\(\dfrac{x}{3}\) = \(\dfrac{y}{2}\) và 2x + 5y = 32
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16 tháng 7 2021
a, \(3x=5y=7z=>\dfrac{3x}{105}=\dfrac{5y}{105}=\dfrac{7z}{105}=>\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}\)
áp dụng tính chất dãy tỉ số = nhau
\(=>\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}=\dfrac{x+y+z}{35+21+15}=\dfrac{10}{71}\)
\(=>\dfrac{x}{35}=\dfrac{10}{71}=>x=\dfrac{350}{71}\)
\(=>\dfrac{y}{21}=\dfrac{10}{71}=>y=\dfrac{210}{71}\)
\(=>\dfrac{z}{15}=\dfrac{10}{71}=>z=\dfrac{150}{71}\)
b, \(\)\(6x=5y=>\dfrac{x}{5}=\dfrac{y}{6}=>\dfrac{x}{20}=\dfrac{y}{24}\)
có \(7y=8z=>\dfrac{y}{8}=\dfrac{z}{7}=>\dfrac{y}{24}=\dfrac{z}{21}\)
\(=>\dfrac{x}{20}=\dfrac{y}{24}=\dfrac{z}{21}=>\dfrac{3x}{60}=\dfrac{2y}{48}=\dfrac{4z}{84}\)
áp dụng t/c dãy tỉ số = nhau
\(=>\dfrac{3x}{60}=\dfrac{2y}{48}=\dfrac{4z}{84}=\dfrac{3x+2y+4z}{60+48+84}=\dfrac{12}{192}=\dfrac{1}{16}\)
\(=>\dfrac{3x}{60}=\dfrac{1}{16}=>x=1,25\)
\(=>\dfrac{2y}{48}=\dfrac{1}{16}=>y=1,5\)
\(=>\dfrac{4z}{84}=\dfrac{1}{16}=>z=1,3125\)
c, \(x:y:z=1:2:3=>\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\)
\(=>x=\dfrac{y}{2},z=\dfrac{3y}{2}\)
thay x,z vào \(x^3+y^3+z^3=36=>\left(\dfrac{y}{2}\right)^3+y^3+\left(\dfrac{3y}{2}\right)^3=36\)
\(=>y=2\)
\(=>x=\dfrac{y}{2}=\dfrac{2}{2}=1,z=\dfrac{3y}{2}=\dfrac{3.2}{2}=3\)
d, \(\dfrac{x}{2}=\dfrac{y}{3}=>x=\dfrac{2y}{3}\)
thay x vào \(3x^3+y^3=51=>3.\left(\dfrac{2y}{3}\right)^3+y^3=51=>y=3\)
\(=>x=\dfrac{2.3}{3}=2\)
16 tháng 7 2021
c, từ đoạn này á
\(\left(\dfrac{y}{2}\right)^3+y^3+\left(\dfrac{3y}{2}\right)^3=36\)
\(< =>\dfrac{y^3}{8}+\dfrac{8y^3}{8}+\dfrac{27y^3}{8}=36\)
\(=>\dfrac{36y^3}{8}=36=>36y^3=8.36=>y^3=8=>y=2\)
MH
15 tháng 9 2021
\(2x=3y\text{⇒}\dfrac{x}{3}=\dfrac{y}{2}\text{⇒}\dfrac{x}{21}=\dfrac{y}{14}\)
\(5y=7z\text{⇒}\dfrac{y}{7}=\dfrac{z}{5}\text{⇒}\dfrac{y}{14}=\dfrac{z}{10}\)
⇒\(\dfrac{x}{21}=\dfrac{y}{14}=\dfrac{z}{10}\)⇒\(\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{3x}{63}=\dfrac{7y}{98}=\dfrac{5z}{50}=\dfrac{3x-7y+5z}{63-98+50}=\dfrac{30}{15}=2\)
⇒x=42,y=28,z=20
MH
15 tháng 9 2021
\(\dfrac{x}{3}=\dfrac{y}{2}\)⇒\(\dfrac{x}{15}=\dfrac{y}{10}\)
\(\dfrac{x}{5}=\dfrac{z}{7}\text{⇒}\dfrac{x}{15}=\dfrac{z}{21}\)
⇒\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{21}\)⇒\(\dfrac{x}{15}=\dfrac{2y}{20}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{15}=\dfrac{2y}{20}=\dfrac{x+2y}{15+20}=\dfrac{-112}{35}=\dfrac{-16}{5}\)
⇒x=48,y=32,z=336/5
11 tháng 11 2021
4: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y-z}{8-12-15}=\dfrac{38}{-19}=-2\)
Do đó: x=-16; y=-24; z=-30
11 tháng 10 2021
a) \(\dfrac{2x+3}{24}=\dfrac{3x-1}{32}\)
\(\Rightarrow32\left(2x+3\right)=24\left(3x-1\right)\)
\(\Rightarrow64x+96=72x-24\)
\(\Rightarrow8x=120\Rightarrow x=15\)
b) \(\dfrac{13x-2}{2x+5}=\dfrac{76}{17}\)
\(\Rightarrow17\left(13x-2\right)=76\left(2x+5\right)\)
\(\Rightarrow221x-34=152x+380\)
\(\Rightarrow69x=414\Rightarrow x=6\)
AH
Akai Haruma
Giáo viên
16 tháng 7 2018
Lời giải
a)
\(\left(\frac{3}{2x-y}-\frac{2}{2x+y}-\frac{1}{2x-5y}\right).\frac{4x^2-y^2}{y^2}\)
\(=\frac{3(4x^2-y^2)}{(2x-y)y^2}-\frac{2(4x^2-y^2)}{(2x+y)y^2}-\frac{4x^2-y^2}{(2x-5y)y^2}\)
\(=\frac{3(2x-y)(2x+y)}{(2x-y)y^2}-\frac{2(2x-y)(2x+y)}{(2x+y)y^2}-\frac{4x^2-y^2}{(2x-5y)y^2}\)
\(=\frac{3(2x+y)-2(2x-y)}{y^2}-\frac{4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}\)
\(=\frac{2x+5y}{y^2}-\frac{4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}\)
\(=\frac{(2x+5y)(2x-5y)-4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}\)
\(=\frac{4x^2-25y^2-4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}=\frac{-25}{2x-5y}+\frac{1}{2x-5y}=\frac{-24}{2x-5y}\)
Ta có đpcm.
b)
\(\frac{x^2-x+1}{x^2+x}.\frac{x+1}{3x-2}.\frac{9x-6}{x^2-x+1}\)
\(=\frac{(x^2-x+1)(x+1).3(3x-2)}{x(x+1)(3x-2)(x^2-x+1)}\)
\(=\frac{3}{x}\) (đpcm)
MH
24 tháng 9 2021
a) \(2x=5y\)⇒\(x=\dfrac{5}{2}y\)⇒\(xy=\dfrac{5}{2}y^2\)
Thay \(xy=250\), ta có:
\(250=\dfrac{5}{2}y^2\)
⇒\(y^2=100\)⇒\(y=+-10\)
+) \(y=10\text{⇒}x=250:10=25\)
+) \(y=-10\text{⇒}x=250:-10=-25\)
24 tháng 9 2021
\(a,2x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{2}=k\\ \Rightarrow x=5k;y=2k\\ xy=250\Rightarrow5k\cdot2k=250\Rightarrow k^2=25\Rightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=25;y=10\\x=-25;y=-10\end{matrix}\right.\\ b,\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{4}=a\Rightarrow x=3a;y=2a;z=4a\\ xyz=192\Rightarrow24a^3=192\Rightarrow a^3=8\Rightarrow a=2\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=4\\z=8\end{matrix}\right.\\ c,\Rightarrow\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{-3}=q\Rightarrow x=5q;y=2q;z=-3q\\ xyz=240\Rightarrow-30q^3=240\Rightarrow q^3=-8\Rightarrow q=-2\\ \Rightarrow\left\{{}\begin{matrix}x=-10\\y=-4\\z=6\end{matrix}\right.\)
\(\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{y}=\dfrac{3}{2}\Rightarrow x=1,5y\)
\(2x+5y=32\)
\(2\times1,5y+5y=32\)
\(3y+5y=32\)
\(\left(3+5\right)y=32\)
\(8y=32\)
\(y=32:8=4\)
⇒ \(x=4\times\dfrac{3}{2}=6\)
Vậy \(x=6\) ; \(y=4\)