mình cũng đang bí bài này

C=3/(1²+2² )+ 11/(2²+3²⁾+....+19799/(99²+100²⁾

Suy ra C có 99 số .

C=(1-2/5)+(1-2/13)+...+(1-2/19801)

C=99-(2/5+2/13+...+2/19801)

bây giờ chúng ta phải chứng minh biểu thức trong ngoặc nhỏ hơn 1.

Xét B=2/5+2/13+...+2/19801

B<2/4+2/12+...+2/19800

B<1/2+1/6+...+1/9900

B<1/1x2+1/2x3+...+1/99x100

B<1-1/100<1

Vậy B < 1 thì C> 99-1=98

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a ) \(-\frac{13}{30}+\frac{11}{20}-\frac{7}{15}\)

\(=-\frac{26}{60}+\frac{33}{60}-\frac{28}{60}\)

\(=\frac{-26+33-28}{60}=-\frac{7}{20}\)

b ) \(-\frac{5}{72}:\left(\frac{3}{8}.\frac{7}{9}\right)\)

 \(=-\frac{5}{72}:\frac{3.7}{8.9}\)

\(=-\frac{5}{72}:\frac{7}{24}\)

\(=-\frac{5}{72}.\frac{24}{7}=-\frac{5}{21}\)

a) \(\frac{-13}{30}+\frac{11}{20}-\frac{7}{15}=\frac{-26}{60}+\frac{33}{60}-\frac{28}{60}=-\frac{21}{60}=-\frac{7}{20}\)

b) \(\frac{-5}{72}:\left(\frac{3}{8}.\frac{7}{9}\right)=\frac{-5}{72}:\frac{7}{24}=-\frac{5}{21}\)

c) \(\frac{-23}{25}.\frac{10}{13}+\frac{-23}{25}.\frac{3}{13}+\frac{-27}{25}\)

\(=\frac{-23}{25}.\left(\frac{10}{13}+\frac{3}{13}\right)+\frac{-27}{25}\)

\(=\frac{-23}{25}+\frac{-27}{25}\)

\(=\frac{-50}{25}=-2\)

a)

\(\begin{array}{l}A = \frac{5}{{11}}.\left( {\frac{{ - 3}}{{23}}} \right).\frac{{11}}{5}.\left( { - 4,6} \right)\\A = \frac{5}{{11}}.\left( {\frac{{ - 3}}{{23}}} \right).\frac{{11}}{5}.\frac{{ - 23}}{5}\\A = \frac{{5.\left( { - 3} \right).11.\left( { - 23} \right)}}{{11.23.5.5}}\\A = \frac{3}{5}\end{array}\)  

b)

\(\begin{array}{l}B = \left( {\frac{{ - 7}}{9}} \right).\frac{{13}}{{25}} - \frac{{13}}{{25}}.\frac{2}{9}\\B = \frac{{13}}{{25}}.\left( {\frac{{ - 7}}{9} - \frac{2}{9}} \right)\\B = \frac{{13}}{{25}}.(-1)\\B = \frac{{-13}}{{25}}.\end{array}\)

1)-73

2)-5

3)-64

4)-56

5)9

6)-20

7)6

8)-2

Tick nha

mk ra kết quả thôi nha

đăng tùng câu  một thui bạn 

bạn làm thế chắc tụi mk xỉu gần hết 

a) \(\dfrac{7}{30}+\dfrac{\left(-12\right)}{37}+\dfrac{23}{30}+\dfrac{\left(-25\right)}{37}=\left(\dfrac{7}{30}+\dfrac{23}{30}\right)+\left(\dfrac{-12}{37}+\dfrac{-25}{37}\right)=1+\left(-1\right)=0\)

b) \(\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}=\dfrac{5}{7}\cdot\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}\cdot\left(-\dfrac{7}{11}\right)=-\dfrac{5}{11}\)

c) \(\dfrac{\left(-5\right)}{7}\cdot\dfrac{3}{13}-\dfrac{5}{7}\cdot\dfrac{10}{13}+1\dfrac{5}{7}=\dfrac{5}{7}\cdot\dfrac{-3}{13}-\dfrac{5}{7}\cdot\dfrac{10}{13}+\dfrac{12}{7}=\dfrac{5}{7}\cdot\left(\dfrac{-3}{13}-\dfrac{10}{13}\right)+\dfrac{12}{7}=\dfrac{5}{7}\cdot\left(-1\right)+\dfrac{12}{7}=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=\dfrac{7}{7}=1\)

câu 1 = -68

câu 2 =-2500

câu 3=168

câu 4 =4571

câu 6=124

Bài 1:

a; \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{7}{21}\) + (- \(\dfrac{10}{36}\) + \(\dfrac{11}{19}\)  + \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)

=  \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{1}{3}\) -\(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\) - \(\dfrac{5}{8}\)

= (\(\dfrac{5}{18}\) - \(\dfrac{10}{36}\)) + (\(\dfrac{8}{19}\) + \(\dfrac{11}{19}\)) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)

= (\(\dfrac{5}{18}\) - \(\dfrac{5}{18}\)) + \(\dfrac{19}{19}\) - 0 - \(\dfrac{5}{8}\)

= 0 + 1 - \(\dfrac{5}{8}\)

= \(\dfrac{3}{8}\)

b; \(\dfrac{1}{13}\) + (\(\dfrac{-5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\)) - (\(\dfrac{12}{17}\) - \(\dfrac{5}{18}\) + \(\dfrac{7}{5}\))

= \(\dfrac{1}{13}\) - \(\dfrac{5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{12}{17}\) + \(\dfrac{5}{18}\) - \(\dfrac{7}{5}\)

= (\(\dfrac{1}{13}\) - \(\dfrac{1}{13}\)) + (\(\dfrac{12}{17}\) - \(\dfrac{12}{17}\)) + (-\(\dfrac{5}{18}\) + \(\dfrac{5}{18}\)) - \(\dfrac{7}{5}\)

= 0 + 0 + 0 - \(\dfrac{7}{5}\)

= - \(\dfrac{7}{5}\)

Bài 1 c;

   \(\dfrac{15}{14}\) - (\(\dfrac{17}{23}\) - \(\dfrac{80}{87}\) + \(\dfrac{5}{4}\)) + (\(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\))

=  \(\dfrac{15}{14}\) - \(\dfrac{17}{23}\) + \(\dfrac{80}{87}\) - \(\dfrac{5}{4}\) + \(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\)

= (\(\dfrac{15}{14}-\dfrac{15}{14}\)) + (\(-\dfrac{17}{23}+\dfrac{17}{23}\)) - (\(\dfrac{5}{4}\) - \(\dfrac{1}{4}\)) + \(\dfrac{80}{87}\)

= 0 + 0 - 1 + \(\dfrac{80}{87}\)

= - \(\dfrac{7}{87}\)

Bài 1 d; 

   \(\dfrac{1}{25}\) - \(\dfrac{4}{27}\) + (-\(\dfrac{23}{27}\) + \(\dfrac{-1}{25}\) - \(\dfrac{5}{43}\)) + \(\dfrac{5}{43}\) - \(\dfrac{4}{7}\)

= \(\dfrac{1}{25}\) - \(\dfrac{4}{27}\) - \(\dfrac{23}{27}\) - \(\dfrac{1}{25}\) - \(\dfrac{5}{43}\) + \(\dfrac{5}{43}\) - \(\dfrac{4}{7}\)

= (\(\dfrac{1}{25}\) - \(\dfrac{1}{25}\)) - (\(\dfrac{4}{27}\) + \(\dfrac{23}{27}\)) - (\(\dfrac{5}{43}\) - \(\dfrac{5}{43}\)) - \(\dfrac{4}{7}\)

= 0 - 1 - 0 - \(\dfrac{4}{7}\)

= \(\dfrac{-11}{7}\)