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Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a^2}{b^2}=\dfrac{b^2}{c^2}=\dfrac{a^2+b^2}{b^2+c^2}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{a^2+b^2}{b^2+c^2}\)
\(\Rightarrow\dfrac{a}{b}.\dfrac{b}{c}=\dfrac{a^2+b^2}{b^2+c^2}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{a^2+b^2}{b^2+c^2}\)
Vậy nếu \(\dfrac{a}{b}=\dfrac{b}{c}\) thì \(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\left(đpcm\right)\)
may mà m đăng để t đỡ phải viết (mỏi tay) Nguyen Ngoc Linh
\(A=\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+2^{12}}\)
\(=\dfrac{2^{19}.3^9+5.2^{18}.3^9}{2^9.3^{19}+2^{12}}=\dfrac{2^{10}+5.2^8}{3^{10}+2^3}=\dfrac{2^7+5.2^5}{3^{10}}\)
Đặt: \(A=\frac{3^6.21^{12}}{175^9.7^3}=\frac{3^{18}.7^{12}}{7^{12}.25^9}=\frac{3^{18}}{5^{18}}=\left(\frac{3}{5}\right)^{18}\)
\(B=\frac{3^{10}.6^7.4}{10^9.5^8}=\frac{3^{10}.2^7.3^7.2^2}{2^9.5^9.5^8}=\frac{3^{17}.2^9}{2^9.5^{17}}=\left(\frac{3}{5}\right)^{17}\)
Vì: \(\left(\frac{3}{5}\right)^{18}< \left(\frac{3}{5}\right)^{17}\Rightarrow A< B\)
a, Ta có :\(A=\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}+\dfrac{1}{2^{50}}\\ \Rightarrow2A=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}\\ \Rightarrow2A-A=\left(1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}\right)-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{50}}\right)\\ \Rightarrow A=1-\dfrac{1}{2^{50}}< 1\\ \Rightarrow A< 1\) Vậy \(A< 1\)
b, Ta có :
\(B=\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\\ \Rightarrow3B=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\\ \Rightarrow3B-B=\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\right)\\ \Rightarrow2B=1-\dfrac{1}{3^{100}}< 1\\ \Rightarrow B< \dfrac{1}{2}\)Vậy \(B< \dfrac{1}{2}\)
c, Ta có :
\(C=\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{1000}}\\ \Rightarrow4C=1+\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{999}}\\\Rightarrow4C-C=\left(1+\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{999}}\right)-\left(\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{1000}}\right)\\ \Rightarrow3C=1-\dfrac{1}{4^{1000}}< 1\\ \Rightarrow C< \dfrac{1}{3}\)Vậy \(C< \dfrac{1}{3}\)
Sai đề nhé: Sửa đề:
\(A=\dfrac{5^{100}+2}{5^{102}+2};B=\dfrac{5^{101}+2}{5^{103}+2}\)
nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(B=\dfrac{5^{101}+2}{5^{103}+2}< 1\)
\(B< \dfrac{5^{101}+2+8}{5^{103}+2+8}\Rightarrow B< \dfrac{5^{101}+10}{5^{103}+10}\Rightarrow B< \dfrac{5\left(5^{100}+2\right)}{5\left(5^{101}+2\right)}\Rightarrow B< \dfrac{5^{100}+2}{5^{101}+2}=A\)
\(B< A\)
nè bạn ơi bạn nhập nó vânx như cũ mà có khác các gì đâu bạn!
a, Ta có :
\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)...........\left(\dfrac{1}{10}-1\right)\)
\(=\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\left(\dfrac{1}{3}-\dfrac{3}{3}\right).........\left(\dfrac{1}{10}-\dfrac{10}{10}\right)\)
\(=\dfrac{-1}{2}.\dfrac{-2}{3}...............\dfrac{-9}{10}\)
\(=\dfrac{-1.\left(-2\right)............\left(-9\right)}{2.3........9.10}\)
\(=\dfrac{-1}{10}< \dfrac{-1}{9}\)
\(\Leftrightarrow A< \dfrac{-1}{9}\)
b, \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)..........\left(\dfrac{1}{100}-1\right)\)
\(=\left(\dfrac{1}{4}-\dfrac{4}{4}\right)\left(\dfrac{1}{9}-\dfrac{9}{9}\right).........\left(\dfrac{1}{100}-\dfrac{100}{100}\right)\)
\(=\dfrac{-3}{4}.\dfrac{-8}{9}..............\dfrac{-99}{100}\)
\(=\dfrac{1.\left(-3\right).2\left(-4\right)............9\left(-11\right)}{2^2.3^2.......10^2}\)
\(=\dfrac{1.2.3........9}{2.3.......10}.\dfrac{\left(-3\right)\left(-4\right)....\left(-11\right)}{2.3...10}\)
\(=\dfrac{1}{10}.\dfrac{-11}{1}\)
\(=\dfrac{-11}{10}>\dfrac{-11}{21}\)
\(\Leftrightarrow B>\dfrac{-11}{21}\)
1) Tìm x, biết :
a) \(\dfrac{x-1}{3}=\dfrac{x+1}{5}\)
=> \(5\left(x-1\right)=3\left(x+1\right)\)
=> \(5x-5=3x+3\)
=> \(5x-5-3=3x\)
=> \(5x-8=3x\)
=> \(8=5x-3x\)
=> \(8=2x\)
=> x = 8 : 2
=> x = 4
ko giải đâu
đùa thôi =)