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\(\left\{{}\begin{matrix}3x^2+xz-yz+y^2=2\left(1\right)\\y^2+xy-yz+z^2=0\left(2\right)\\x^2-xy-xz-z^2=2\left(3\right)\end{matrix}\right.\)
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 : x+xy+y=1
<=> x(y+1) + (y+1)=2
<=> (x+1)(y+1)=2
tương tự(y+1)(z+1)=5
(x+1)(z+1)=10
ta đc hệ pt............
đặt x+1=a,y+1=b,z+1=c
ta có : ab=2 (1) , bc=5 (2) , ac=10
=> abc=2c , abc=5a , abc= 10b
=> 5a=10b=2c
+ 5a=10b
=> a=2b . (1)=> 2b^2=1=> b=1 hoặc b=-1
=> a=2 hoăc a=-2 . (2)=> c=5 hoăc c=-5
like nha :))
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
1/Liên hợp đi cho nó nhẹ:D
ĐKXĐ: \(x\ge16\)
PT \(\Leftrightarrow\sqrt{x+24}-7+\sqrt{x-16}-3=0\)
\(\Leftrightarrow\frac{x-25}{\sqrt{x+24}+7}+\frac{x-25}{\sqrt{x-16}+3}=0\)
\(\Leftrightarrow\left(x-25\right)\left(\frac{1}{\sqrt{x+24}+7}+\frac{1}{\sqrt{x-16}+3}\right)=0\)
\(\Leftrightarrow x=25\)
\(\left\{{}\begin{matrix}xy+y+x+1=10\\yz+y+z+1=5\\zx+x+z+1=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(y+1\right)=10\\\left(y+1\right)\left(z+1\right)=5\\\left(z+1\right)\left(x+1\right)=2\end{matrix}\right.\)
\(\Rightarrow\left[\left(x+1\right)\left(y+1\right)\left(z+1\right)\right]^2=100\)
\(\Rightarrow\left(x+1\right)\left(y+1\right)\left(z+1\right)=10\)
\(\Rightarrow\left\{{}\begin{matrix}z+1=1\\x+1=2\\y+1=5\end{matrix}\right.\)
bn ơi tại sao z+1=1; x+1=2 và y+1=5 z?