}\)

\(\Leftrightarrow\frac{1}{x-28}-\frac{1}{x-4}-\frac{1}{x-28}=-\frac{5}{2}\)

\(\Leftrightarrow\frac{1}{x-4}=\frac{5}{2}\)

=> 5x - 20 = 2

=> 5x = 22 

\(\Rightarrow x=\frac{22}{5}=4,4\)

Vậy, x = 4,4