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a.Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) => \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) (1)
\(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\dfrac{k^2\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\)(2)
Từ (1) và (2) suy ra: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
b.M = \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{50^2}\right)\)
= \(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{2499}{2500}\)
= \(\dfrac{1.3.2.4.3.5...49.51}{2^2.3^2.4^2...50^2}\)
\(\dfrac{51}{2.50}=\dfrac{51}{100}\)
Lời giải:
a)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)
\(\Rightarrow \left(\frac{a}{b}\right)^2=\left(\frac{b}{d}\right)^2=\frac{(a+c)^2}{(b+d)^2}(1)\)
Mặt khác, \(\frac{a}{b}=\frac{c}{d}\Rightarrow \frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}(2)\) (áp dụng tính chất dãy tỉ số bằng nhau)
Từ \((1),(2)\Rightarrow \frac{(a+c)^2}{(b+d)^2}=\frac{a^2+c^2}{b^2+d^2}\)
b) Vì \(1-\frac{1}{2^2};1-\frac{1}{3^2};...;1-\frac{1}{50^2}<1\) nên:
\(\left\{\begin{matrix} \left \{ 1-\frac{1}{2^2} \right \}=1-\frac{1}{2^2}\\ \left \{ 1-\frac{1}{3^2} \right \}=1-\frac{1}{3^2}\\ ....\\ \left \{ 1-\frac{1}{50^2} \right \}=1-\frac{1}{50^2}\end{matrix}\right.\)
\(\Rightarrow M=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)....\left(1-\frac{1}{50^2}\right)\)
\(\Leftrightarrow M=\frac{(2^2-1)(3^2-1)(4^2-1)....(50^2-1)}{(2.3....50)^2}\)
\(\Leftrightarrow M=\frac{[(2-1)(3-1)...(50-1)][(2+1)(3+1)...(50+1)]}{(2.3.4...50)^2}\)
\(\Leftrightarrow M=\frac{(2.3...49)(3.4.5...51)}{(2.3.4...50)^2}=\frac{(2.3.4...49)^2.50.51}{2.(2.3....49)^2.50^2}=\frac{50.51}{2.50^2}=\frac{51}{100}\)
a)hình như đề sai thì phải
sửa lại
\(\left(\dfrac{1}{7}-\dfrac{2}{5}\right).\dfrac{2016}{2017}+\left(\dfrac{13}{7}+\dfrac{2}{5}\right).\dfrac{2016}{2017}\)
=\(\dfrac{2016}{2017}.\left(\dfrac{1}{7}-\dfrac{2}{5}+\dfrac{13}{7}+\dfrac{2}{5}\right)\)
=\(\dfrac{2016}{2017}.2=\dfrac{4032}{2017}\)
b,
\(B=\frac{1}{2000.1999}-\frac{1}{1999.1998}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(1-\frac{1}{1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\frac{1998}{1999}\)
\(\Rightarrow B=\frac{1}{1999}-\frac{1}{2000}-\frac{1998}{1999}\)
\(\Rightarrow B=\left(\frac{1}{1999}-\frac{1998}{1999}\right)-\frac{1}{2000}\)
\(\Rightarrow B=\frac{-1997}{1999}-\frac{1}{2000}\)
1, \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...........\left(1-\dfrac{1}{n+1}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right)...........\left(\dfrac{n+1}{n+1}-\dfrac{1}{n+1}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}..............\dfrac{n}{n+1}\)
\(=\dfrac{1.2.3........n}{2.3.......\left(n+1\right)}\)
\(=\dfrac{1}{n+1}\)
2, \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...........+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+............+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
C=\(-66\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)
=\(-66.\left(\dfrac{5}{66}\right)+124\left(-37-63\right)=-5+124.\left(-100\right)\)
=-12405
a) \(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}+\dfrac{2}{3}-1\dfrac{15}{17}\)
\(=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{2}{3}\right)-1\dfrac{15}{17}\)
\(=1+1-1\dfrac{15}{17}=\dfrac{2}{17}\)
a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)
=>x+1=0
hay x=-1
b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
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, \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)..........\left(\dfrac{1}{100}-1\right)\)
\(=\left(\dfrac{1}{4}-\dfrac{4}{4}\right)\left(\dfrac{1}{9}-\dfrac{9}{9}\right).........\left(\dfrac{1}{100}-\dfrac{100}{100}\right)\)
\(=\dfrac{-3}{4}.\dfrac{-8}{9}..............\dfrac{-99}{100}\)
\(=\dfrac{1.\left(-3\right).2\left(-4\right)............9\left(-11\right)}{2^2.3^2.......10^2}\)
\(=\dfrac{1.2.3........9}{2.3.......10}.\dfrac{\left(-3\right)\left(-4\right)....\left(-11\right)}{2.3...10}\)
\(=\dfrac{1}{10}.\dfrac{-11}{1}\)
\(=\dfrac{-11}{10}>\dfrac{-11}{21}\)
\(\Leftrightarrow B>\dfrac{-11}{21}\)
thanks nha bạn