6/7 x y + 14/36 + 1/7 x y = 7/18
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7 tháng 6 2017
( 2 x y + 2/15 ) x 3 = 4/5
( 2 x y + 2/15 ) = 4/5 : 3
( 2 x y + 2/15 ) = 4/15
2 x y = 4/15 - 2/15
2 x y = 2/15
y = 2/15 :2
y = 1/15
DP
7 tháng 6 2017
(2 x y + 2/15) x 3 = 4/5
2 x y + 2/15) = 4/5 : 3
2 x y + 2/15 = 4/15
2 x y = 4/15 - 2/15
2 x y = 2/15
y = 2/15 : 2
y = 1/15
7/9 x (2 - 1/3 x y) = 14/15
(2 - 1/3 x y) = 14/15 : 7/9
(2 - 1/3 x y) = 6/5
2 - y = 6/5 x 1/3
2 - y = 2/5
y = 2/5 + 2
y = 12/5
4/21 + 5 x y - 8/7 = 1/3
4/21 + 5 x y = 1/3 + 8/7
4/21 + 5 x y = 31/21
5 x y = 31/21 - 4/21
5 x y = 9/7
y = 9/7 : 5
y = 9/35
7/12 x y - 3/12 x y = 5
y x (7/12 - 3/12) = 5
y x 1/3 = 5
y = 5 : 1/3
y = 15
7 tháng 12 2021
\(1,\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{21}{7}=3\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=15\end{matrix}\right.\\ 2,7x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{16}{-4}=-4\\ \Rightarrow\left\{{}\begin{matrix}x=-12\\y=-28\end{matrix}\right.\\ 3,\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=-\dfrac{9}{2}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{45}{2}\\y=-27\\z=-\dfrac{63}{2}\end{matrix}\right.\\ 4,x:y:z=3:5:7\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)
7 tháng 12 2021
3. Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=\dfrac{-9}{2}\)
\(x=\dfrac{-45}{2}\)
\(y=-27\)
\(z=\dfrac{-63}{2}\)
H9
HT.Phong (9A5)
CTVHS
20 tháng 11 2023
\(\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{2x+6}{x-1}+\dfrac{3y+14}{y+3}=18\end{matrix}\right.\left(x\ne1;y\ne-3\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{2x-2+8}{x-1}+\dfrac{3y+9+5}{y+3}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{2\left(x-1\right)}{x-1}+\dfrac{8}{x-1}+\dfrac{3\left(y+3\right)}{y+3}+\dfrac{5}{y+3}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\2+\dfrac{8}{x-1}+3+\dfrac{5}{y+3}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{8}{x-1}+\dfrac{5}{y+3}=13\end{matrix}\right.\) (I)
Đặt: \(\left\{{}\begin{matrix}u=\dfrac{1}{x-1}\\v=\dfrac{1}{y+3}\end{matrix}\right.\)
Hệ (I) trở thành:
\(\Leftrightarrow\left\{{}\begin{matrix}12u+7v=19\\8u+5v=13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}24u+14v=38\\24u+15v=39\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}12u+7=19\\v=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}12u=12\\v=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}u=1\\v=1\end{matrix}\right.\)
Trả ẩn phụ:
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-1}=1\\\dfrac{1}{y+3}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y+3=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\left(tm\right)\)
Vậy hệ pt có 1 cặp nghiệm duy nhất là: (2;-2)
MK
20 tháng 11 2023
⎪ ⎪⎨⎪ ⎪ ⎪⎩12x−1+7y+3=192x+6x−1+3y+14y+3=18(x≠1;y≠−3){12�−1+7�+3=192�+6�−1+3�+14�+3=18(�≠1;�≠−3)
⇔⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩12x−1+7y+3=192x−2+8x−1+3y+9+5y+3=18⇔{12�−1+7�+3=192�−2+8�−1+3�+9+5�+3=18
⇔⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩12x−1+7y+3=192(x−1)x−1+8x−1+3(y+3)y+3+5y+3=18⇔{12�−1+7�+3=192(�−1)�−1+8�−1+3(�+3)�+3+5�+3=18
⇔⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩12x−1+7y+3=192+8x−1+3+5y+3=18⇔{12�−1+7�+3=192+8�−1+3+5�+3=18
⇔⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩12x−1+7y+3=198x−1+5y+3=13⇔{12�−1+7�+3=198�−1+5�+3=13 (I)
Đặt: ⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩u=1x−1v=1y+3{�=1�−1�=1�+3
Hệ (I) trở thành:
⇔{12u+7v=198u+5v=13⇔{12�+7�=198�+5�=13
⇔{24u+14v=3824u+15v=39⇔{24�+14�=3824�+15�=39
⇔{12u+7=19v=1⇔{12�+7=19�=1
⇔{12u=12v=1⇔{12�=12�=1
⇔{u=1v=1⇔{�=1�=1
Trả ẩn phụ:
⇔⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩1x−1=11y+3=1⇔{1�−1=11�+3=1
⇔{x−1=1y+3=1⇔{�−1=1�+3=1
⇔{x=2y=−2(tm)⇔{�=2�=−2(��)
Vậy hệ pt có 1 cặp nghiệm duy nhất là: (2;-2)
LV
2
10 tháng 3 2022
a: =y x15/5=3y
b: =(4/6+14/6)+(7/13+19/13)+(17/9-8/9)
=3+2+1
=6
CQ
1
28 tháng 2 2021
`-7/6=x/18`
`=>-21/18=x/18`
`=>x=-21(TM\ x in Z)`
`-7/6=-98/y`
`=>-98/84=-98/y`
`=>y=84(TM\ y in Z)`
`-7/6=-14/z`
`=>-14/12=-14/z`
`=>z=12(TM\ z in Z)`
`-7/6=t/102`
`=>-119/102=t/102`
`=>t=-119(TM\ t in Z)`
Vậy `(x,y,z,t)=(-21,84,12,-119)`
25 tháng 8 2018
a) ADTCDTSBN
có: \(\frac{x}{2}=\frac{z}{4}=\frac{x+z}{2+4}=\frac{18}{6}=3.\)
=> x/2 = 3 => x = 6
y/3 = 3 => y = 9
z/4 = 3 => z = 12
KL:...
b,c làm tương tự nha
d) ta có: \(\frac{x}{5}=\frac{y}{-6}=\frac{z}{7}=\frac{2x}{10}\)
ADTCDTSBN
có: \(\frac{2x}{10}=\frac{y}{-6}=\frac{z}{7}=\frac{2x+y-z}{10+\left(-6\right)-7}=\frac{49}{-3}\)
=>...
25 tháng 8 2018
e) ADTCDTSBN
có: \(\frac{x+1}{2}=\frac{y+2}{3}=\frac{z+3}{4}=\frac{x+1+y+2+z+3}{2+3+4}=\frac{\left(x+y+z\right)+\left(1+2+3\right)}{9}\)
\(=\frac{21+6}{9}=\frac{27}{9}=3\)
=>...
g) ta có: \(\frac{x}{4}=\frac{y}{3}=k\Rightarrow\hept{\begin{cases}x=4k\\y=3k\end{cases}}\)
mà xy = 12 => 4k.3k = 12
12.k2 = 12
k2 = 1
=> k = 1 hoặc k = -1
=> x = 4.1 = 4
y = 3.1 = 3
x=4.(-1) = -4
y=3.(-1) = -3
KL:...
h) ta có: \(\frac{x}{5}=\frac{y}{3}\Rightarrow\frac{x^2}{25}=\frac{y^2}{9}\)
ADTCDTSBN
có: \(\frac{x^2}{25}=\frac{y^2}{9}=\frac{x^2-y^2}{25-9}=\frac{16}{16}=1\)
=>...
H
2
18 tháng 2 2019
\(\frac{-7}{6}=\frac{x}{18}=\frac{-98}{y}=\frac{-14}{z}=\frac{t}{102}=\frac{u}{-18}\)
\(\text{+) }\frac{-7}{6}=\frac{x}{18}\)
\(\Rightarrow\left(-7\right).18=6.x\)
\(\Rightarrow-126=6.x\)
\(\Rightarrow x=-21\)
\(\text{+) }\frac{-7}{6}=\frac{-98}{y}\)
\(\Rightarrow\left(-7\right).y=6.\left(-98\right)\)
\(\Rightarrow y=\frac{6.\left(-98\right)}{-7}\)
\(\Rightarrow y=-84\)
\(\text{+) }\frac{-7}{6}=\frac{-14}{z}\)
\(\Rightarrow\left(-7\right).z=\left(-14\right).6\)
\(\Rightarrow\left(-7\right).z=-84\)
\(\Rightarrow z=12\)
\(\frac{-7}{6}=\frac{t}{102}\)
\(\Rightarrow\left(-7\right).102=t.6\)
\(\Rightarrow t=\frac{\left(-7\right).102}{6}=\frac{\left(-7\right).17.6}{6}\)
\(\Rightarrow t=-119\)
\(\frac{-7}{6}=\frac{u}{-18}\)
\(\Rightarrow\frac{-7}{6}=\frac{-u}{18}\)
\(\Rightarrow\left(-7\right).18=\left(-u\right).6\)
\(\Rightarrow-126=\left(-u\right).6\)
\(\Rightarrow-u=-21\)
\(\Rightarrow u=21\)
\(\frac{6}{7}.y+\frac{7}{18}+\frac{1}{7}.y=\frac{7}{18}\)
\(\frac{6}{7}.y+\frac{1}{7}.y+\frac{7}{18}=\frac{7}{18}\)
\(\frac{6}{7}.y+\frac{1}{7}.y=\frac{7}{18}-\frac{7}{18}\)
\(y.\left(\frac{6}{7}+\frac{1}{7}\right)=0\)
\(y.1=0\)
\(y=0:1\)
\(y=0\)