\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)

\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)

\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)

\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)

\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)

\(A=\dfrac{1}{2}-\dfrac{1}{4}\)

\(A=\dfrac{2}{4}-\dfrac{1}{4}\)

\(A=\dfrac{1}{4}\)

\(D=6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)

\(D=\dfrac{77}{12}:\dfrac{13}{4}+\dfrac{45}{4}.\dfrac{2}{15}\)

\(D=\dfrac{77}{39}+\dfrac{3}{2}\)

\(D=\dfrac{271}{78}\)

\(C=\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)

\(C=\dfrac{5}{16}:0,125-\left(\dfrac{9}{4}-0,6\right).\dfrac{10}{11}\)

\(C=\dfrac{5}{16}:0,125-\dfrac{33}{20}.\dfrac{10}{11}\)

\(C=\dfrac{5}{2}-\dfrac{3}{2}\)

\(C=1\)

a) \(\left(\frac{7}{8}-\frac{3}{4}\right)\cdot\frac{1}{3}-\frac{2}{7}\cdot\left(3,5\right)=\left(\frac{7}{8}-\frac{3}{4}\right)\cdot\frac{1}{3}-\frac{2}{7}\cdot\frac{7}{2}\)

\(=\left(\frac{7}{8}-\frac{6}{8}\right)\cdot\frac{1}{3}-1=\frac{1}{8}\cdot\frac{1}{3}-1=\frac{1}{24}-\frac{24}{24}=-\frac{23}{24}\)

b) \(\left(\frac{3}{5}+0,415-\frac{3}{200}\right)\cdot2\frac{2}{3}\cdot0,25\)

\(=\left(\frac{3}{5}+\frac{83}{200}-\frac{3}{200}\right)\cdot\frac{8}{3}\cdot\frac{1}{4}\)

\(=\left(\frac{3}{5}+\frac{80}{200}\right)\cdot\frac{8}{3}\cdot\frac{1}{4}=\left(\frac{3}{5}+\frac{2}{5}\right)\cdot\frac{8}{3}\cdot\frac{1}{4}=1\cdot\frac{8}{3}\cdot\frac{1}{4}=1\cdot\frac{2}{3}\cdot\frac{1}{1}=\frac{2}{3}\)

c) \(\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right)\cdot\frac{10}{11}\)

\(=\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right)\cdot\frac{10}{11}\)

\(=\frac{5}{16}\cdot8-\frac{33}{20}\cdot\frac{10}{11}=\frac{5}{2}-\frac{3}{2}=1\)

d) \(0,25:\left(10,3-9,8\right)-\frac{3}{4}=\frac{1}{4}:\left(\frac{103}{10}-\frac{98}{10}\right)-\frac{3}{4}\)

\(=\frac{1}{4}:\frac{1}{2}-\frac{3}{4}=\frac{1}{4}\cdot2-\frac{3}{4}=\frac{2}{4}-\frac{3}{4}=-\frac{1}{4}\)

Câu cuối tương tự

              Bài làm :

\(\text{a)}=2,5-1,65.\frac{10}{11}=2,5-1.5=1\)

\(b\text{)}=\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{2015-2012}{2012.2015}\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-...+\frac{1}{2012}-\frac{1}{2015}\)

\(=\frac{1}{5}-\frac{1}{2015}\)

\(=\frac{402}{2015}\)

\(a,\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right).\frac{10}{11}\)

\(=\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right).\frac{10}{11}\)

 \(=\frac{5}{2}-\frac{33}{20}.\frac{10}{11}\)

\(=\frac{5}{2}-\frac{3}{2}=\frac{2}{2}=1\)

Bài 1:

5/16:0,125-(2 1/4-0,6).10/11

=5/16:5/4-(9/4-3/5).10/11

=1/4-33/20.10/11

=1/4-3/2

=-5/4

\(\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right).\frac{10}{11}\)

\(=\frac{5}{16}.8-\frac{33}{20}.\frac{10}{11}\)

\(=\frac{5}{2}-\frac{3}{2}=\frac{2}{2}=1\)

<br class="Apple-interchange-newline"><div></div>49823 −(5732 +14823 )

\(49\frac{8}{23}-19\frac{417}{736}\)

\(=\frac{953}{12}\)

Study well 

\(\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right).\frac{10}{11}\)ư

= \(\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right).\frac{10}{11}\)

= \(\frac{5}{16}.8-\left(\frac{9}{4}-\frac{3}{5}\right).\frac{10}{11}\)

= \(\frac{5}{2}-\left(\frac{45}{20}-\frac{12}{20}\right).\frac{10}{11}\)

= \(\frac{5}{2}-\frac{33}{20}.\frac{10}{11}\)

= \(\frac{5}{2}-\frac{3}{2}\)

= \(1\)

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

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

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

= \(35-\frac{167}{32}\)

= \(\frac{1120}{32}-\frac{167}{32}\)

= \(\frac{953}{32}\)

Chúc bạn học tốt !!! 

Bài 8:

a: \(\left(\dfrac{2}{5}+\dfrac{3}{4}\right)^2=\left(\dfrac{8+15}{20}\right)^2=\left(\dfrac{23}{20}\right)^2=\dfrac{529}{400}\)

b: \(\left(\dfrac{5}{4}-\dfrac{1}{6}\right)^2=\left(\dfrac{15}{12}-\dfrac{2}{12}\right)^2=\left(\dfrac{13}{12}\right)^2=\dfrac{169}{144}\)

`a)1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10`

`=(1/10+9/10)+(2/10+8/10)+(3/10+7/10)+(4/10+6/10)+5/10`

`=10/10 + 10/10+10/10+10/10+1/2`

`=1+1+1+1+1/2`

`=2+2+1/2`

`=4+1/2`

`=8/2+1/2`

`=9/2`

__

`13,25:0,5+13,25:0,25+13,25:0,125+13,25×6`

`=13,25×1/(0,5)+13,25×1/(0,25)+13,25×1/(0,125)+13,25×6`

`=13,25×(1/(0,5)+1/(0,25)+1/(0,125)+6)`

`=13,25×(2+4+8+6)`

`=13,25×20`

`=265`

`#QiN`

\(a,\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right)\cdot\frac{10}{11}\)

\(=\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right)\cdot\frac{10}{11}\)

\(=\frac{5}{2}-\frac{33}{20}\cdot\frac{10}{11}\)

\(=\frac{5}{2}-\frac{3}{2}\)

\(=1\)

\(\frac{5}{16}:0,125-\left(2\frac{1}{4}-0,6\right).\frac{10}{11}\)

\(\Rightarrow\frac{5}{16}:\frac{125}{1000}-\left(\frac{9}{4}-\frac{6}{10}\right).\frac{10}{11}\)

\(\Rightarrow\frac{5}{16}:\frac{1}{8}-\left(\frac{9}{4}-\frac{3}{5}\right).\frac{10}{11}\)

\(\Rightarrow\frac{5}{16}.\frac{8}{1}-\left(\frac{45}{20}-\frac{12}{20}\right).\frac{10}{11}\)

\(\Rightarrow\frac{5}{2}-\frac{33}{20}.\frac{10}{11}\)

\(\Rightarrow\frac{5}{2}-\frac{3.1}{2.1}\)

\(\Rightarrow\frac{5}{2}-\frac{3}{2}=\frac{2}{2}=1\)

b: \(x+\dfrac{5}{6}=\dfrac{3}{8}\)

=>\(x=\dfrac{3}{8}-\dfrac{5}{6}\)

=>\(x=\dfrac{9}{24}-\dfrac{20}{24}=-\dfrac{11}{24}\)

c: \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{5}{6}\)

=>\(\dfrac{2}{3}x=\dfrac{5}{6}+\dfrac{1}{2}=\dfrac{8}{6}=\dfrac{4}{3}\)

=>2x=4

=>x=4/2=2

d: \(\dfrac{x}{7}=\dfrac{6}{-21}\)

=>\(\dfrac{x}{7}=\dfrac{-2}{7}\)

=>x=-2

e: \(\left(\dfrac{7}{3}x-0,6\right):\dfrac{3^2}{5}=1\)

=>\(\dfrac{7}{3}x-0,6=\dfrac{3^2}{5}=1,8\)

=>\(\dfrac{7}{3}x=2,4\)

=>\(x=2,4:\dfrac{7}{3}=2.4\cdot\dfrac{3}{7}=\dfrac{7.2}{7}=\dfrac{36}{35}\)

f: \(\dfrac{x}{45}=\dfrac{5}{6}+\dfrac{-29}{30}\)

=>\(\dfrac{x}{45}=\dfrac{25}{30}-\dfrac{29}{30}=-\dfrac{4}{30}=-\dfrac{2}{15}\)

=>\(x=-\dfrac{2}{15}\cdot45=-6\)

g: \(\left(4,5-2x\right)\cdot\left(-\dfrac{1^4}{7}\right)=\dfrac{11}{14}\)

=>\(4,5-2x=\dfrac{11}{14}:\dfrac{-1}{7}=\dfrac{-11}{2}\)

=>\(2x=4,5+\dfrac{11}{2}=\dfrac{20}{2}=10\)

=>x=10/2=5

h: \(-\dfrac{2}{7}+\dfrac{4}{7}x=\dfrac{5}{7}\)

=>\(\dfrac{4}{7}x=\dfrac{5}{7}+\dfrac{2}{7}=\dfrac{7}{7}\)

=>4x=7

=>\(x=\dfrac{7}{4}\)