`A=(8 2/7-4 2/7)-3 4/9`

`=8+2/7-4-2/7-3-4/9`

`=4-3-4/9`

`=1-4/9=5/9`

`B=(10 2/9-6 2/9)+2 3/5`

`=10+2/9-6-2/9+2+3/5`

`=4+2+3/5`

`=6+3/5=33/5`

Bài 2:

`a)5 1/2*3 1/4`

`=11/2*13/4`

`=143/8`

`b)6 1/3:4 2/9`

`=19/3:38/9`

`=19/3*9/38=3/2`

`c)4 3/7*2`

`=31/7*2`

`=62/7`

Bài 1:

\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\) 

\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\) 

\(A=4-\dfrac{31}{9}\) 

\(A=\dfrac{5}{9}\) 

\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\) 

\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\) 

\(B=4+\dfrac{13}{5}\) 

\(B=\dfrac{33}{5}\)

Bài 2:

a) \(5\dfrac{1}{2}.3\dfrac{1}{4}=\dfrac{11}{2}.\dfrac{13}{4}=\dfrac{11.13}{2.4}=\dfrac{143}{8}\) 

b) \(6\dfrac{1}{3}:4\dfrac{2}{9}=\dfrac{19}{3}:\dfrac{38}{9}=\dfrac{19}{3}.\dfrac{9}{38}=\dfrac{3}{2}\) 

c) \(4\dfrac{3}{7}.2=\dfrac{31}{7}.2=\dfrac{31.2}{7}=\dfrac{62}{7}\)

Mik làm Bài 2 nhé ~

Bài 2 :

a) \(x-\dfrac{1}{2}=-\dfrac{1}{10}\)

\(x=-\dfrac{1}{10}+\dfrac{1}{2}\)

\(x=\dfrac{2}{5}\)

b) \(\dfrac{2}{3}x-\dfrac{7}{6}=\dfrac{5}{2}\)

\(\dfrac{2}{3}x=\dfrac{5}{2}+\dfrac{7}{6}\)

\(\dfrac{2}{3}x=\dfrac{11}{3}\)

\(x=\dfrac{11}{3}:\dfrac{2}{3}\)

\(x=\dfrac{11}{3}.\dfrac{3}{2}\)

\(x=\dfrac{11}{2}\)

c) \(2,5-\left(\dfrac{1}{8}x+\dfrac{1}{2}\right)=\dfrac{3}{4}\)

\(\left(\dfrac{1}{8}x+\dfrac{1}{2}\right)=2,5-\dfrac{3}{4}\)

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

\(\dfrac{1}{8}x+\dfrac{1}{2}=\dfrac{7}{4}\)

\(\dfrac{1}{8}x=\dfrac{7}{4}-\dfrac{1}{2}\)

\(\dfrac{1}{8}x=\dfrac{5}{4}\)

\(x=10\)

Bài 1:

a) \(\dfrac{-4}{11}.\dfrac{7}{9}+\dfrac{-4}{11}.\dfrac{2}{9}-\dfrac{7}{11}\) 

\(=\dfrac{-4}{11}.\left(\dfrac{7}{9}+\dfrac{2}{9}\right)-\dfrac{7}{11}\) 

\(=\dfrac{-4}{11}.1-\dfrac{7}{11}\) 

\(=\dfrac{-4}{11}-\dfrac{7}{11}\) 

\(=-1\) 

b) \(\dfrac{3}{5}:\dfrac{-7}{10}+0,5-\left(\dfrac{-9}{14}\right)\) 

\(=\dfrac{-6}{7}+\dfrac{1}{2}+\dfrac{9}{14}\) 

\(=\dfrac{2}{7}\) 

c) \(\dfrac{3}{5}-\dfrac{8}{5}:\left(5,25+75\%\right)\) 

\(=\dfrac{3}{5}-\dfrac{8}{5}:\left(\dfrac{21}{4}+\dfrac{3}{4}\right)\) 

\(=\dfrac{3}{5}-\dfrac{8}{5}:6\) 

\(=\dfrac{3}{5}-\dfrac{4}{15}\) 

\(=\dfrac{1}{3}\)

Bài 2:

a) \(x-\dfrac{1}{2}=\dfrac{-1}{10}\) 

            \(x=\dfrac{-1}{10}+\dfrac{1}{2}\) 

            \(x=\dfrac{2}{5}\) 

b) \(\dfrac{2}{3}x-\dfrac{7}{6}=\dfrac{5}{2}\) 

            \(\dfrac{2}{3}x=\dfrac{5}{2}+\dfrac{7}{6}\) 

            \(\dfrac{2}{3}x=\dfrac{11}{3}\) 

               \(x=\dfrac{11}{3}:\dfrac{2}{3}\) 

               \(x=\dfrac{11}{2}\) 

c) \(2,5-\left(\dfrac{1}{8}x+\dfrac{1}{2}\right)=\dfrac{3}{4}\) 

                   \(\dfrac{1}{8}x+\dfrac{1}{2}=\dfrac{5}{2}-\dfrac{3}{4}\) 

                   \(\dfrac{1}{8}x+\dfrac{1}{2}=\dfrac{7}{4}\) 

                          \(\dfrac{1}{8}x=\dfrac{7}{4}-\dfrac{1}{2}\) 

                          \(\dfrac{1}{8}x=\dfrac{5}{4}\) 

                             \(x=\dfrac{5}{4}:\dfrac{1}{8}\) 

                             \(x=10\)

a: \(=\dfrac{14-2+9}{32}\cdot\dfrac{4}{5}=\dfrac{21}{5}\cdot\dfrac{1}{8}=\dfrac{21}{40}\)

b: \(=10+\dfrac{2}{9}+2+\dfrac{3}{5}+6+\dfrac{2}{9}=18+\dfrac{47}{45}=\dfrac{857}{45}\)

c: \(=\dfrac{3}{10}-\dfrac{12}{5}+\dfrac{1}{10}=\dfrac{4}{10}-\dfrac{12}{5}=\dfrac{2}{5}-\dfrac{12}{5}=-2\)

d: \(=\dfrac{-25}{30}\left(\dfrac{37}{44}+\dfrac{13}{44}-\dfrac{6}{44}\right)=\dfrac{-25}{30}\cdot1=-\dfrac{5}{6}\)

\(a,\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right).\)

\(=\left(\dfrac{21}{60}+\dfrac{44}{60}-\dfrac{75}{60}\right):\left(\dfrac{99}{180}-\dfrac{104}{180}\right).\)

\(=\left(\dfrac{65}{60}-\dfrac{75}{60}\right):\left(-\dfrac{5}{180}\right).\)

\(=-\dfrac{10}{60}:\left(-\dfrac{5}{180}\right).\)

\(=-\dfrac{1}{6}:\left(-\dfrac{1}{36}\right).\)

\(=-\dfrac{1}{6}.\left(-36\right).\)

\(=\dfrac{-1.\left(-36\right)}{6}=\dfrac{36}{6}=6.\)

Vậy......

\(b,\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}.\)

\(=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}:\dfrac{15\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}{16\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}.\)

\(=\dfrac{5}{8}:\dfrac{15}{16}.\)

\(=\dfrac{5}{8}.\dfrac{16}{15}=\dfrac{5.16}{8.15}=\dfrac{1.2}{1.3}=\dfrac{2}{3}.\)

Vậy......

c, (làm tương tự câu b).

~ Học tốt!!! ~

Câu a)

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\geq \frac{9}{a+2b}\) (1)

\(\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\geq \frac{9}{b+2c}\)(2)

\(\frac{1}{c}+\frac{1}{a}+\frac{1}{a}\geq \frac{9}{c+2a}\) (3)

Lấy \((1)+2.(2)+3.(3)\) ta có:

\(\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{2}{b}+\frac{2}{c}+\frac{2}{c}+\frac{3}{c}+\frac{3}{a}+\frac{3}{a}\geq 9\left(\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\right)\)

\(\Leftrightarrow \frac{7}{a}+\frac{4}{b}+\frac{7}{c}\geq 9\left(\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\right)\)

Ta có đpcm

Dấu bằng xảy ra khi \(a=b=c\)

Câu b)

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{1}{a}+\frac{4}{b}\geq \frac{(1+2)^2}{a+b}=\frac{9}{a+b}\)

\(\Rightarrow \frac{1}{3a}+\frac{4}{3b}\geq \frac{3}{a+b}(1)\)

\(\frac{1}{3b}+\frac{1}{2c}+\frac{1}{2c}\geq \frac{9}{3b+4c}\)

\(\Rightarrow \frac{2}{3b}+\frac{2}{c}\geq \frac{18}{3b+4c}\) (2)

\(\frac{1}{c}+\frac{1}{3a}+\frac{1}{3a}\geq \frac{9}{c+6a}\) (3)

Từ (1); (2); (3) cộng theo vế:

\(\Rightarrow \frac{1}{a}+\frac{2}{b}+\frac{3}{c}\geq \frac{3}{a+b}+\frac{18}{3b+4c}+\frac{9}{c+6a}\)

(đpcm)

Dấu bằng xảy ra khi \(a=\frac{b}{2}=\frac{c}{3}\)

Câu c)

BĐT cần chứng minh tương đương với:
\(\frac{b+c+a}{a}+\frac{2a+c}{b}+\frac{4(a+b)}{a+c}\geq 10\) (*)

Áp dụng BĐT AM-GM:

\(\text{VT}=\frac{b}{a}+\frac{c+a}{2a}+\frac{c+a}{2a}+\frac{a}{b}+\frac{a+c}{2b}+\frac{a+c}{2b}+\frac{a+b}{a+c}+\frac{a+b}{a+c}+\frac{a+b}{a+c}+\frac{a+b}{a+c}\)

\(\geq 10\sqrt[10]{\frac{ba(c+a)^4(a+b)^4}{16a^3b^3(a+c)^4}}=10\sqrt[10]{\frac{(a+b)^4}{16a^2b^2}}\)

Theo AM-GM: \((a+b)^2\geq 4ab\Rightarrow (a+b)^4\geq 16a^2b^2\)

\(\Rightarrow \text{VT}\geq 10\sqrt[10]{\frac{(a+b)^4}{16a^2b^2}}\geq 10\)

Vậy (*) được cm. Ta có đpcm. Dấu bằng xảy ra khi a=b=c

Bài 2: 

a: \(A=11+\dfrac{3}{13}-2-\dfrac{4}{7}-5-\dfrac{3}{13}\)

\(=4-\dfrac{4}{7}=\dfrac{24}{7}\)

b: \(B=6+\dfrac{4}{9}+3+\dfrac{7}{11}-4-\dfrac{4}{9}\)

\(=5+\dfrac{7}{11}=\dfrac{62}{11}\)

c: \(C=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1+\dfrac{5}{7}=1\)

d: \(D=\dfrac{7}{10}\cdot\dfrac{8}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)

\(=\dfrac{20}{10}\cdot7\cdot\dfrac{8}{3}\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}=2\cdot\dfrac{5}{4}=\dfrac{5}{2}\)

a: \(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\)

\(=-\dfrac{1}{10}\)

9<10

=>1/9>1/10

=>\(-\dfrac{1}{9}< -\dfrac{1}{10}\)

=>\(A>-\dfrac{1}{9}\)

b: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{10}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{11}{10}\)

\(=\dfrac{-1}{10}\cdot\dfrac{11}{2}=\dfrac{-11}{20}\)

20<21

=>\(\dfrac{11}{20}>\dfrac{11}{21}\)

=>\(-\dfrac{11}{20}< -\dfrac{11}{21}\)

=>\(B< -\dfrac{11}{21}\)

2:

a: 2/9-x=-5/9

=>x=2/9+5/9=7/9

b: x-7/13=1/2

=>x=1/2+7/13=27/26

câu a

\(\dfrac{7}{4}+\dfrac{3}{2}+\dfrac{-9}{16}\\ =\dfrac{28}{16}+\dfrac{24}{16}-\dfrac{9}{16}=\dfrac{43}{16}\)

câu b

\(-\dfrac{2}{7}+\dfrac{3}{5}+\dfrac{9}{7}+\dfrac{-18}{5}\\ =-\dfrac{10}{35}+\dfrac{21}{35}+\dfrac{45}{35}-\dfrac{126}{35}\\ =-\dfrac{70}{35}=-2\)

câu c

\(-\dfrac{5}{13}+\dfrac{11}{10}-\dfrac{-9}{10}+\dfrac{-8}{13}\\ =-\dfrac{5}{13}+\dfrac{11}{10}+\dfrac{9}{10}-\dfrac{8}{13}\\ =-\dfrac{50}{130}+\dfrac{143}{130}+\dfrac{117}{130}-\dfrac{80}{130}\\ =\dfrac{130}{130}=1\)

bài 2

câu a

\(\dfrac{2}{9}-x=-\dfrac{5}{9}\\ x=\dfrac{2}{9}-\dfrac{-5}{9}\\ x=\dfrac{7}{9}\)

câu b

\(x+\dfrac{-7}{13}=\dfrac{1}{2}\\ x=\dfrac{1}{2}-\dfrac{-7}{13}\\ x=\dfrac{13}{26}+\dfrac{14}{26}\\ x=\dfrac{17}{26}\)