Cho \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\). Tìm x và y
Tìm x và y
\(\dfrac{x}{3}=\dfrac{y}{4}\)và\(x+y=14\)
\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\)
\(\dfrac{x}{3}=\dfrac{y}{4}\)
Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{x+y}{3+4}=\dfrac{14}{7}\)=2
* \(\dfrac{x}{3}=2=>x=6\)
*\(\dfrac{y}{4}=2=>y=8\)
Vậy( x, y) ∈{ 6, 8}
Kiểm tra lại nhaa
áp dụng tính chất dãy tỉ số bằng nhau
Ta có:x/3=y/4=x+y/3+4=14/7=2
Vậy x=2.3=6
y=2.4=8
Tìm x,y biết: \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\)
Cho \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\). Tìm x, y
+) Xét \(2x+3y-1=0\) có:
\(\Rightarrow\left\{{}\begin{matrix}2x+1=0\\3y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{2}\\y=\dfrac{2}{3}\end{matrix}\right.\)
+) Xét \(2x+3y-1\ne0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{12}=\dfrac{2x+3y-1}{6x}\)
\(\Rightarrow6x=12\Rightarrow x=2\)
\(\Rightarrow y=3\)
Vậy...
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+1+3y-2}{5+7}=\dfrac{2x+3y-1}{12}=\dfrac{2x+3y-1}{6x}\)
\(\Rightarrow6x=12\Rightarrow x=2\)
Thay vào biểu thức, ta có:
\(2.2+\dfrac{1}{5}=\dfrac{3y-2}{7}\Rightarrow1=\dfrac{3y-2}{7}\Rightarrow3y-2=7\)
\(\Rightarrow3y=9\Rightarrow y=3\)
Vậy \(\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có
\(\dfrac{2x+1}{5}\)=\(\dfrac{3y-2}{7}\)=\(\dfrac{2x+3y-1}{6x}\)=\(\dfrac{2x+1+3y-2}{5+7}\)=\(\dfrac{2x+3y-1}{12}\)=
\(\dfrac{2x+3y-1}{6x}\)(1)
TH1: 2x+3y-1≠0
Từ (1) ⇒6x=12⇔x=2
Thầy x=2vào biểu thức trên ta có
\(\dfrac{2\cdot2+3y-1}{12}\)=\(\dfrac{3y-2}{7}\)⇔y=\(\dfrac{1}{5}\)
TH2: 2x+3y-1=0
⇒2x+1=0 và 3y-2=0
⇔x=\(\dfrac{-1}{2}\);y=\(\dfrac{2}{3}\)
Vậy (x;y)∈{(2;\(\dfrac{1}{5}\));(\(\dfrac{-1}{2}\);\(\dfrac{2}{3}\))}
a,\(\dfrac{x+1}{x-3}+\dfrac{-2x^2+2x}{x^2-9}+\dfrac{x-1}{x+3}\)
b,\(\dfrac{1-2x}{6x^3y}+\dfrac{3+2y}{6x^3y}+\dfrac{2x-4}{6x^3y}\)
c,\(\dfrac{5}{2x^2y}+\dfrac{3}{5xy^2}+\dfrac{x}{3y^3}\)
d,\(\dfrac{5}{4\left(x+2\right)}+\dfrac{8-x}{4x^2+8x}\)
c,\(\dfrac{x^2+2}{x^3+1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
Tìm x ; y biết : \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\)
Áp dụng dãy tỉ số bằng nhau là ra mà.
Giải:
Áp dụng dãy tỉ số bằng nhau,ta có:
\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y+1-2}{5+7}\)\(=\dfrac{2x+3y-1}{12}\) (1)
\(\Rightarrow\dfrac{2x+3y-1}{12}=\dfrac{2x+3y-1}{6x}\)\(\Rightarrow6x=12\Rightarrow x=2\)
Thay vào (1), ta được:
\(\dfrac{2.2+1}{5}=\dfrac{3y-2}{7}\Rightarrow1=\dfrac{3y-2}{7}\) \(\Rightarrow3y-2=7\Rightarrow y=3\)
Vậy x=2 , y=3
Tìm 3 số x,y,biết :
\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\)
Tìm x,y,z biết :
1) \(\dfrac{x}{5}=\dfrac{y}{4}\)và x2 - y2=4
2) \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\)
3) \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)và 2x + 3y - z = 95
1.
Đặt \(\dfrac{x}{5}=\dfrac{y}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\end{matrix}\right.\)
\(\Rightarrow x^2-y^2=\left(5k\right)^2-\left(4k\right)^2=25k^2-16k^2=9k^2=4\)
\(\Rightarrow k^2=\dfrac{4}{9}\Rightarrow k=\pm\dfrac{2}{3}\)
\(\circledast k=\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10}{3}\\y=\dfrac{8}{3}\end{matrix}\right.\)
\(\circledast k=-\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{3}\\y=-\dfrac{8}{3}\end{matrix}\right.\)
2.
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+1+3y-2}{5+7}=\dfrac{2x+3y-1}{12}=\dfrac{2x+3y-1}{6x}\)
\(\Rightarrow6x=12\Rightarrow x=2\)
\(\Rightarrow y=\dfrac{\dfrac{2\cdot2+1}{5}\cdot7+2}{3}=3\)
3.
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Leftrightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-\left(z-3\right)}{4+9-4}=\dfrac{95-8+3}{9}=10\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10\cdot4+2}{2}=21\\y=\dfrac{10\cdot9+6}{3}=32\\z=10\cdot4+3=43\end{matrix}\right.\)
Tìm x , y ,z :
a, \(\dfrac{x+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
b, 10x = 6y và \(2x^2-y^2=-28\)
c, \(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
d, \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\)
Ta có : 2x+1 /5 = 3y-2/7 = 2x+3y -1 /6x
=> 2x+1+3y-2 / 5+7 = 2x+3y-1 /6x
=> 2x+3y-1 / 12 = 2x+3y-1 / 6x
=> 12 = 6x => x =2
A = \(\dfrac{5xy^2-3z}{3xy}+\dfrac{4x^2y+3z}{3xy}\)
B = \(\dfrac{3y+5}{y-1}+\dfrac{-y^2-4y}{1-y}+\dfrac{y^2+y+7}{y-1}\)
C = \(\dfrac{6x}{x^2-9}+\dfrac{5x}{x-3}+\dfrac{x}{x+3}\)
D = \(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\)
E = \(\dfrac{x^3+2x}{x^3+1}+\dfrac{2x}{x^2-x+1}+\dfrac{1}{x+1}\)
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)