\(\dfrac{72}{x}+\dfrac{54}{y}=6\)
x-y=6
giải/hệ/pt
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TA
31 tháng 5 2021
Đặt \(\left[{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\).
Ta có hệ: \(\left[{}\begin{matrix}a+b=\dfrac{1}{16}\\3a+6b=\dfrac{1}{4}\end{matrix}\right.\)
`<=>` \(\left[{}\begin{matrix}a=\dfrac{1}{24}\\b=\dfrac{1}{48}\end{matrix}\right.\)
`=>` \(\left[{}\begin{matrix}x=24\\y=48\end{matrix}\right.\)
Vậy `(x;y)=(24;48)`.
1 tháng 1 2019
Câu a : \(4\sqrt{x+1}=x^2-5x+14\)
\(\Leftrightarrow x^2-5x+14-4\sqrt{x+1}=0\)
\(\Leftrightarrow\left(x^2-6x+9\right)+\left(x+1-4\sqrt{x+1}+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)^2=0\\\left(\sqrt{x+1}-2\right)^2=0\end{matrix}\right.\Leftrightarrow x=3\)
Câu b : \(\left\{{}\begin{matrix}y=x^2\\z=xy\\\dfrac{1}{x}=\dfrac{1}{y}+\dfrac{6}{z}\end{matrix}\right.\) ( ĐK : \(x,y,z\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=x^2\left(1\right)\\z=x^3\left(2\right)\\\dfrac{1}{x}=\dfrac{1}{x^2}+\dfrac{6}{x^3}\left(3\right)\end{matrix}\right.\)
Xét phương trình (3) :
\(\left(3\right)\Leftrightarrow x^2=x+6\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Thay từng giá trị của x vào pt (1) và (2) . Ta được những cặp nghiệm :
\(\left\{{}\begin{matrix}\left(x;y;z\right)=\left(-2;4;-8\right)\\\left(x;y;z\right)=\left(3;9;27\right)\end{matrix}\right.\)
2
2
7 tháng 10 2023
ĐKXĐ: x<>0; y<>0
\(\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{3}{y}=1\\\dfrac{2}{x}+\dfrac{1}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{3}{y}=1\\\dfrac{6}{x}+\dfrac{3}{y}=-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{1}{x}=4\\\dfrac{2}{x}+\dfrac{1}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\\dfrac{1}{y}=-1-\dfrac{2}{x}=-1-2:\dfrac{-1}{4}=-1+8=7\end{matrix}\right.\)
=>x=-1/4 và y=1/7
D
datcoder
CTVVIP
7 tháng 10 2023
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}a=\dfrac{1}{x}\\b=\dfrac{1}{y}\end{matrix}\right.\)
Hệ phương trình trở thành \(\left\{{}\begin{matrix}5a+3b=1\\2a+b=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}b=-1-2a\\5a+3\left(-1-2a\right)=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=-1-2a\\-a-3=1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a=-4\\b=-1-2.\left(-4\right)\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=-4\\b=7\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}a=\dfrac{1}{x}=-4\\b=\dfrac{1}{y}=7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\left(tm\right)\\y=\dfrac{1}{7}\left(tm\right)\end{matrix}\right.\)
Vậy HPT có nghiệm \(x=-\dfrac{1}{4}\) và \(y=\dfrac{1}{7}\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 12 2020
1. Với mọi số thực x;y;z ta có:
\(x^2+y^2+z^2+\dfrac{1}{2}\left(x^2+1\right)+\dfrac{1}{2}\left(y^2+1\right)+\dfrac{1}{2}\left(z^2+1\right)\ge xy+yz+zx+x+y+z\)
\(\Leftrightarrow\dfrac{3}{2}P+\dfrac{3}{2}\ge6\)
\(\Rightarrow P\ge3\)
\(P_{min}=3\) khi \(x=y=z=1\)
1.1
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}=a>0\\\dfrac{1}{\sqrt{y}}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+\sqrt{2-b^2}=2\\b+\sqrt{2-a^2}=2\end{matrix}\right.\)
\(\Rightarrow a-b+\sqrt{2-b^2}-\sqrt{2-a^2}=0\)
\(\Leftrightarrow a-b+\dfrac{\left(a-b\right)\left(a+b\right)}{\sqrt{2-b^2}+\sqrt{2-a^2}}=0\)
\(\Leftrightarrow a=b\Leftrightarrow x=y\)
Thay vào pt đầu:
\(a+\sqrt{2-a^2}=2\Rightarrow\sqrt{2-a^2}=2-a\) (\(a\le2\))
\(\Leftrightarrow2-a^2=4-4a+a^2\Leftrightarrow2a^2-4a+2=0\)
\(\Rightarrow a=1\Rightarrow x=y=1\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 12 2020
2.
\(\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\x^2-xy+y^2=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x^2+3xy+3y^2=21\\7x^2-7xy+7y^2=21\end{matrix}\right.\)
\(\Rightarrow4x^2-10xy+4y^2=0\)
\(\Leftrightarrow2\left(2x-y\right)\left(x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2x\\y=\dfrac{1}{2}x\end{matrix}\right.\)
Thế vào pt đầu
...
13 tháng 2 2022
\(\left(x\ne-y;x>\dfrac{y}{2}\right)\Rightarrow\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2x-y}}-\dfrac{21}{x+y}=\dfrac{1}{2}\\\dfrac{3}{\sqrt{2x-y}}+\dfrac{7-\left(x+y\right)}{x+y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2x-y}}-\dfrac{21}{x+y}=\dfrac{1}{2}\\\dfrac{3}{\sqrt{2x-y}}+\dfrac{7}{x+y}=2\end{matrix}\right.\)
\(đặt:\dfrac{1}{\sqrt{2x-y}}=a>0;\dfrac{1}{x+y}=b\)
\(\Rightarrow\left\{{}\begin{matrix}4a-21b=\dfrac{1}{2}\\3a+7b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\left(tm\right)\\b=\dfrac{1}{14}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{2x-y}}=\dfrac{1}{2}\\\dfrac{1}{x+y}=\dfrac{1}{14}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=4\\x+y=14\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=8\end{matrix}\right.\)(thỏa)
23 tháng 1 2022
ĐK: \(x\ne0\) ; \(y\ne0\)
Hệ phương trình tương đương với:
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)=4\\\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2=8\end{matrix}\right.\)
Đặt \(S=\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)\)
\(P=\left(x+\dfrac{1}{x}\right)\left(y+\dfrac{1}{y}\right)\)
Mà \(S^2\ge4P\)
Ta có: \(\left\{{}\begin{matrix}S=4\\S^2-2P=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}S=4\\P=4\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)=4\\\left(x+\dfrac{1}{x}\right)\left(y+\dfrac{1}{y}\right)=4\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=2\\y+\dfrac{1}{y}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
16 tháng 12 2021
Lời giải:
Đặt $\frac{1}{x-y+2}=a;\frac{1}{x+y-1}=b$ thì HPT trở thành cơ bản:
\(\left\{\begin{matrix}
14a-10b=9\\
3a+2b=4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
14a-10b=9\\
15a+10b=20\end{matrix}\right.\)
$\Rightarrow (14a-10b)+(15a+10b)=9+20$
$\Leftrightarrow 29a=29\Leftrightarrow a=1$.
$b=\frac{4-3a}{2}=\frac{1}{2}$
Vậy: \(\left\{\begin{matrix} \frac{1}{x-y+2}=1\\ \frac{1}{x+y-1}=\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x-y+2=1\\ x+y-1=2\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x-y=-1\\ x+y=3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1\\ y=2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{72}{x}+\dfrac{54}{y}=6\\x-y=6\end{matrix}\right.\left(x,y>0\right)< =>\left\{{}\begin{matrix}x=6+y\left(1\right)\\\dfrac{72}{\left(6+y\right)}+\dfrac{54}{y}=6\left(2\right)\end{matrix}\right.\)(x,y>0,y\(\ne-6\))
giải pt(2) \(\dfrac{72}{\left(6+y\right)}+\dfrac{54}{y}=6< =>\dfrac{72y+54\left(6+y\right)}{y\left(6+y\right)}=6\)
\(< =>\dfrac{126y+324}{y\left(6+y\right)}=6=>126y+324=6y\left(6+y\right)\)
\(< =>126y+324=36y+6y^2\)
\(< =>-6y^2+90y+324=0\)
\(\Delta=90^2-4\left(-6\right).324=15876>0\)
=>x1=\(\dfrac{-90+\sqrt{15876}}{2\left(-6\right)}=-3\left(loai\right)\)
x2=\(\dfrac{-90-\sqrt{15876}}{2\left(-6\right)}=18\left(TM\right)\)
=>x=x2=18 thay vào pt(1)=>x=6+18=24
vậy (x,y)=(24,18)