Cho \(\dfrac{3}{5}\cdot x=\dfrac{2}{3}\cdot y\) và \(x^2-y^2=8\) tính x;y
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21 tháng 5 2022
Câu 1:
c: 2x=3y
nên x/3=y/2
=>x/9=y/6
5y=3z
nên y/3=z/5
=>y/6=z/10
=>x/9=y/6=z/10
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{10}=\dfrac{3x+3y-7z}{3\cdot9+3\cdot6-7\cdot10}=\dfrac{35}{-25}=-\dfrac{7}{5}\)
Do đó: x=-63/5; y=-42/5; z=-14
Bài 2:
Gọi ba số lần lượt là a,b,c
Theo đề, ta có: 4/3a=b=3/4c
\(\Leftrightarrow\dfrac{a}{\dfrac{3}{4}}=\dfrac{b}{1}=\dfrac{c}{\dfrac{4}{3}}\)
\(\Leftrightarrow\dfrac{a}{9}=\dfrac{b}{12}=\dfrac{c}{16}\)
Đặt \(\dfrac{a}{9}=\dfrac{b}{12}=\dfrac{c}{16}=k\)
=>a=9k; b=12k; c=16k
Theo đề, ta có: \(a^2+b^2+c^2=481\)
\(\Leftrightarrow81k^2+144k^2+256k^2=481\)
=>k2=1
Trường hợp 1: k=1
=>a=9; b=12; c=16
Trường hợp 2: k=-1
=>a=-9; b=-12; c=-16
TN
23 tháng 7 2023
\(a,\dfrac{8y}{3x^2}.\dfrac{9x^2}{4y^2}=\dfrac{72x^2y}{12x^2y^2}=\dfrac{6}{y}\\b,\dfrac{3x+x^2}{x^2+x+1}.\dfrac{3x^3-3}{x+3}=\dfrac{x\left(x+3\right)3\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x+3\right)}=3x\left(x-1\right)=3x^2-3x \)
\(c,\dfrac{2x^2+4}{x-3}.\dfrac{3x+1}{x-1}.\dfrac{6-2x}{x^2+2}=\dfrac{2\left(x^2+2\right)\left(3x+1\right)2\left(3-x\right)}{\left(x-3\right)\left(x-1\right)\left(x^2+2\right)}=\dfrac{-4\left(3x+1\right)}{x-1}=\dfrac{-12x-4}{x-1}\)
\(d,\dfrac{2x^2}{3y^3}:\left(-\dfrac{4x^3}{21y^2}\right)=\dfrac{-2x^2.21y^2}{3y^3.4x^3}=\dfrac{-42x^2y^2}{12x^3y^3}=\dfrac{-7}{2xy}\)
\(e,\dfrac{2x+10}{x^3-64}:\dfrac{\left(x+5\right)^2}{2x-8}=\dfrac{2\left(x+5\right)}{\left(x-4\right)\left(x^2+4x+16\right)}.\dfrac{2\left(x-4\right)}{\left(x+5\right)^2}=\dfrac{4}{\left(x+5\right)\left(x^2+4x+16\right)}=\dfrac{4}{x^3+9x^2+16x+80}\)
\(f,\dfrac{1}{x+y}\left(\dfrac{x+y}{xy}-x-y\right)-\dfrac{1}{x^2}:\dfrac{y}{x}=\dfrac{1}{x+y}\left(\dfrac{\left(x+y\right)\left(1-xy\right)}{xy}\right)-\dfrac{x}{x^2y}=\dfrac{1-xy}{xy}-\dfrac{x}{x^2y}=\dfrac{-x^2y}{x^2y}=-1\)
2 tháng 1 2023
Ta có: \(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}=\dfrac{y-x+x-z}{\left(x-y\right)\left(x-z\right)}\)\(=\dfrac{y-x}{\left(x-y\right)\left(x-z\right)}+\dfrac{x-z}{\left(x-y\right)\left(x-z\right)}\) \(=\dfrac{1}{z-x}+\dfrac{1}{x-y}\)
Tương tự:
\(\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}=\dfrac{1}{x-y}+\dfrac{1}{y-z}\)
\(\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}=\dfrac{1}{y-z}+\dfrac{1}{z-x}\)
\(\Rightarrow\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\) \(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\) \(\left(đpcm\right)\)
AD
23 tháng 7 2023
\(a,=\dfrac{4y.5x^3}{3x^2.2y^3}=\dfrac{20x^3y}{6x^2y^3}=\dfrac{10x}{3y^2}\\ b,=\dfrac{\left(x-1\right)^2.x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x-1\right)}=\dfrac{\left(x-1\right)^2.x.\left(x+1\right)}{\left(x-1\right)^2.\left(x+1\right)}=x\)
\(c,=\dfrac{x\left(2+x\right).3\left(x^3+1\right)}{\left(x^2-x+1\right).3.\left(x+2\right)}=\dfrac{3x.\left(x+2\right).\left(x+1\right)\left(x^2-x+1\right)}{\left(x^2-x+1\right).3\left(x+2\right)}=x\left(x+1\right)\)
DM
1
26 tháng 9 2021
\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}\Rightarrow y^2=\dfrac{16x^2}{9}\)
Ta có: \(x.y^2=384\Rightarrow x.\dfrac{16x^2}{9}=384\)
\(\Rightarrow x^3=216\Rightarrow x=6\)
\(\Rightarrow y=\dfrac{4x}{3}=\dfrac{4.6}{3}=8\)
6 tháng 1 2018
\(A=\left(\dfrac{-3}{7}.x^3.y^2\right).\left(\dfrac{-7}{9}.y.z^2\right).\left(6.x.y\right)\)
\(A=\left(\dfrac{-3}{7}x^3y^2\right).\left(\dfrac{-7}{9}yz^2\right).6xy\)
\(A=\left(\dfrac{-3}{7}.\dfrac{-7}{9}.6\right).\left(x^3.x\right)\left(y^2.y.y\right).z^2\)
\(A=2x^4y^4z^2\)
\(B=-4.x.y^3\left(-x^2.y\right)^3.\left(-2.x.y.z^3\right)^2\)
\(B=\left[\left(-4\right).\left(-2\right)\right].\left(x.x^6.x^2\right)\left(y^3.y^3.y^2\right)\left(z^6\right)\)
\(B=8x^7y^{y^8}z^6\)
YN
21 tháng 6 2022
\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
TN
Tìm x
a) 4/5 ( x - 1/3) - \(3\dfrac{1}{2}\)= 50%
b) \(8\dfrac{3}{5}\cdot x-2\dfrac{1}{5}\cdot x=16\%\)
1
MH
15 tháng 4 2022
a) \(\dfrac{4}{5}\left(x-\dfrac{1}{3}\right)-3\dfrac{1}{2}=50\%\)
\(\dfrac{4}{5}x-\dfrac{4}{15}-\dfrac{7}{2}=\dfrac{1}{2}\)
\(\dfrac{4}{5}x=\dfrac{1}{2}+\dfrac{7}{2}+\dfrac{4}{15}\)
\(\dfrac{4}{5}x=\dfrac{64}{15}\)
\(x=\dfrac{64}{15}:\dfrac{4}{5}\\ x=\dfrac{16}{3}\)
b) \(8\dfrac{3}{5}.x-2\dfrac{1}{5}.x=16\%\)
\(\dfrac{43}{5}x-\dfrac{11}{5}x=\dfrac{4}{25}\)
\(\dfrac{32}{5}x=\dfrac{4}{25}\)
\(x=\dfrac{4}{25}:\dfrac{32}{5}\)
\(x=\dfrac{1}{40}\)
\(\dfrac{3}{5}.x=\dfrac{2}{3}.y\\ \Rightarrow\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{5}}\\ \Rightarrow\dfrac{x^2}{\dfrac{4}{9}}=\dfrac{y^2}{\dfrac{9}{25}}=\dfrac{x^2-y^2}{\dfrac{4}{9}-\dfrac{9}{25}}=\dfrac{8}{\dfrac{19}{225}}=\dfrac{1800}{19}\\ \)