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Câu hỏi của Phan Thúy Vy - Toán lớp 7 - Học toán với OnlineMath
1) x2 = \(\frac{3^2}{5^2^{ }}\)
x = \(\frac{3}{5}\)
x2 = 0.09
x2 = \(\frac{9}{100}\)
x2 = \(\frac{3^2}{10^2}\)
x = \(\frac{3}{10}\)
1. \(x^2=\frac{9}{25}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{-3}{5}\\x=\frac{3}{5}\end{cases}}\)
Vậy \(x=\frac{-3}{5}\)hoặc \(x=\frac{3}{5}\)
2. \(x^2=0,09\)\(\Rightarrow x^2=\frac{9}{100}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{-3}{10}\\x=\frac{3}{10}\end{cases}}\)
Vậy \(x=\frac{-3}{10}\)hoặc \(x=\frac{3}{10}\)
3. \(\sqrt{2}.x=2\)\(\Rightarrow\left(\sqrt{2}.x\right)^2=2^2\)
\(\Rightarrow2x^2=4\)\(\Rightarrow x^2=2\)\(\Rightarrow x=\pm\sqrt{2}\)
Vì \(\sqrt{2}>0\); \(2>0\)\(\Rightarrow\)Để \(\sqrt{2}.x=2\)thì \(x>0\)
\(\Rightarrow x=\sqrt{2}\)
Vậy \(x=\sqrt{2}\)
\(\dfrac{1}{2019^2}-\dfrac{1}{2020^2}=\dfrac{2020^2-2019^2}{2019^2\cdot2020^2}\\ =\dfrac{\left(2020-2019\right)\left(2020+2019\right)}{2019^2\cdot2020^2}=\dfrac{4039}{2019^2\cdot2020^2}\)
\(a,\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\left(\frac{3}{7}\right)^2\right)^{10}.\frac{3}{7}:\left(\frac{9}{49}\right)^6=\left(\frac{9}{49}\right)^{10}.\frac{3}{7}:\left(\frac{9}{49}\right)^6\)
\(=\left(\left(\frac{9}{49}\right)^{10}:\left(\frac{9}{49}\right)^6\right).\frac{3}{7}=\left(\frac{9}{49}\right)^{10-6}.\frac{3}{7}=\left(\frac{9}{49}\right)^4.\frac{3}{7}=\left(\left(\frac{3}{7}\right)^2\right)^4.\frac{3}{7}\)
\(=\left(\frac{3}{2}\right)^8.\frac{3}{7}=\left(\frac{3}{2}\right)^9\)
\(b,3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2=3-1+\left(\frac{1}{2}\right)^2.\frac{1}{2}=2+\left(\frac{1}{2}\right)^3=2+\frac{1}{6}=2\frac{1}{6}\)
A = 1/1^2 + 1/2^2 + 1/3^2 + ... + 1/2020^2
1/2^2 < 1/1.2
1/3^2 < 1/2.3
...
1/2020^2 < 1/2019.2020
=> A < 1 + 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/2019*2020
=> A < 1 + 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2019 - 1/2020
=> A < 2 - 1/2020
=> A < 4039/2020 < 7/4
=> a < 7/4