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- \(\dfrac{5}{7}\) - (-\(\dfrac{5}{67}\)) + \(\dfrac{13}{10}\) + \(\dfrac{1}{2}\) + (- 1\(\dfrac{5}{6}\)) + 1\(\dfrac{3}{14}\) - (-\(\dfrac{2}{5}\))
= (- \(\dfrac{5}{7}\) + 1\(\dfrac{3}{14}\)+ \(\dfrac{1}{2}\))+ \(\dfrac{5}{67}\) - \(\dfrac{11}{6}\) + (\(\dfrac{13}{10}\) + \(\dfrac{2}{5}\))
= (-\(\dfrac{10}{14}\) + \(\dfrac{17}{14}\)+ \(\dfrac{1}{2}\)) + \(\dfrac{5}{67}\) - \(\dfrac{11}{6}\) + (\(\dfrac{13}{10}\) + \(\dfrac{4}{10}\))
= (\(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)) + \(\dfrac{5}{67}\) - \(\dfrac{11}{6}\)+ \(\dfrac{17}{10}\)
= 1 + \(\dfrac{5}{67}\) - \(\dfrac{11}{6}\) + \(\dfrac{17}{10}\)
= \(\dfrac{2010}{2010}\) + \(\dfrac{150}{2010}\) - \(\dfrac{3685}{2010}\) + \(\dfrac{3417}{2010}\)
= \(\dfrac{1892}{2010}\)
= \(\dfrac{946}{1005}\)
Đặt 1/5 + 1/14+1/28 +1/44 + 1/61+1/85+1/97 =A
Ta có : A = 1/5 + ( 1/14+1/28+1/44)+(1/61+1/85+1/97)
A < 1/5 ( 1/14.3)+(1/61.3)
A < 1/5+ 3/14+3/61
A < 1/5 + 3/12+1/20
A < 1/5+1/4+1/20
A < 1/2 ( đpcm)
hellu cả lò nhà mik ạ giải giúp mik vs cả lò nhà mik ui
-5/7-(-5/67)+13/30+1/2+(-1 5/6)+1 3/14-(-2/5)
=13/30-1-5/6+2/5+17/14+1/2-5/7+5/67
=-1+17/14-3/14+5/67
=-1+1+5/67
=5/67
7) 5x=4y ⇒\(\dfrac{x}{4}=\dfrac{y}{5}\)
Nhân cả hai vế với \(\dfrac{x}{4}\), ta có: \(\left(\dfrac{x}{4}\right)^2=\dfrac{x}{4}.\dfrac{y}{5}=\dfrac{xy}{20}=\dfrac{20}{20}=1\)
\(\left(\dfrac{x}{4}\right)^2=1\Rightarrow\left[{}\begin{matrix}\dfrac{x}{4}=1\\\dfrac{x}{4}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}y=5\\y=-5\end{matrix}\right.\)
4) áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{0,5}=\dfrac{y}{0,3}=\dfrac{z}{0,2}=\dfrac{z-y+x}{0,2-0,3+0,5}=\dfrac{1}{\dfrac{2}{5}}=\dfrac{5}{2}\)
\(\dfrac{x}{0,5}=\dfrac{5}{2}\Rightarrow x=\dfrac{5}{4}\)
\(\dfrac{y}{0,3}=\dfrac{5}{2}\Rightarrow y=\dfrac{3}{4}\)
\(\dfrac{z}{0,2}=\dfrac{5}{2}\Rightarrow z=\dfrac{1}{2}\)
6) áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x+11}{13}=\dfrac{y+12}{14}=\dfrac{z+13}{15}=\dfrac{x+11+y+12+z+13}{13+14+15}=\dfrac{42}{42}=1\)
\(\dfrac{x+11}{13}=1\Rightarrow x=2\)
\(\dfrac{y+12}{13}=1\Rightarrow y=1\)
\(\dfrac{z+13}{15}=1\Rightarrow z=2\)
7) \(5x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{5}=k\)
\(\Rightarrow x=4k,y=5k\)
\(x.y=20\\ \Rightarrow4k.5k=20\\ \Rightarrow20k^2=20\\ \Rightarrow k^2=1\\ \Rightarrow\left[{}\begin{matrix}k=-1\\k=1\end{matrix}\right.\)
\(x=4k\Rightarrow\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)
\(y=5k\Rightarrow\left[{}\begin{matrix}y=-5\\y=5\end{matrix}\right.\)
Vậy \(\left(x,y\right)=\left\{\left(-4;-5\right);\left(4;5\right)\right\}\)
\(9^{3x}:9^7-14=67\)
\(\Rightarrow9^{3x-7}=67+14\)
\(\Rightarrow9^{3x-7}=81\)
\(\Rightarrow9^{3x-7}=9^2\)
\(\Rightarrow3x-7=2\)
\(\Rightarrow3x=9\)
\(\Rightarrow x=\dfrac{9}{3}\)
\(\Rightarrow x=3\)