Bài 1:

\(A=\left(\frac{-5}{11}+\frac{7}{22}-\frac{4}{33}-\frac{5}{44}\right):\left(38\frac{1}{122}-39\frac{7}{22}\right)\)

\(=\frac{-49}{132}:\left(-\frac{879}{671}\right)=\frac{2989}{105408}\)

Bài 2:

\(\frac{4}{5}-\left(\frac{-1}{8}\right)=\frac{7}{8}-x\)

<=>  \(\frac{7}{8}-x=\frac{27}{40}\)

<=>  \(x=\frac{7}{8}-\frac{27}{40}=\frac{1}{5}\)

Vậy...

bài 2 mình tính sai, sửa

.......

<=>  \(\frac{7}{8}-x=\frac{37}{40}\)

<=>  \(x=\frac{7}{8}-\frac{37}{40}=\frac{-1}{20}\)

Vậy....

\(\frac{4}{5}-\frac{-1}{8}=\frac{7}{8}-x\)

\(\Leftrightarrow x=\frac{7}{8}-\frac{4}{5}-\frac{1}{8}\)

\(\Leftrightarrow x=\frac{-1}{20}\)

a)  \(3\left(4-2x\right)-2\left(x+3\right)=12-7x\)

\(\Leftrightarrow\)\(12-6x-2x-6=12-7x\)

\(\Leftrightarrow\)\(6-8x=12-7x\)

\(\Leftrightarrow\)\(x=-6\)

Vậy...

b)  \(\left|16+\right|3\left(x-2\right)||-5=20\)

\(\Leftrightarrow\)\(\left|16+\right|3\left(x-2\right)||=25\)(1)

Ta thấy:  \(\left|3\left(x-2\right)\right|\ge0\)\(\Rightarrow\)\(16+\left|3\left(x-2\right)\right|>0\)

nên từ (1)   \(\Rightarrow\)  \(16+\left|3\left(x-2\right)\right|=25\)

                   \(\Leftrightarrow\)\(\left|3\left(x-2\right)\right|=9\)

                   \(\Leftrightarrow\)  \(\orbr{\begin{cases}3\left(x-2\right)=9\\3\left(x-2\right)=-9\end{cases}}\)

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

Vậy....

c)   \(\left|-5-3^2\right|-||3x+5|-7.2^3|=3^9:3^7\)

\(\Leftrightarrow\)\(14-||3x+5|-56|=9\)

\(\Leftrightarrow\)\(||3x+5|-56|=5\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}\left|3x+5\right|-56=5\\\left|3x+5\right|-56=-5\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}\left|3x+5\right|=61\\\left|3x+5\right|=51\end{cases}}\)

đến đây bn giải tiếp nhé

a,32

b,\(-\frac{1}{10}\)

c,-1000000

d,\(\frac{9}{16}\)

a)  \(\left(\frac{2}{5}-\frac{1}{2}\right)^2-\frac{11}{5}:\frac{-11}{5}=\left(-\frac{1}{10}\right)^2+1=1\frac{1}{100}\)

b)  \(\left(-\frac{5}{7}\right)^2+8.\left(0,5\right)^2+\left(-1\right)^{2010}=\frac{25}{49}+2+1=3\frac{25}{49}\)

c)  \(\frac{9999^2}{3333^2}+\left(0,5\right)^2.\left(-2\right)^4-\left(-\frac{4}{3}\right)^2=9+1-\frac{16}{9}=8\frac{2}{9}\)

d) \(\left|-\frac{2}{5}+\frac{1}{7}\right|:\frac{-3}{35}+\frac{-3}{7}.\frac{7}{5}=\frac{9}{35}.\frac{35}{-3}-\frac{3}{5}=-3\frac{3}{5}\)

e) \(\frac{1}{2}-\left(-0,4\right)+\frac{1}{3}+\frac{1}{5}-\frac{-1}{6}+\frac{-4}{35}+\frac{1}{41}\)

\(=\frac{1}{2}+\frac{2}{5}+\frac{1}{3}+\frac{1}{5}+\frac{1}{6}-\frac{4}{35}+\frac{1}{41}=1\frac{732}{1435}\)

a)\(A=12-\left|x-3\right|-\left|y+7\right|\)

\(-\left|x-3\right|\le0;-\left|y+7\right|\le0\)

\(\Rightarrow A\le12-0-0=12\)

Vậy Max A = 12 <=> x = 3 ; y = -7

b)\(B=-\left(x-2018\right)^6-1\)

\(-\left(x-2018\right)^6\le0\)

\(B\le0-1=-1\)

Vậy Max B = -1 <=> x = 2018

a)  \(A=12-\left|x-3\right|-\left|y+7\right|\)

Nhận thấy: \(\left|x-3\right|\ge0;\)\(\left|y+7\right|\ge0\)

suy ra:  \(A=12-\left|x-3\right|-\left|y+7\right|\le12\)

Vậy MIN A = 12

Dấu "=" xảy ra <=> \(x=3;y=-7\)

b) \(B=-\left(x-2018\right)^6-1\)

Nhận thấy:  \(\left(x-2018\right)^6\ge0\)

suy ra:  \(B=-\left(x-2018\right)^2-1\le-1\)

Vậy MIN B = -1

Dấu "=" xảy ra  <=>   \(x=2018\)

c) \(C=\frac{20}{7}-\left|x+8\right|-\left(3y+7\right)^{2016}\)

Nhận thấy:  \(\left|x+8\right|\ge0\)    \(\left(3y+7\right)^{2016}\ge0\)

suy ra:  \(C=\frac{20}{7}-\left|x+8\right|-\left(3y+7\right)^{2016}\le\frac{20}{7}\)

Vậy MIN  C = 20/7

Dấu "=" xảy ra <=>  \(x=-8;y=-\frac{7}{3}\)

xin lỗi mk nhầm, mẫy cái kết luận sửa lại 

MIN thành MAX hộ mk nhé

p/s: xin lỗi bạn T.T

\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)

\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)

\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)

\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)

\(A=\left[35-0\right]-5\frac{7}{32}\)

\(A=35-5\frac{7}{32}\)

\(A=\frac{953}{32}\)

\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)

\(B=71\frac{38}{45}-\frac{36377}{855}\)

\(B=\frac{1670}{57}\)

\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)

\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)

\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)

\(C=\frac{153}{14}:\frac{4}{5}\)

\(C=\frac{765}{56}\)

\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)

\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)

\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)

\(D=0-\frac{1}{4}\)

\(D=-\frac{1}{4}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)

\(\)\(E=\frac{22}{45}\)

CHUC BAN HOC TOT >.<

bạn có ghi nhầm đề ko ?

1/h=1/2(1/a+1/b)=1/2a+1/2b=(a+b)/2ab

=>(a+b/)2ab-1/h=0

quy dong len ta co

(a+b)h/2abh-2ab/2abh=0=> (ah+bh-2ab)/2abh=0 =>ah+bh-2ab=0

                                                                       =>ah+bh-ab-ab=0

                                                                         =>a(h-b)-b(a-h)=0  

                                                                           =>a(h-b)=b(a-h)

                                                                              =>a/b=(a-h)(h-b)