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a)\(pt\left(2\right)\Leftrightarrow\left(5t-z\right)^2=0\Rightarrow z=5t\)
\(pt\left(3\right)\Leftrightarrow\left(x-2y\right)^2+\left(y-2z\right)^2=0\Rightarrow....\)
b)vĩ đại vậy chắc xài BĐT thôi, loanh quanh C-S và AM-GM 3 số
\(\left\{{}\begin{matrix}3x^2+xz-yz+y^2=2\left(1\right)\\y^2+xy-yz+z^2=0\left(2\right)\\x^2-xy-xz-z^2=2\left(3\right)\end{matrix}\right.\)
Lấy (2) cộng (3) ta được
\(x^2+y^2-yz-zx=2\) (4)
Lấy (1) - (4) ta được
\(2x\left(x+z\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-z\end{matrix}\right.\)
Xét 2 TH rồi thay vào tìm được y và z
1. \(\left\{{}\begin{matrix}6xy=5\left(x+y\right)\\3yz=2\left(y+z\right)\\7zx=10\left(z+x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{6}{5}\\\dfrac{y+z}{yz}=\dfrac{3}{2}\\\dfrac{z+x}{zx}=\dfrac{7}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{5}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{3}{2}\\\dfrac{1}{z}+\dfrac{1}{x}=\dfrac{7}{10}\end{matrix}\right.\)
Đến đây thì dễ rồi nhé
Ta có:
\(\sqrt{2x\left(x+y\right)^3}+y\sqrt{2\left(x^2+y^2\right)}\)
\(=\sqrt{\left(2x^2+2xy\right)\left(x^2+2xy+y^2\right)}+\sqrt{2}y.\sqrt{x^2+y^2}\)
\(\le\sqrt{\left(2x^2+2xy+2y^2\right)\left(x^2+2xy+y^2+x^2+y^2\right)}=2\left(x^2+xy+y^2\right)\)
\(\Rightarrow3\left(x^2+y^2\right)\le2\left(x^2+xy+y^2\right)\)
\(\Rightarrow\left(x-y\right)^2\le0\)
\(\Rightarrow x=y\)
Thế vào pt đầu:
\(x^2+3x+1=\left(x+3\right)\sqrt{x^2+1}\)
Đặt \(\sqrt{x^2+1}=t\Rightarrow t^2-\left(x+3\right)t+3x=0\)
\(\Delta=\left(x+3\right)^2-12x=\left(x-3\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{x+3-\left(x-3\right)}{2}=3\\t=\dfrac{x+3+x-3}{2}=x\end{matrix}\right.\)
\(\Rightarrow...\)
2. 4 biến xét dài quá, để người khác
b, Ta có : \(\left\{{}\begin{matrix}x+y+z=3\\y+z+t=4\\z+t+x=5\\t+x+y=6\end{matrix}\right.\)
=> \(x+y+z+y+z+t+z+t+x+t+x+y=18\)
=> \(3\left(x+y+z+t\right)=18\)
=> \(x+y+z+t=6\)
=> \(x+y+z+t=x+y+t\)
=> \(z=0\)
=> \(\left\{{}\begin{matrix}x+y=3\\y+t=4\\x+t=5\\x+y+t=6\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x+y=3\\y+t=4\\x+t=5\\y+5=6\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x+1=3\\t+1=4\\x+t=5\\y=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=2\\t=3\\x+t=5\\y=1\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=2\\y=1\\z=0\\t=3\end{matrix}\right.\)
a, Ta có : \(\left\{{}\begin{matrix}7xy=12\left(x+y\right)\\9yz=20\left(y+z\right)\\8zx=15\left(z+x\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}7xy-12x-12y=0\\9yz-20y-20z=0\\8zx-15z-15x=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{12y}{7y-12}\\y=\frac{20z}{9z-20}\\x=\frac{15z}{8z-15}\end{matrix}\right.\)
=> \(12y\left(8z-15\right)=15z\left(7y-12\right)\)
=> \(96yz-180y=105yz-180z\)
=> \(105yz-96yz=-180y+180z\)
=> \(9yz=-180y+180z\)
=> \(180z-180y=20y+20z\)
=> \(180z-20z=180y+20y=160z=200y\)
=> \(y=\frac{4}{5}z\)
=> \(\frac{20z}{9z-20}=\frac{4z}{5}\)
=> \(4z\left(9z-20\right)=100z\)
=> \(36z^2-180z=0\)
=> \(\left[{}\begin{matrix}z=5\\z=0\end{matrix}\right.\)
TH1 : z = 0 .
=> \(x=y=z=0\)
TH2 : z = 5 .
=> \(\left\{{}\begin{matrix}7xy=12\left(x+y\right)\\45y=20\left(y+5\right)\\40x=15\left(5+x\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=4\\x=3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=3\\y=4\\z=5\end{matrix}\right.\)
\(hpt\left\{{}\begin{matrix}3xy=2\left(x+y\right)\\5yz=6\left(y+z\right)\\4zx=3\left(x+z\right)\end{matrix}\right.\)\(\Rightarrow x=y=z=0\) \(là\) \(nghiệm\)
\(x=y=z\ne0\Rightarrow hpt\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2\left(x+y\right)}{2xy}=\dfrac{3xy}{2xy}\\\dfrac{6\left(y+z\right)}{6yz}=\dfrac{5yz}{6yz}\\\dfrac{3\left(x+z\right)}{3zx}=\dfrac{4xz}{3zx}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{2}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{5}{6}\\\dfrac{1}{x}+\dfrac{1}{z}=\dfrac{4}{3}\end{matrix}\right.\)\(ddặt\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)=\left(a;b;c\right)\Rightarrow\left\{{}\begin{matrix}a+b=\dfrac{3}{2}\\b+c=\dfrac{5}{6}\\a+c=\dfrac{4}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=1=\dfrac{1}{x}\Leftrightarrow x=1\left(tm\right)\\b=\dfrac{1}{2}=\dfrac{1}{y}\Leftrightarrow y=2\left(tm\right)\\c=\dfrac{1}{3}\Leftrightarrow z=3\left(tm\right)\end{matrix}\right.\)
TK
Hệ có nghiệm là x = y = z = 0
Với xyz ≠ 0 thì (I) được viết lại
\(\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{3}{2}\\\dfrac{y+z}{yz}=\dfrac{5}{6}\\\dfrac{z+x}{zx}=\dfrac{4}{3}\end{matrix}\right.\Leftrightarrow\left(II\right)\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{3}{2}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{5}{6}\\\dfrac{1}{z}+\dfrac{1}{x}=\dfrac{4}{3}\end{matrix}\right.\)
Cộng 3 phương trình của hệ (II) theo vế ta được
\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{11}{3}\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{11}{6}\)
Trừ phương trình trên cho từng phương trình của hệ (II) theo vế ta lần lượt có \(x=1,y=2,z=3\)
Vậy hệ phương trình có hai nghiệm \(\left(0;0;0\right)\&\left(1;2;3\right)\)