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a
\(5\frac{4}{7}:x+=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\frac{4}{7}x=\frac{9}{8}-0,125\)
\(\frac{4}{7}x=1\)
\(x=1:\frac{4}{7}\)
\(x=\frac{7}{4}=1\frac{3}{4}\)
a) \(\frac{4}{3}x-1=\frac{x}{5}\)
=> \(\frac{4}{3}x-\frac{1}{5}x=1\)
=> \(\frac{17}{15}x=1\)
=> \(x=1:\frac{17}{15}=\frac{15}{17}\)
b) \(x+50\%=\frac{4\left(x+1\right)}{3}-\frac{1}{3}\)
=> \(x+\frac{1}{2}=\frac{4x+4-1}{3}\)
=> \(\frac{2x+1}{2}=\frac{4x+3}{3}\)
=> \(\left(2x+1\right).3=2.\left(4x+3\right)\)
=> \(6x+3=8x+6\)
=> \(6x-8x=6-3\)
=> \(-2x=3\)
=> \(x=3:\left(-2\right)=-\frac{3}{2}\)
\(a,\frac{4}{3}x-1=\frac{x}{5}\)
\(\Rightarrow\frac{4}{3}x=\frac{x}{5}+1\)
\(\Rightarrow\frac{4x}{3}=\frac{x+5}{5}\)
\(\Rightarrow20x=3x+15\)
\(\Rightarrow17x=15\)
\(\Rightarrow x=\frac{15}{17}\)
\(b,x+50\%=\frac{4\left(x+1\right)}{3}-\frac{1}{3}\)
\(\Rightarrow x+\frac{1}{2}=\frac{4x+4-1}{3}\)
\(\Rightarrow\frac{2x+1}{2}=\frac{4x+3}{3}\)
\(\Rightarrow3\left(2x+1\right)=\left(4x+3\right).2\)
\(\Rightarrow6x+3=8x+6\)
\(\Rightarrow2x=-3\)
\(\Rightarrow x=-\frac{3}{2}\)
\(c,-200\%.x+\frac{4}{3}=\frac{7}{4}\left(x+1\right)\)
\(\Rightarrow-2x+\frac{4}{3}=\frac{7}{4}x+\frac{7}{4}\)
\(\Rightarrow\frac{-6x+4}{3}=\frac{7x+7}{4}\)
\(\Rightarrow4\left(-6x+4\right)=\left(7x+7\right)3\)
\(\Rightarrow-24x+16=21x+21\)
\(\Rightarrow45x=-5\)
\(\Rightarrow x=-\frac{1}{7}\)
Ta có : \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{2016}{2017}\)
\(=\frac{1.2.3......2016}{2.3.4.......2017}\)
\(=\frac{1}{2017}\)
mk muốn xem bài của mk đúng hay sai thôi !
chứ làm thì mk làm xong rồi !
a) \(\left(4,5-2x\right)\cdot1\frac{4}{7}=\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)\cdot\frac{11}{7}=\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)=\frac{11}{7}\div\frac{11}{7}\)
\(\left(\frac{9}{2}-2x\right)=1\)
\(2x=\frac{9}{2}-1\)
\(x=\frac{7}{2}\div2\)
\(x=\frac{7}{4}\)
b) \(|\frac{3}{4}\cdot x-\frac{1}{2}|-1=\frac{1}{4}\)
\(|\frac{3}{4}\cdot x-\frac{1}{2}|=\frac{1}{4}+1\)
\(|\frac{3}{4}\cdot x|=\frac{5}{4}+\frac{1}{2}\)
\(x=\frac{7}{4}\div\frac{3}{4}\)
\(x=\frac{7}{3}\)
c) \(\frac{1}{4}-|3-x|=-\frac{3}{4}\)
\(|3-x|=\frac{1}{4}-\left(-\frac{3}{4}\right)\)
\(|3-x|=1\)
\(x=3-1\)
\(\Rightarrow x=2\)
d) \(4\cdot\left(x-\frac{6}{7}\right)-\frac{3}{5}=1,4\)
\(4\cdot\left(x-\frac{6}{7}\right)-\frac{3}{5}=\frac{7}{5}\)
\(4\cdot\left(x-\frac{6}{7}\right)=\frac{7}{5}+\frac{3}{5}\)
\(4\cdot\left(x-\frac{6}{7}\right)=2\)
\(\left(x-\frac{6}{7}\right)=2\div4\)
\(x=\frac{1}{2}+\frac{6}{7}\)
\(x=\frac{19}{14}\)
\(\)
Ta có
\(S=3.\left(\frac{1}{1}.4\right)+3.\left(\frac{1}{4}.7\right)+...+3.\left(\frac{1}{197}.200\right)\)
\(S=3.\left(\frac{1}{1}.4+\frac{1}{4}.7+\frac{1}{7}.10+...+\frac{1}{197}.200\right)\)