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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 .
=> \(\left\{{}\begin{matrix}7xy=12\left(x+y\right)\\45y=20\left(y+5\right)\\40x=15\left(5+x\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=4\\x=3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=3\\y=4\\z=5\end{matrix}\right.\)
Từ pt (2) \(\Rightarrow t=15-y-z\) thay xuống 2 pt dưới:
\(\left\{{}\begin{matrix}x+y+z=10\\x+z+15-y-z=14\\x+y+15-y-z=12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=10\\x-y=-1\\x-z=-3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y=3\\z=5\\t=15-y-z=7\end{matrix}\right.\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}2x+2y+4z=8\\2x-y+3z=6\\2x-6y+8z=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y+z=2\\8y-4z=1\\x+y+2z=4\end{matrix}\right.\)
=>y=9/20; z=13/20; x=4-y-2z=9/4
b: \(\Leftrightarrow\left\{{}\begin{matrix}z=23-x-y\\z=31-y-t\\z=27-t-x\\x+y+t=33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x-y+23=-y-t+31\\-y-t-31=-x-t+27\\x+y+t=33\\z=23-x-y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x+t=8\\x-y=58\\x+y+t=33\\z=23-x-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}t=x+8\\y=x-58\\x-58+x+8+x=33\\z=23-x-y\end{matrix}\right.\)
=>x=83/3; t=107/3; y=-91/3; z=23-83/3+91/3=77/3
\(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y+1=2\\yz+y+z+1=5\\zx+z+x+1=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(y+1\right)=2\\\left(y+1\right)\left(z+1\right)=5\\\left(z+1\right)\left(x+1\right)=10\end{matrix}\right.\) (1)
Nhân vế với vế: \(\left[\left(x+1\right)\left(y+1\right)\left(z+1\right)\right]^2=100\)
\(\Leftrightarrow\left(x+1\right)\left(y+1\right)\left(z+1\right)=10\) (2)
Chia vế cho vế của (2) cho từng pt của (1):
\(\Rightarrow\left\{{}\begin{matrix}z+1=5\\x+1=2\\y+1=1\end{matrix}\right.\) \(\Rightarrow\left(x;y;z\right)=\left(1;0;4\right)\) (loại)
Hệ vô nghiệm do \(y>0\)
Ta có:
\(\left(1.x+3.y+4.z+1.t\right)^2\le\left(1^2+3^2+4^2+1^2\right)\left(x^2+y^2+z^2+t^2\right)\)
\(\Leftrightarrow\left(x+3y+4z+t\right)^2\le27\left(x^2+y^2+z^2+t^2\right)\)
Dấu "=" xảy ra khi và chỉ khi: \(x=\frac{y}{3}=\frac{z}{4}=t\Leftrightarrow\left\{{}\begin{matrix}y=3x\\z=4x\\t=x\end{matrix}\right.\)
Thay vào pt dưới:
\(x^3+27x^3+64x^3+x^3=93\)
\(\Leftrightarrow x=1\Rightarrow\left\{{}\begin{matrix}y=3\\z=4\\t=1\end{matrix}\right.\)
b) Áp dụng bđt Svac-xơ:
\(\dfrac{1}{x}+\dfrac{9}{y}+\dfrac{16}{z}\ge\dfrac{\left(1+3+4\right)^2}{x+y+z}\ge\dfrac{64}{4}=16>9\)
=> hpt vô nghiệm
c) Ở đây x,y,z là các số thực dương
Áp dụng cosi: \(x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\ge xyz\left(x+y+z\right)=3xyz\)
Dấu = xảy ra khi \(x=y=z=\dfrac{3}{3}=1\)
Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix} (x+y)+(z+t)=4(1)\\ (x+y)-(z+t)=8(2)\\ (x-y)+(z-t)=12(3)\\ (x-y)-(z-t)=16(4)\end{matrix}\right.\)
Lấy \((1)+(2)\Rightarrow 2(x+y)=12\Rightarrow x+y=6(5)\)
Lấy \((3)+(4)\Rightarrow 2(x-y)=28\Rightarrow x-y=14(6)\)
Lấy \((5)+(6)\Rightarrow 2x=20\Rightarrow x=10\Rightarrow y=6-10=-4\)
Lấy \((1)-(2)\Rightarrow 2(z+t)=-4\Rightarrow z+t=-2(7)\)
Lấy \((3)-(4)\Rightarrow 2(z-t)=-4\Rightarrow z-t=-2(8)\)
Lấy \((7)+(8)\Rightarrow 2z=-4\Rightarrow z=-2\Rightarrow t=-2-z=0\)
Vậy \((x,y,z,t)=(10,-4,-2,0)\)