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a) \(\frac{x}{4}=\frac{16}{x^2}\)\(=>x^3=16.4\)\(=>x^3=64\)\(=>x=4\)
b) \(\frac{4}{3}:\frac{4}{5}=\frac{2}{3}.\left(\frac{1}{10}.x\right)\)\(=>\frac{4}{3}.\frac{5}{4}=\frac{2}{3}\left(\frac{1}{10}x\right)\)\(=>\frac{5}{3}=\frac{2}{3}\left(\frac{1}{10}x\right)\)\(=>\frac{5}{3}:\frac{2}{3}=\frac{1}{10}x\)\(=>\frac{5}{3}.\frac{3}{2}=\frac{1}{10}x\)\(=>\frac{5}{2}=\frac{1}{10}x\)\(=>x=\frac{5}{2}:\frac{1}{10}\)\(=>x=\frac{5}{2}.10\)\(=>x=25\)
vậy x=25
1.
a) \(\frac{x}{4}=\frac{16}{x^2}\)
\(\Rightarrow x^3=64\)
\(\Rightarrow x^3=4^3\)
\(\Rightarrow x=4\)
b) \(1\frac{1}{3}:0,8=\frac{2}{3}.\left(0,1.x\right)\)
\(\frac{5}{3}=\frac{2}{3}.\frac{x}{10}\)
\(\frac{x}{10}=\frac{5}{2}\)
\(\Rightarrow x=\frac{5.10}{2}=25\)
2.
\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\)
\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)\)
\(2A=1-\frac{1}{3^{99}}< 1\)
\(\Rightarrow A=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)
1) \(+2x+3y⋮17\)
\(\Rightarrow26x+39y⋮17\)
\(\Rightarrow\left(9x+5y\right)+17x+34y⋮17\)
Mà \(17x+34y⋮17\)
\(\Rightarrow9x+5y⋮17\)
\(+9x+5y⋮17\)
\(\Rightarrow36x+20y⋮17\)
\(\Rightarrow\left(2x+3y\right)+34x+17y⋮17\)
Mà \(34x+17y⋮17\)
\(\Rightarrow2x+3y⋮17\)
\(\Rightarrow A=\frac{1}{\frac{\left(2+1\right).2}{2}}+\frac{1}{\frac{\left(3+1\right).3}{2}}+\frac{1}{\frac{\left(4+1\right).4}{2}}+...+\frac{1}{\frac{\left(99+1\right).99}{2}}+\frac{1}{50}\)
\(=\frac{2}{\left(2+1\right).2}+\frac{2}{\left(3+1\right).3}+\frac{2}{\left(4+1\right).4}+...+\frac{2}{\left(99+1\right).99}+\frac{1}{50}\)
\(=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\right)+\frac{1}{50}\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\right)+\frac{1}{50}\)
\(=2.\left(\frac{1}{2}-\frac{1}{100}\right)+\frac{1}{50}\)
\(=2.\frac{49}{100}+\frac{1}{50}\)
\(=\frac{49}{50}+\frac{1}{50}=\frac{50}{50}=1\)
Vậy A=1.
Cái này có trong violympic vòng 10..bạn nhớ ôn cho kĩ nếu như bạn thi violympic!
Ta có: \(y=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{99}}\Leftrightarrow3y=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{98}}\)
\(\Leftrightarrow3y-y=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{99}\right)\)
\(\Leftrightarrow2y=1-\frac{1}{3^{99}}<1\Leftrightarrow y<\frac{1}{2}\)
Phần b tương tự
tick cho mình nha
Ta có \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+....+\left(\frac{1}{99}-\frac{1}{100}\right)\)
\(\frac{1}{2}-\frac{1}{100}=\frac{49}{100}< \frac{3}{4}\left(đpcm\right)\)
bạn điền thêm vào như thế này:
...................
A= 1-1/2^99 <1
Hay A<1
Vậy.........
Có. Chúng ta lí luận:
Vì \(1-\frac{1}{2^{99}}>1\)
\(\Rightarrow A>1\)