\(\text{a)Để C đạt GTNN}\)

\(\Rightarrow\hept{\begin{cases}\left(x+2\right)^2\\\left(y-\frac{1}{5}\right)^2\end{cases}\ge0}\)

\(\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2\ge0\)

\(\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\ge0-10\)

\(\Rightarrow C\ge-10\)

\(\text{Vậy minC=-10 khi x=-2;y= }\frac{1}{5}\)

b)\(\text{Để D đạt GTLN}\)

=>(2x-3)²+5 đạt GTNN

Mà (2x-3)²\(\ge\)5

\(\Rightarrow GTLN\)của \(A=\frac{4}{5}\)khi \(x=\frac{3}{2}\)

Pk tìm GTLN chứ

Ta có: \(\left|5x+7\right|\ge0\)

\(\Rightarrow4\left|5x+7\right|\ge0\)

\(\Rightarrow4\left|5x+7\right|+24\ge24\)

\(\Rightarrow\frac{-8}{4\left|5x+7\right|+24}\le\frac{-1}{3}\)

\(\Rightarrow5+\frac{-8}{4\left|5x+7\right|+24}\le\frac{14}{3}\)

Vậy Amax_{\(=\frac{14}{3}\Leftrightarrow5x+7=0\Leftrightarrow x=\frac{-7}{5}\)}

ko ghi lại đề 

\(C=\frac{-15|x+7|}{3|x+7|}\)

\(C=\frac{-15}{3}+\frac{-68}{12}\)

\(C=\frac{-15}{3}+\frac{-17}{3}\)

\(C=\frac{-32}{3}\)

Vì \(\hept{\begin{cases}3\left|x-3y\right|\ge0\\\left|2x+1\right|\ge0\end{cases}}\)

\(\Rightarrow3\left|x-3y\right|+\left|2x+1\right|+35\ge35\)

\(\Rightarrow\frac{28}{3\left|x-3y\right|+\left|2x+1\right|}\le\frac{4}{5}\)

\(\Rightarrow-\frac{28}{3\left|x-3y\right|+\left|2x+1\right|}\ge-\frac{4}{5}\)

\(\Rightarrow\frac{15}{12}-\frac{28}{3\left|x-3y\right|+\left|2x+1\right|}\ge\frac{9}{20}\)

Vậy \(C_{min}=\frac{9}{20}\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=\frac{-1}{6}\end{cases}}\)

Chỉ tìm được với điều kiện x;a;b dương, còn bất kì thì chắc là chịu

\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)

\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)

\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)

\(=3-\left(-1\right)\)

\(=4\)

b)   \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)

       \(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)

     \(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)

      \(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)

    \(=\frac{199}{16}:\left(12-2\right)\)

\(=\frac{199}{16}:10\)

\(=\frac{199}{160}\)

c)   \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)

\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)

\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)

\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)

\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)

giờ mk phải đi ngủ r mai mk làm cho 

nhớ làm giúp mình nhá mai mình phải đi học r :<

khi x>0, ta có

x - 38/7*x - 3/4 = 2*x + (-8/7)

-45/7*x=-11/28

x=-11/180( ko thoả )

khi x<0 có

-x -38/7x - 3/4 = -2x -8/7

bạn​ tự​ giải​ nhé​ rồi​ ktra dđiều​ kiện​ nhé !

 Cũng dễ

Bạn chỉ cần xét từng trường hợp thôi

bạn giải ra đi!    

\(\frac{7}{8}.(\frac{2}{12}+\frac{4}{10})\)

\(\Rightarrow\frac{7}{8}.(\frac{10+24}{60})\)

\(\Rightarrow\frac{7}{8}.\frac{34}{60}=\frac{238}{480}\)

bt2

\(2.x-\frac{5}{4}=\frac{20}{15}\)

\(\Leftrightarrow2x=\frac{20}{15}+\frac{5}{4}\)

\(\Leftrightarrow2x=\frac{80+75}{60}\)

\(\Leftrightarrow2x=2,5\)

\(\Leftrightarrow x=1,25\)

.7/8.(1/6+2/5)=7/8.17/30=119/240

3/2-5/6:1/4+\(\sqrt{4}\)=3/2-10/3+2=1/6

2x=20/15+5/4

2x=31/12

x=31/12:2

x=31/24

ko bt nha thông cảm

(2x+9)/(x+1)(x+8)-(2x+15)/(x+8)(x+7)+(2x+10)/(x+7)(x+3)=4/3

(x+1+x+8)/(x+1)(x+8)-(x+8+x+7)/(x+8)(x+7)+(x+7+x+3)/(x+7)(x+3)=4/3

1/(x+8)+1/(x+1)-1/(x+7)-1/(x+8)+1/(x+7)+1/(x+3)=4/3

1/(x+1)+1/(x+3)=4/3

(x+3+x+1)/(x+3)(x+1)=4/3

(2x+4)/(x+3)(x+1)=4/3

=>(2x+4).3=(x+3)(x+1).4

6(x+2)=4(x+3)(x+1)

3(x+2)=2(x+3)(x+1)

3x+6=2(x^2+4x+3)

3x+6=2x^2+8x+6

2x^2+8x+6-3x-6=0

2x^2+5x=0

x(2x+5)=0

=> x=0 hoac 2x+5=0

=> x=0 hoac x=-5/2