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Lời giải:
$x,y,z>0$ thì $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ mới xác định.
Áp dụng BĐT AM-GM:
$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 3\sqrt[3]{xyz}.3\sqrt[3]{\frac{1}{xyz}}=9$
Dấu "=" xảy ra khi $x=y=z$. Thay vào pt $(2)$:
$x^3=x^2+x+2$
$\Leftrightarrow x^3-x^2-x-2=0$
$\Leftrightarrow x^2(x-2)+x(x-2)+(x-2)=0$
$\Leftrightarrow (x^2+x+1)(x-2)=0$
Dễ thấy $x^2+x+1>0$ với mọi $x>0$ nên $x-2=0$
$\Rightarrow x=2$
Vậy hpt có nghiệm $(x,y,z)=(2,2,2)$
Giải hệ sau :
Câu a :
\(\left\{{}\begin{matrix}x+y=-1\\2x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\-x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\x=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=-3\\x=2\end{matrix}\right.\)
Vậy ...........................
Câu b :
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\) . Ta có :
\(\left\{{}\begin{matrix}a+b=\dfrac{1}{5}\\3a+4b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b=\dfrac{3}{5}\\3a+4b=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-b=-\dfrac{7}{5}\\3a+4b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{7}{5}\\a=-\dfrac{6}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{7}{5}\\\dfrac{1}{y}=-\dfrac{6}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{7}\\y=-\dfrac{5}{6}\end{matrix}\right.\)
Vậy..................
\(a,\left\{{}\begin{matrix}2x-y=4\\x+5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=4\\2x+10y=6\end{matrix}\right.\left\{{}\begin{matrix}11y=2\\2x+10y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{11}\\2x+10.\dfrac{2}{11}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{11}\\2x=\dfrac{46}{11}\end{matrix}\right.\left\{{}\begin{matrix}y=\dfrac{2}{11}\\x=\dfrac{23}{11}\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=-3\\\dfrac{3}{x}-\dfrac{2}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y}=-10\\\dfrac{1}{x}+\dfrac{1}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)
`{(x+3y=x(5y-1)),(1/x-3/y=-2):}` `ĐK: x; y ne 0`
`<=>{(x+3y=5xy-x),(-3x+y=-2xy):}`
`<=>{(5xy-2x=3y),(-3x+y=-2xy):}`
`<=>{(x(5y-2)=3y),(-3x+y=-2xy):}`
`<=>{(x=[3y]/[5y-2]),(-3x+y=-2xy):}` `ĐK: y ne 2/5` (TH `y=2/5` ko t/m)
`<=>{(x=[3y]/[5y-2]),(-3[3y]/[5y-2]+y=-2[3y]/[5y-2]y):}`
`<=>{(x=[3y]/[5y-2]),(-9y+5y^2-2y=-6y^2):}`
`<=>{(x=[3y]/[5y-2]),(11y^2-11y=0):}`
`<=>{(x=[3y]/[5y-2]),([(y=0(ko t//m)),(y=1(t//m)):}):}`
`<=>{(x=[3. 1]/[5.1-2]=1),(y=1):}` (t/m)
a: \(\left\{{}\begin{matrix}\dfrac{x}{35}-y=2\\y-\dfrac{x}{50}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x-35y}{35}=2\\\dfrac{50y-x}{50}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-35y=70\\-x+50y=50\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}15y=120\\x-35y=70\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=8\\x=70+35y=70+35\cdot8=350\end{matrix}\right.\)
b: ĐKXĐ: x<>0 và y<>0
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}\\\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=\dfrac{3}{16}\\\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{y}=\dfrac{3}{16}-\dfrac{1}{4}=\dfrac{-1}{16}\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=48\\\dfrac{1}{x}=\dfrac{1}{16}-\dfrac{1}{48}=\dfrac{2}{48}=\dfrac{1}{24}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=24\\y=48\end{matrix}\right.\left(nhận\right)\)
a: \(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}+1\\5x-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\5\cdot\left(\dfrac{2}{3}y+2\right)-8y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\\dfrac{10}{3}y+10-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{14}{3}y=-7\\x=\dfrac{2}{3}y+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=7:\dfrac{14}{3}=7\cdot\dfrac{3}{14}=\dfrac{3}{2}\\x=\dfrac{2}{3}\cdot\dfrac{3}{2}+2=1+2=3\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}3x+2y=2\\6x-3y=18\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=2-2y\\2\cdot3x-3y=18\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=2-2y\\2\left(2-2y\right)-3y=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4-7y=18\\3x=2-2y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7y=-14\\3x=2-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\3x=2-2\cdot\left(-2\right)=6\end{matrix}\right.\)
=>x=2 và y=-2
Còn lại tương tự
Lời giải:
Phương hướng giải là bạn sử dụng phương pháp thế, biểu diễn $x$ theo $y$ qua 1 trong 2 PT, sau đó thế vô PT còn lại giải PT 1 ẩn $y$
a) \(\left\{\begin{matrix}
x-6y=17\\
5x+y=23\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x=17+6y\\
5x+y=23\end{matrix}\right.\)
\(\Rightarrow 5(17+6y)+y=23\)
\(\Leftrightarrow 31y=-62\Leftrightarrow y=-2\)
$x=17+6y=17+6(-2)=5$
Vậy $(x,y)=(5,-2)$
Các phần còn lại bạn giải tương tự
b) $(x,y)=(\frac{1}{4}, 0)$
c) $(x,y)=(3, 4)$
d) $(x,y)=(\frac{79}{21}, \frac{44}{21})$
\(\Leftrightarrow\dfrac{x^2-1}{x}\ge0\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x}\ge0\)
Trường hợp 1:
\(\left\{{}\begin{matrix}\left(x-1\right)\left(x+1\right)\ge0\\x>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\\x>0\end{matrix}\right.\Leftrightarrow x\ge1\)
Trường hợp 2:
\(\left\{{}\begin{matrix}\left(x-1\right)\left(x+1\right)\le0\\x< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-1\le x\le1\\x< 0\end{matrix}\right.\Leftrightarrow-1\le x< 0\)
Vậy hệ có nghiệm \(S=[1;+\infty)\cup [-1;0)\)