a,\(\dfrac{x}{-4}=\dfrac{-9}{x}\)
b,\(\dfrac{x-1}{-15}=\dfrac{-60}{x-1}\)
c,\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\)
d,\(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)
Ai tìm x giúp mình với!!Mình sẽ tích điểm cho!!
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 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
1. tìm x
a.\(x+\dfrac{1}{2}=5^3:5^2\)
b. \(\dfrac{x-1}{-15}\)=\(\dfrac{-60}{x-1}\)
c. \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\)
a: =>x+1/2=5
=>x=9/2
b: =>(x-1)^2=900
=>x-1=30 hoặc x-1=-30
=>x=-29 hoặc x=31
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: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
Tìm x,y,z biết :
1) \(\dfrac{x}{-7}=\dfrac{y}{4}\) và \(2x-3y=-78\)
2) \(\dfrac{x}{y}=\dfrac{9}{7};\dfrac{y}{z}=\dfrac{7}{3}\) và \(x-y+z=-15\)
1. Ta có: \(\dfrac{x}{-7}=\dfrac{y}{4}\Rightarrow\dfrac{2x}{-14}=\dfrac{3y}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{2x-3y}{-14-12}=\dfrac{-78}{-26}=3\)
=> \(\left\{{}\begin{matrix}x=-21\\y=12\end{matrix}\right.\)
2. Ta có:
- \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)
- \(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)
=> \(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\dfrac{x-y+z}{9-7+3}=\dfrac{-15}{5}=-3\)
=> \(\left\{{}\begin{matrix}x=-27\\y=-21\\z=-9\end{matrix}\right.\)
Giải hpt
a)\(\left\{{}\begin{matrix}\dfrac{4}{2x-3y}+\dfrac{5}{3x+y}=-2\\\dfrac{3}{3x+y}-\dfrac{5}{2x-3y}=21\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{7}{x-y+2}-\dfrac{5}{x+y-1}=\dfrac{9}{2}\\\dfrac{3}{x-y+2}+\dfrac{2}{x+y-1}=4\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\dfrac{3}{2x-y}-\dfrac{6}{x+y}=-1\\\dfrac{1}{2x-y}-\dfrac{1}{x+y}=0\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{4}{x+y-1}-\dfrac{5}{2x-y+3}=\dfrac{5}{2}\\\dfrac{3}{x+y-1}+\dfrac{1}{2x-y+3}=\dfrac{7}{5}\end{matrix}\right.\)
e)\(\left\{{}\begin{matrix}\dfrac{6}{x-2y}+\dfrac{2}{x+2y}=3\\\dfrac{3}{x-2y}+\dfrac{4}{x+2y}=-1\end{matrix}\right.\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x< >\dfrac{3}{2}y\\x< >-\dfrac{y}{3}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{4}{2x-3y}+\dfrac{5}{3x+y}=-2\\\dfrac{-5}{2x-3y}+\dfrac{3}{3x+y}=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{20}{2x-3y}+\dfrac{25}{3x+y}=-10\\-\dfrac{20}{2x-3y}+\dfrac{12}{3x+y}=84\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{37}{3x+y}=74\\-\dfrac{5}{2x-3y}+\dfrac{3}{3x+y}=21\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x+y=\dfrac{1}{2}\\-\dfrac{5}{2x-3y}+3:\dfrac{1}{2}=21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+y=\dfrac{1}{2}\\\dfrac{-5}{2x-3y}=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x+y=\dfrac{1}{2}\\2x-3y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+3y=\dfrac{3}{2}\\2x-3y=-\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}11x=\dfrac{7}{6}\\2x-3y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{66}\\3y=2x+\dfrac{1}{3}=\dfrac{7}{33}+\dfrac{1}{3}=\dfrac{6}{11}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{7}{66}\\y=\dfrac{2}{11}\end{matrix}\right.\)(nhận)
b: ĐKXĐ: \(\left\{{}\begin{matrix}x< >y-2\\x< >-y+1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{7}{x-y+2}-\dfrac{5}{x+y-1}=\dfrac{9}{2}\\\dfrac{3}{x-y+2}+\dfrac{2}{x+y-1}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{14}{x-y+2}-\dfrac{10}{x+y-1}=9\\\dfrac{15}{x-y+2}+\dfrac{10}{x+y-1}=20\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{29}{x-y+2}=29\\\dfrac{3}{x-y+2}+\dfrac{2}{x+y-1}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-y+2=1\\3+\dfrac{2}{x+y-1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\\dfrac{2}{x+y-1}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-y=-1\\x+y-1=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x=2\\x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)(nhận)
c:
ĐKXĐ: \(\left\{{}\begin{matrix}y< >2x\\y< >-x\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3}{2x-y}-\dfrac{6}{x+y}=-1\\\dfrac{1}{2x-y}-\dfrac{1}{x+y}=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{2x-y}-\dfrac{6}{x+y}=-1\\\dfrac{3}{2x-y}-\dfrac{3}{x+y}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{x+y}=-1\\\dfrac{1}{2x-y}-\dfrac{1}{x+y}=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y=3\\2x-y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=6\\2x-y=3\end{matrix}\right.\)
=>x=2 và y=2x-3=4-3=1(nhận)
d:ĐKXĐ: \(\left\{{}\begin{matrix}x< >-y+1\\x< >\dfrac{1}{2}y-\dfrac{3}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{4}{x+y-1}-\dfrac{5}{2x-y+3}=\dfrac{5}{2}\\\dfrac{3}{x+y-1}+\dfrac{1}{2x-y+3}=\dfrac{7}{5}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{4}{x+y-1}-\dfrac{5}{2x-y+3}=\dfrac{5}{2}\\\dfrac{15}{x+y-1}+\dfrac{5}{2x-y+3}=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{19}{x+y-1}=\dfrac{19}{2}\\\dfrac{15}{x+y-1}+\dfrac{5}{2x-y+3}=7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y-1=2\\\dfrac{15}{2}+\dfrac{5}{2x-y+3}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\\dfrac{5}{2x-y+3}=7-\dfrac{15}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+y=3\\2x-y+3=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\2x-y=-13\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=-10\\x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{3}\\y=3-x=3+\dfrac{10}{3}=\dfrac{19}{3}\end{matrix}\right.\left(nhận\right)\)
e:
ĐKXĐ: \(x\ne\pm2y\)
\(\left\{{}\begin{matrix}\dfrac{6}{x-2y}+\dfrac{2}{x+2y}=3\\\dfrac{3}{x-2y}+\dfrac{4}{x+2y}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{6}{x-2y}+\dfrac{2}{x+2y}=3\\\dfrac{6}{x-2y}+\dfrac{8}{x+2y}=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{6}{x+2y}=5\\\dfrac{3}{x-2y}+\dfrac{4}{x+2y}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+2y=-\dfrac{6}{5}\\\dfrac{3}{x-2y}+4:\dfrac{-6}{5}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+2y=-\dfrac{6}{5}\\\dfrac{3}{x-2y}=-1+4\cdot\dfrac{5}{6}=-1+\dfrac{10}{3}=\dfrac{7}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+2y=-\dfrac{6}{5}\\x-2y=\dfrac{9}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{3}{35}\\x-2y=\dfrac{9}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{3}{70}\\2y=x-\dfrac{9}{7}=-\dfrac{87}{70}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{70}\\y=-\dfrac{87}{140}\end{matrix}\right.\left(nhận\right)\)
1/ Tìm x,y biết:
a/ \(\dfrac{x}{2}\) = \(\dfrac{y}{5}\) và x+y=-21
b/ 7x = 3y và x-y=16
c/ \(\dfrac{x}{y}\) = \(\dfrac{5}{9}\) và 3x+2x=66
d/ \(\dfrac{x}{15}\) = \(\dfrac{y}{7}\) và x-2y=16
e/ \(\dfrac{x}{5}\) = \(\dfrac{y}{2}\) và x × y = 1000
2/ Tìm x,y,z biết
\(\dfrac{x}{13}\) = \(\dfrac{y}{7}\) = \(\dfrac{z}{5}\) và x-y-z=6
a. Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{x}{2}=\frac{y}{5}=\frac{x+y}{2+5}=\frac{-21}{7}=-3$
$\Rightarrow x=2(-3)=-6; y=5(-3)=-15$
b. Áp dụng tính chất dãy tỉ số bằng nhau:
$7x=3y=\frac{x}{\frac{1}{7}}=\frac{y}{\frac{1}{3}}=\frac{x-y}{\frac{1}{7}-\frac{1}{3}}=\frac{16}{\frac{-4}{21}}=-84$
$\Rightarrow x=(-84):7=-12; y=-84:3=-28$
c. $\frac{x}{y}=\frac{5}{9}\Rightarrow \frac{x}{5}=\frac{y}{9}$
Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{x}{5}=\frac{y}{9}=\frac{3x}{15}=\frac{2y}{18}=\frac{3x+2y}{15+18}=\frac{66}{33}=2$
$\Rightarrow x=2.5=10; y=9.2=18$
d. Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{x}{15}=\frac{y}{7}=\frac{2y}{14}=\frac{x-2y}{15-14}=\frac{16}{1}=16$
$\Rightarrow x=16.15=240; y=7.16=112$
e.
Đặt $\frac{x}{5}=\frac{y}{2}=k\Rightarrow x=5k ; y=2k$
Khi đó: $xy=5k.2k=10k^2=1000\Rightarrow k^2=100\Rightarrow k=\pm 10$
Với $k=10$ thì $x=5k=50; y=2k=20$
Với $k=-10$ thì $x=5k=-50; y=2k=-20$
Bài 2:
Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{x}{13}-\frac{y}{7}-\frac{z}{5}=\frac{x-y-z}{13-7-5}=\frac{6}{1}=6$
$\Rightarrow x=13.6=78; y=7.6=42; z=5.6=30$
Giải các hệ phương trình sau
f.{ (2x - y) (x + 3y) = 4
{ (5x + y) (x + 3y) = 24
g.{ \(\dfrac{8x-5y-3}{7}+\dfrac{11y-4x-7}{5}=12\)
{ \(\dfrac{9x+4y-13}{5}+\dfrac{3\left(x-2\right)}{4}=15\)
h.{\(\dfrac{1}{x}+\dfrac{1}{y}=2\)
{\(\dfrac{3}{x}-\dfrac{4}{y}=-1\)
h) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{3}{x}-\dfrac{4}{y}=-1\end{matrix}\right.\)\(\left(1\right)\)\(\left(đk:x,y\ne0\right)\)
Đặt \(a=\dfrac{1}{x},b=\dfrac{1}{y}\)
\(\left(1\right)\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=2\\3a-4b=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6\\3a-4b=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\7b=7\end{matrix}\right.\)\(\Leftrightarrow a=b=1\)
Thay a,b:
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{y}=1\Leftrightarrow x=y=1\left(tm\right)\)
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