\(\frac{3}{5}\)+ \(\frac{4}{7}\)+ \(\frac{6}{11}\)=
giải cả cách làm hộ mình nhé .
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\(\frac{3}{4}+\frac{2}{3}-\frac{1}{6}\)
\(=\frac{3}{4}+\frac{2}{3}\)
\(=\frac{17}{12}\)
\(=\frac{17}{12}-\frac{1}{6}\)
\(=\frac{90}{72}\)
Ta có : \(\frac{\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}}{\frac{6}{5}+\frac{6}{7}-\frac{2}{3}+\frac{6}{11}}=\frac{\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}}{2\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{3}+\frac{3}{11}\right)}=\frac{1}{2}\)
Lại có : \(\frac{\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{20}\right).2021}{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}}=\frac{0.2021}{\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}}=0\)
Khi đó \(B=\frac{1}{2}+0=\frac{1}{2}\)
Ta có :
\(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\)\(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(=\)\(\frac{2}{7}-\frac{1}{\frac{7}{2}}\)
\(=\)\(\frac{2}{7}-\frac{2}{7}\)
\(=\)\(0\)
Chúc bạn học tốt ~
\(\frac{17}{2}-\left|2x-\frac{5}{2}\right|=-\frac{7}{6}\)
\(\left|2x-\frac{5}{2}\right|=\frac{17}{2}-\frac{-7}{6}\)
\(\left|2x-\frac{5}{2}\right|=\frac{51}{6}+\frac{7}{6}\)
\(\left|2x-\frac{5}{2}\right|=\frac{29}{3}\)
\(2x-\frac{5}{2}=\frac{29}{3}\)hoặc \(2x-\frac{5}{2}=\frac{-29}{3}\)
Trường hợp 1:
\(2x-\frac{5}{2}=\frac{29}{3}\)
\(2x=\frac{29}{3}+\frac{5}{2}\)
\(2x=\frac{73}{6}\)
\(x=\frac{73}{6}:2\)
\(x=\frac{73}{12}\)
Trường hợp 2:
\(2x-\frac{5}{2}=\frac{-29}{3}\)
\(2x=\frac{-29}{3}+\frac{5}{2}\)
\(2x=\frac{-43}{6}\)
\(x=\frac{-43}{6}:2\)
\(x=\frac{-43}{12}\)
Vậy \(x=\frac{73}{12}\)hoặc \(x=\frac{-43}{12}\)
\(a,\left(\frac{6}{11}+\frac{5}{11}\right)\times\frac{3}{7}\)
Cách 1: \(\left(\frac{6}{11}+\frac{5}{11}\right)\times\frac{3}{7}=1\times\frac{3}{7}=\frac{10}{7}\)
Cách 2: \(\left(\frac{6}{11}+\frac{5}{11}\right)\times\frac{3}{7}=\frac{6}{11}\times\frac{3}{7}+\frac{5}{11}\times\frac{3}{7}=\frac{18}{77}+\frac{15}{77}=\frac{33}{77}=\frac{3}{7}\)
\(b,\frac{3}{5}\times\frac{7}{9}+\frac{3}{5}\times\frac{2}{9}\)
Cách 1: \(\frac{3}{5}\times\frac{7}{9}+\frac{3}{5}\times\frac{2}{9}=\frac{7}{15}+\frac{2}{15}=\frac{9}{15}\)
Cách 2: \(\frac{3}{5}\times\frac{7}{9}+\frac{3}{5}\times\frac{2}{9}=\frac{3}{5}\times1=\frac{3}{5}\)
P/s: Ý B có vấn đề thì phải
\(a,\left(\frac{6}{11}+\frac{5}{11}\right)x\frac{3}{7}\)
\(=\frac{11}{11}=1x\frac{3}{7}\)
\(=\frac{3}{7}\)
cach 2
\(\frac{6}{11}x\frac{3}{7}+\frac{5}{11}x\frac{3}{7}\)
\(=\frac{18}{77}+\frac{15}{77}\)
\(=\frac{33}{77}=\frac{3}{7}\)
\(b,\frac{3}{5}x\frac{7}{9}+\frac{3}{5}x\frac{2}{9}\)
\(=\frac{21}{45}+\frac{6}{45}\)
\(=\frac{27}{45}=\frac{3}{5}\)
cách 2 :
\(\frac{3}{5}x\left(\frac{7}{9}+\frac{2}{9}\right)\)
\(=\frac{3}{5}x\frac{9}{9}\)
\(=\frac{27}{45}=\frac{3}{5}\)
A=\(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+\frac{3}{5}-\frac{1}{3}\)
A=[ \(\frac{1}{3}-\frac{1}{3}\)] + [ \(-\frac{3}{5}+\frac{3}{5}\)] + [ \(-\frac{5}{7}+\frac{5}{7}\)] + [ \(-\frac{7}{9}+\frac{7}{9}\)] + [ \(-\frac{9}{11}+\frac{9}{11}\)] \(-\frac{11}{13}\)
Các bạn tự làm tiếp nhé!Sorry
3/5+4/7+6/11=41/35+6/11
=661/385
3/5 + 4/7 + 6/11 = 661/385