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Do \(\left|a\right|\ge0\) nên:
a) \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\ge0\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\right)=101x\) (100 số hạng x)
\(\Leftrightarrow100x+5050=101x\Leftrightarrow201x=5050\Leftrightarrow x=\frac{5050}{201}\)
b) Đề sai nhé!
+) \(40\div\left\{\left[11+\left(26-27\right)\right]\times2\right\}\)
\(=40\div\left(10\times2\right)=2\)
+) \(\left|\frac{3}{5}\right|\times4\frac{1}{7}+\left|\frac{-6}{7}\right|\times\frac{3}{5}\)
\(=\frac{3}{5}\times\frac{29}{7}+\frac{6}{7}\times\frac{3}{5}=\frac{3}{5}\times\left(\frac{29}{7}+\frac{6}{7}\right)=\frac{3}{5}\times5=3\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{9}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{8}{9}\)
\(A=\frac{1}{9}\)
\(\Rightarrow\)A= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}.\frac{7}{8}\frac{8}{9}\)
\(\Rightarrow\)A=\(\frac{1.2.3.4.5.6.7.8}{2.3.4.5.6.7.8.9}\)
\(\Rightarrow\)A=\(\frac{1}{9}\)
HỌC TỐT!!!
Dùng tích chất kết hợp cho nó lẹ
a/\(\left(\frac{-2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(\frac{-1}{3}+\frac{4}{7}\right):\frac{4}{5}=\left(\frac{-2}{3}+\frac{3}{7}+\frac{-1}{3}+\frac{4}{7}\right):\frac{4}{5}=\left(-1+1\right):\frac{4}{5}=0\)
b/\(\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{2}{3}\right)=\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}+\frac{1}{15}-\frac{2}{3}\right)=\frac{5}{9}:\left(\frac{-3}{22}+\frac{-3}{5}\right)=\frac{-5}{3\left(\frac{1}{22}+\frac{1}{5}\right)}=\frac{-550}{81}\)
Mà hình như câu b mình làm sai
b/\(\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}\right)+\frac{5}{9}:\left(\frac{1}{15}-\frac{2}{3}\right)=\frac{5}{9}:\frac{-3}{22}+\frac{5}{9}:\frac{-3}{5}=\frac{5.22}{9.-3}+\frac{5.5}{9.-3}=\frac{-\left(5.22+5.5\right)}{27}=-5\)
\(\text{A}=\left(1\times2\right)^{-1}+\left(2\times3\right)^{-1}+(3\times4)^{-1}+...+\left(9\times10\right)^{-1}\)
\(\text{A}=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)
\(\text{A}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(\text{A}=1-\frac{1}{10}=\frac{9}{10}\).