Giải hẹ phương trình:
\(\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\)
\(\dfrac{2x+6}{x-1}+\dfrac{3y+14}{y+3}=18\)
K
Khách
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28 tháng 9 2021
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:
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{y}=1\Leftrightarrow x=y=1\left(tm\right)\)
MH
11 tháng 10 2023
\(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=3\\\dfrac{x}{x+1}+\dfrac{3y}{y+1}=-1\end{matrix}\right.\)\(\left(Đk:x,y\ne-1\right)\)
\(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=3\\\dfrac{2x}{x+1}+\dfrac{6y}{y+1}=-2\end{matrix}\right.\)
\(\Rightarrow\dfrac{5y}{y+1}=-5\)
\(\Leftrightarrow5y=-5y-5\)
\(\Leftrightarrow10y=-5\)
\(\Leftrightarrow y=-\dfrac{1}{2}\Rightarrow x=-2\)
11 tháng 10 2023
<=>\(\dfrac{2x}{x+1}-\dfrac{x}{x+1}=4< =>x-4x=4< =>x=-\dfrac{4}{3}\Rightarrow y=-\dfrac{1}{4}\)
R
5 tháng 2 2022
TK
https://lazi.vn/edu/exercise/giai-phuong-trinh-4x-5-x-1-2-x-x-1-7-x-2-3-x-5
5 tháng 2 2022
a: \(\Leftrightarrow4x-5=2x-2+x\)
=>4x-5=3x-2
=>x=3(nhận)
b: =>7x-35=3x+6
=>4x=41
hay x=41/4(nhận)
c: \(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)
\(\Leftrightarrow\dfrac{28}{6\left(x-4\right)}-\dfrac{6\left(x+2\right)}{6\left(x-4\right)}=\dfrac{-9}{6\left(x-4\right)}-\dfrac{5\left(x-4\right)}{6\left(x-4\right)}\)
\(\Leftrightarrow28-6x-12=-9-5x+20\)
=>-6x+16=-5x+11
=>-x=-5
hay x=5(nhận)
d: \(\Leftrightarrow x^2+2x+1-\left(x^2-2x+1\right)=16\)
\(\Leftrightarrow4x=16\)
hay x=4(nhận)
13 tháng 10 2021
\(\dfrac{x}{3}=\dfrac{y}{4}\)
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
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:
\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y-z}{5-6-7}=\dfrac{36}{-8}=\dfrac{-9}{2}\)
\(x=\dfrac{-45}{2}\)
\(y=-27\)
\(z=\dfrac{-63}{2}\)
6 tháng 10 2021
a) \(\left\{{}\begin{matrix}5a+b=5\\b-10a=-19\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}5a+b=5\\15a=24\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{8}{5}\\b=-3\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}=17\\\dfrac{5}{x}+\dfrac{6}{y}=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}=17\\\dfrac{6}{x}=30\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{1}{2}\end{matrix}\right.\)
29 tháng 8 2021
Bài 2:
Ta có: \(A=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)
\(\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{2x+6}{x-1}+\dfrac{3y+14}{y+3}=18\end{matrix}\right.\left(x\ne1;y\ne-3\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{2x-2+8}{x-1}+\dfrac{3y+9+5}{y+3}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{2\left(x-1\right)}{x-1}+\dfrac{8}{x-1}+\dfrac{3\left(y+3\right)}{y+3}+\dfrac{5}{y+3}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\2+\dfrac{8}{x-1}+3+\dfrac{5}{y+3}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x-1}+\dfrac{7}{y+3}=19\\\dfrac{8}{x-1}+\dfrac{5}{y+3}=13\end{matrix}\right.\) (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)
⎪ ⎪⎨⎪ ⎪ ⎪⎩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)