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\(\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{299.302}\)
\(=2\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+..+\frac{3}{299.302}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{299}-\frac{1}{302}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{302}\right)=2.\frac{75}{151}=\frac{150}{151}\)
\(A=\frac{54.107-53}{53.107+54}=\frac{\left(53+1\right).107-53}{53.107+54}\)
\(=\frac{53.107+107-53}{53.107+54}=\frac{53.107+54}{53.107+54}=1\)
\(B=\frac{135.269-133}{134.269+135}=\frac{\left(134+1\right)269-133}{134.269+135}\)
\(=\frac{134.269+269-133}{134.269+135}=\frac{134.269+136}{134.269+135}>1\)
\(\Rightarrow B>A\)
\(A=\frac{54\cdot107-53}{53\cdot107+54}=\frac{\left(53+1\right)107-53}{53\cdot107+54}=\frac{53\cdot107+107-53}{53\cdot107+54}=\frac{53\cdot107+54}{53\cdot107+54}=1\)
\(B=\frac{135\cdot268-133}{134\cdot269+135}=\frac{\left(134+1\right)\cdot268-133}{134\cdot269+135}=\frac{134\cdot268+268-133}{34\cdot269+135}=\frac{134\cdot268+135}{134\cdot269+135}=1\)
Vì 1=1 nên A=B
\(A=\frac{54.107-53}{53.107+54}=\frac{54.107+54-107}{\left(54-1\right).107+54}=\frac{54.\left(107+1\right)-107}{54.107-107+54}=\frac{54.108-107}{54.\left(107+1\right)-107}=\frac{54.108-107}{54.108-107}=1\)
\(B=\frac{135.296-133}{134.269+135}=\frac{135.296+135-268}{\left(135-1\right).269+135}=\frac{135.\left(296+1\right)-268}{135.269-269+135}=\frac{135.297-268}{135.\left(269+1\right)-269}=\frac{135.297-268}{135.270-269}>1\)
\(\Rightarrow A< B\)
\(A=\frac{54.107-53}{53.107+54}=\frac{53.107+107-53}{53.107+54}=\frac{53.107+54}{53.107+54}=1\)
\(B=\frac{135.296-133}{134.269+135}=\frac{134.296+296-133}{134.269+135}=\frac{134.296+163}{134.269+135}\)
Vì \(134=134;296>269\Rightarrow134.269< 134.296\)mà\(163>135\Rightarrow134.269+135< 134.296+163\)hay \(B>1\)
Ta có \(A=1;B>1\Rightarrow A< B\)