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Bài 1:
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x+y}{3+5}=\dfrac{16}{8}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=10\end{matrix}\right.\)
Vậy x = 6, y = 10
Bài 2:
Ta có: \(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)
\(\Rightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(\Rightarrow ab-6a+5b-30=ab+6a-5b-30\)
\(\Rightarrow-6a+5b=6a-5b\)
\(\Rightarrow10b=12a\)
\(\Rightarrow6a=5b\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{5}{6}\)
\(\Rightarrowđpcm\)
B1 :
+ Theo bài ra :
\(\dfrac{x}{3}=\dfrac{y}{5}\left(1\right)\)và \(x+y=16\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x+y}{3+5}=\dfrac{16}{8}=2\)
+ Do đó :
\(\dfrac{x}{3}=2\Rightarrow x=2.3=6\)
\(\dfrac{y}{5}=2\Rightarrow y=2.5=10\)
Vậy x = 6 ; y = 10
Chả hỉu gì cả.
Đăng từng bài một thôi bạn!
1)\(\left(-\dfrac{5}{13}\right)^{2017}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(-\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}.\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).1^{2016}\)
\(=-\dfrac{5}{13}\)
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:
\(\left\{\begin{matrix} \left \{ 1-\frac{1}{2^2} \right \}=1-\frac{1}{2^2}\\ \left \{ 1-\frac{1}{3^2} \right \}=1-\frac{1}{3^2}\\ ....\\ \left \{ 1-\frac{1}{50^2} \right \}=1-\frac{1}{50^2}\end{matrix}\right.\)
\(\Rightarrow M=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)....\left(1-\frac{1}{50^2}\right)\)
\(\Leftrightarrow M=\frac{(2^2-1)(3^2-1)(4^2-1)....(50^2-1)}{(2.3....50)^2}\)
\(\Leftrightarrow M=\frac{[(2-1)(3-1)...(50-1)][(2+1)(3+1)...(50+1)]}{(2.3.4...50)^2}\)
\(\Leftrightarrow M=\frac{(2.3...49)(3.4.5...51)}{(2.3.4...50)^2}=\frac{(2.3.4...49)^2.50.51}{2.(2.3....49)^2.50^2}=\frac{50.51}{2.50^2}=\frac{51}{100}\)
1. đề bạn ghi rõ lại giúp mình đc ko r mình giải lại cho
2. Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{2x^2}{2.3^2}=\dfrac{y^2}{5^2}=\dfrac{2x^2-y^2}{18-25}=\dfrac{-28}{-7}=4\)
\(\dfrac{x}{3}=4\Rightarrow x=12\)
\(\dfrac{y}{5}=4\Rightarrow y=20\)
Vậy x=12 và y=20
Mấy dấu chấm là dấu nhân đó
a,Ta có \(\left(3^3\right)^n:3^n=9\Leftrightarrow3^{3n}:3^n=3^2\Leftrightarrow3n-n=2\Leftrightarrow n=1\)
b,TA có \(\dfrac{5^2}{5^n}=5^1\Leftrightarrow2-n=1\Leftrightarrow n=1\)
Các câu sau để bn tự làm
a) 27n : 3n = 9
\(\Leftrightarrow\) (27 : 3)n = 9
\(\Leftrightarrow\) 9n = 9
\(\Leftrightarrow\) n = 1
b) \(\dfrac{25}{5^n}=5\)
\(\Leftrightarrow\dfrac{5^2}{5^n}=5\)
\(\Leftrightarrow5^n.5=5^2\)
\(\Leftrightarrow5^{n+1}=5^2\)
\(\Leftrightarrow n+1=2\)
n = 2 - 1
n = 1
c) \(\dfrac{81}{\left(-3\right)^n}=-243\)
\(\Leftrightarrow\dfrac{\left(-3\right)^4}{\left(-3\right)^n}=\left(-3\right)^5\)
\(\Leftrightarrow\left(-3\right)^n.\left(-3\right)^5=\left(-3\right)^4\)
\(\Leftrightarrow\left(-3\right)^{n+5}=\left(-3\right)^4\)
\(\Leftrightarrow n+5=4\)
n = 4 - 5
n = -1
a) \(\dfrac{12}{25}+\dfrac{5}{13}+\dfrac{13}{25}-\dfrac{18}{13}+\dfrac{3}{5}\)
\(=\left(\dfrac{12}{25}+\dfrac{13}{25}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)+\dfrac{3}{5}\)
\(=1+\left(-1\right)+\dfrac{3}{5}\)
\(=0+\dfrac{3}{5}\)
\(=\dfrac{3}{5}\)
b) \(\dfrac{2}{5}:\left(\dfrac{5}{2}-\dfrac{3}{4}\right)\)
\(=\dfrac{2}{3}:\dfrac{3}{4}\)
\(=\dfrac{8}{21}\)
tích mình nha 
a) \(\dfrac{12}{25}+\dfrac{5}{13}+\dfrac{13}{25}-\dfrac{18}{13}+\dfrac{3}{5}\)
= \(\left(\dfrac{12}{25}+\dfrac{13}{25}\right)+\left(\dfrac{5}{13}-\dfrac{18}{13}\right)+\dfrac{3}{5}\)
= 1 + (-1) + \(\dfrac{3}{5}\)
= 0 + \(\dfrac{3}{5}\) = \(\dfrac{3}{5}\)
b) \(\dfrac{2}{3}:\left(\dfrac{5}{2}-\dfrac{3}{4}\right)\)
=\(\dfrac{2}{3}:\left(\dfrac{10}{4}-\dfrac{3}{4}\right)\)
=\(\dfrac{2}{3}:\dfrac{7}{4}\)
=\(\dfrac{2}{3}.\dfrac{4}{7}\)
=\(\dfrac{8}{21}\)
Chúc bạn học tốt!!! 
a) \(\left(x+\dfrac{1}{2}\right)+\left(x+\dfrac{1}{6}\right)+\left(x+\dfrac{1}{12}\right)+....+\left(x+\dfrac{1}{9900}\right)\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)=1\)
\(\Leftrightarrow50x+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=1\)
\(\Leftrightarrow50x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=1\)
\(\Leftrightarrow50x+\left(1-\dfrac{1}{100}\right)=1\)
\(\Leftrightarrow50x+\dfrac{99}{100}=1\)
\(\Leftrightarrow50x=\dfrac{1}{100}\Rightarrow x=\dfrac{1}{5000}\)
b) \(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{202.205}\)
\(A=\dfrac{3^2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{202}-\dfrac{1}{205}\right)\)
\(A=\dfrac{9}{3}\cdot\left(1-\dfrac{1}{205}\right)\)
\(A=\dfrac{9}{3}\cdot\dfrac{204}{205}=\dfrac{615}{205}\)
a) \(\left(x+\dfrac{1}{2}\right)+\left(x+\dfrac{1}{6}\right)+\left(x+\dfrac{1}{12}\right)+....+\left(x+\dfrac{1}{9900}\right)=1\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)=1\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=1\)
Có tất cả : (99 - 1) : 1 + 1 = 99 (số x)
\(\Rightarrow99x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=1\)
\(\Rightarrow99x+\left(1-\dfrac{1}{100}\right)=1\)
\(\Rightarrow99x+\dfrac{99}{100}=1\Rightarrow99x=1-\dfrac{99}{100}\)
\(\Rightarrow99x=\dfrac{1}{100}\Rightarrow x=\dfrac{1}{100.99}=\dfrac{1}{9900}\)
b) \(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+....+\dfrac{3^2}{202.205}\)
\(A=\dfrac{3^2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{202}-\dfrac{1}{205}\right)\)
\(A=\dfrac{9}{3}\cdot\left(1-\dfrac{1}{205}\right)\)
\(A=3\cdot\dfrac{204}{205}=\dfrac{615}{205}\)