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x,y tỉ lệ thuận với \(\dfrac{3}{4}\) và \(\dfrac{4}{3}\)
\(\Rightarrow\dfrac{x}{\dfrac{3}{4}}=\dfrac{y}{\dfrac{4}{3}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ,ta có :
\(\dfrac{x}{\dfrac{3}{4}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{x+y}{\dfrac{3}{4}+\dfrac{4}{3}}=-\dfrac{50}{\dfrac{25}{12}}=-24\)
\(\dfrac{x}{\dfrac{3}{4}}=-24\Rightarrow x=-18\)
\(\dfrac{y}{\dfrac{4}{3}}=-24\Rightarrow y=-32\)
Vì x tỉ lệ thuận với \(\dfrac{3}{4}\)\(\Rightarrow x=\dfrac{3}{4}.k\)
Vì y tỉ lệ thuận với \(\dfrac{4}{3}\Rightarrow y=\dfrac{4}{3}.k\)
\(\Rightarrow x+y=\dfrac{3}{4}.k+\dfrac{4}{3}.k\)
Mà x+y=50
\(\Rightarrow\dfrac{3}{4}.k +\dfrac{4}{3}.k=-50\)
\(\Rightarrow\left(\dfrac{3}{4}+\dfrac{4}{3}\right).k=-50\)
\(\Rightarrow\dfrac{25}{12}.k=-50\)
\(\Rightarrow k=-50:\dfrac{25}{12}\)
\(\Rightarrow k=-24\)
\(\Rightarrow x=\dfrac{3}{4}.\left(-24\right)=-18\)
Tick mk nha!!!
\(y=\dfrac{4}{3}.\left(-24\right)=-32\)
Vậy \(x=-18,y=-32\)
a)hình như đề sai thì phải
sửa lại
\(\left(\dfrac{1}{7}-\dfrac{2}{5}\right).\dfrac{2016}{2017}+\left(\dfrac{13}{7}+\dfrac{2}{5}\right).\dfrac{2016}{2017}\)
=\(\dfrac{2016}{2017}.\left(\dfrac{1}{7}-\dfrac{2}{5}+\dfrac{13}{7}+\dfrac{2}{5}\right)\)
=\(\dfrac{2016}{2017}.2=\dfrac{4032}{2017}\)
\(A=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(=2.\dfrac{502}{1005}=\dfrac{1004}{1005}\)
\(A=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(=2.\dfrac{502}{1005}=\dfrac{1004}{1005}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\) \(\begin{cases} a = bk \\ c = dk \end{cases}\)
Ta có: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(1\right)\)
\(\dfrac{a.c}{b.d}=\dfrac{bk.dk}{b.d}=\dfrac{k^2.b.d}{b.d}=k^2\left(2\right)\)
Từ (1) và (2) suy ra: \(\dfrac{a.c}{b.d}=\dfrac{a^2+c^2}{b^2+d^2}\) \(\rightarrow đpcm\).
c, \(\left(7-3x\right)\left(2x+1\right)=0\)
=> \(7-3x=0\) hoặc \(2x+1=0\)
\(3x=7-0\) hoặc \(2x=0-1\)
\(3x=7\) hoặc \(2x=-1\)
\(x=7:3\) hoặc \(x=-1:2\)
\(x=\dfrac{7}{3}\) hoặc \(x=-0,5\)
Vậy, \(x\in\left\{\dfrac{7}{3};-0,5\right\}\)
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
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) \(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+....+\dfrac{3^2}{202.205}\)
\(A=\dfrac{3^2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{202}-\dfrac{1}{205}\right)\)
\(A=\dfrac{9}{3}\cdot\left(1-\dfrac{1}{205}\right)\)
\(A=3\cdot\dfrac{204}{205}=\dfrac{615}{205}\)