1)Giải hệ phương trình
\(\left\{{}\begin{matrix}\sqrt{\dfrac{4x}{5y}}=\sqrt{x+y}-\sqrt{x-y}\\\sqrt{\dfrac{5y}{x}}=\sqrt{x+y}+\sqrt{x-y}\end{matrix}\right.\)
2) Tìm MIN MAX
\(P=\dfrac{x}{2+yz}+\dfrac{y}{2+zx}+\dfrac{z}{2+xy}\)
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Những câu hỏi liên quan
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 2 2021
ĐKXĐ: ...
Xét pt đầu: \(\Leftrightarrow\dfrac{x^2-2xy+y^2-1}{xy}-2+\dfrac{2}{x+y}+4=0\)
\(\Leftrightarrow\dfrac{x^2+y^2-1}{xy}+\dfrac{2}{x+y}=0\)
\(\Leftrightarrow\left(x+y\right)\left(x^2+y^2-1\right)+2xy=0\)
\(\Leftrightarrow\left(x+y-1\right)\left(x^2+y^2-1\right)+x^2+y^2-1+2xy=0\)
\(\Leftrightarrow\left(x+y-1\right)\left(x^2+y^2-1\right)+\left(x+y\right)^2-1=0\)
\(\Leftrightarrow\left(x+y-1\right)\left(x^2+y^2-1\right)+\left(x+y-1\right)\left(x+y+1\right)=0\)
\(\Leftrightarrow\left(x+y-1\right)\left(x^2+y^2+x+y\right)=0\)
Từ ĐKXĐ \(x+y-1\ge0\Rightarrow x+y\ge1\Rightarrow x^2+y^2+x+y>0\)
\(\Rightarrow x+y-1=0\Rightarrow y=1-x\)
Thế xuống pt dưới:
\(4x^2-5x+5+6\sqrt{x}=13\)
\(\Leftrightarrow4x^2-4x+1-x+6\sqrt{x}-9=0\)
\(\Leftrightarrow\left(2x-1\right)^2-\left(\sqrt{x}-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\sqrt{x}-3\\2x-1=3-\sqrt{x}\end{matrix}\right.\)
\(\Leftrightarrow...\)
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
...
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 1 2021
a.
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\y\ge3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\5\sqrt{x-2}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\\sqrt{x-2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y-3}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 1 2021
b.
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{4x}{x+1}-\dfrac{10}{y+4}=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{19x}{x+1}=28\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{28}{19}\\\dfrac{1}{y+4}=-\dfrac{4}{19}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}19x=28x+28\\4y+16=-19\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{9}\\y=-\dfrac{35}{4}\end{matrix}\right.\)
H9
HT.Phong (9A5)
CTVHS
13 tháng 7 2023
Ta có: \(\left\{{}\begin{matrix}\dfrac{3}{\sqrt{x+y}}-\dfrac{2}{\sqrt{x-y}}=4\\\dfrac{2}{\sqrt{x+y}}-\dfrac{1}{\sqrt{x-y}}=5\end{matrix}\right.\)
Đặt: \(t=\sqrt{x+y}\) và \(k=\sqrt{x-y}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{t}-\dfrac{2}{k}=4\\\dfrac{2}{t}+\dfrac{1}{k}=5\end{matrix}\right.\)
Ta lại đặt: \(a=\dfrac{1}{t}\) và \(u=\dfrac{1}{k}\)
\(\Leftrightarrow\left\{{}\begin{matrix}3a-2u=4\\2a+u=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3a-2u=4\\4a+2u=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3a-2u=4\\7a=14\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6-2u=4\\a=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}u=1\\a=2\end{matrix}\right.\)
Mà:
\(u=1\Rightarrow\dfrac{1}{k}=1\Rightarrow k=1\)
\(a=2\Rightarrow\dfrac{1}{t}=2\Rightarrow t=\dfrac{1}{2}\)
Ta lại có:
\(k=1\Rightarrow\sqrt{x+y}=1\)
\(t=\dfrac{1}{2}\Rightarrow\sqrt{x-y}=\dfrac{1}{2}\)
Ta có hệ:
\(\left\{{}\begin{matrix}\sqrt{x-y}=1\\\sqrt{x+y}=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=1\\x+y=\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=1\\2x=\dfrac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{8}-y=1\\x=\dfrac{5}{8}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{3}{8}\\x=\dfrac{5}{8}\end{matrix}\right.\)
Vậy \(x-\dfrac{5}{8};y=-\dfrac{3}{8}\)
13 tháng 7 2023
Đặt 1/căn x+y=a; 1/căn x-y=b
Theo đề, ta có hệ:
3a-2b=4 và 2a+b=5
=>a=2 và b=1
=>x+y=1/4 và x-y=1
=>x=5/8 và y=-3/8
24 tháng 1 2021
a) Ta có: \(\left\{{}\begin{matrix}\sqrt{2}x-y=3\\x+\sqrt{2}y=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2}x-y=3\\\sqrt{2}x+2y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-3y=1\\x+\sqrt{2}y=\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\sqrt{2}-\sqrt{2}y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\sqrt{2}-\sqrt{2}\cdot\dfrac{-1}{3}=\dfrac{4\sqrt{2}}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{4\sqrt{2}}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}\dfrac{x}{2}-2y=\dfrac{3}{4}\\2x+\dfrac{y}{3}=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-8y=3\\2x+\dfrac{1}{3}y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{25}{3}y=\dfrac{10}{3}\\2x-8y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{2}{5}\\2x=3+8y=3+8\cdot\dfrac{-2}{5}=-\dfrac{1}{5}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-\dfrac{1}{10}\\y=-\dfrac{2}{5}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{1}{10}\\y=-\dfrac{2}{5}\end{matrix}\right.\)
c) Ta có: \(\left\{{}\begin{matrix}\dfrac{2x-3y}{4}-\dfrac{x+y-1}{5}=2x-y-1\\\dfrac{x+y-1}{3}+\dfrac{4x-y-2}{4}=\dfrac{2x-y-3}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5\left(2x-3y\right)}{20}-\dfrac{4\left(x+y-1\right)}{20}=\dfrac{20\left(2x-y-1\right)}{20}\\\dfrac{4\left(x+y-1\right)}{12}+\dfrac{3\left(4x-y-2\right)}{12}=\dfrac{2\left(2x-y-3\right)}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10x-15y-4x-4y+4=40x-20y-20\\4x+4y-4+12x-3y-6=4x-2y-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-19y+4-40x+20y+20=0\\16x+y-10-4x+2y+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-34x+y=-24\\12x+3y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-102x+3y=-72\\12x+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-114x=-76\\12x+3y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\12\cdot\dfrac{2}{3}+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\3y=4-8=-4\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
13 tháng 12 2022
a: \(\Leftrightarrow\left\{{}\begin{matrix}8x-4y+12-3x+6y-9=48\\9x-12y+9+16x-8y-36=48\end{matrix}\right.\)
=>5x+2y=48-12+9=45 và 25x-20y=48+36-9=48+27=75
=>x=7; y=5
b: \(\Leftrightarrow\left\{{}\begin{matrix}6x+6y-2x+3y=8\\-5x+5y-3x-2y=5\end{matrix}\right.\)
=>4x+9y=8 và -8x+3y=5
=>x=-1/4; y=1
c: \(\Leftrightarrow\left\{{}\begin{matrix}-4x-2+1,5=3y-6-6x\\11,5-12+4x=2y-5+x\end{matrix}\right.\)
=>-4x-0,5=-6x+3y-6 và 4x-0,5=x+2y-5
=>2x-3y=-5,5 và 3x-2y=-4,5
=>x=-1/2; y=3/2
e: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
=>\(x=\sqrt{2};y=\sqrt{3}\)
Câu 1/
\(\left\{{}\begin{matrix}\sqrt{\dfrac{4x}{5y}}=\sqrt{x+y}-\sqrt{x-y}\left(1\right)\\\sqrt{\dfrac{5y}{x}}=\sqrt{x+y}+\sqrt{x-y}\left(2\right)\end{matrix}\right.\)
Lấy (1).(2) vế theo vế được
\(\left(\sqrt{x+y}-\sqrt{x-y}\right)\left(\sqrt{x+y}+\sqrt{x-y}\right)=2\)
\(\Leftrightarrow x+y-\left(x-y\right)=2\)
\(\Leftrightarrow2y=2\)
\(\Leftrightarrow y=1\)
Thế vô tìm được x.
Câu 2/ Đề chưa đủ. x, y, z thuộc R luôn à. Tìm min hay max hay là tìm cả 2.