Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a,\(\left\{{}\begin{matrix}x=35\left(y+2\right)\\x=50\left(y-1\right)\end{matrix}\right.\)
suy ra :35(y+2)=50(y-1)
=>35y+70=50y-50
=>y=8
=>x=350
vậy :\(\left\{{}\begin{matrix}x=350\\y=8\end{matrix}\right.\)
b.\(\left\{{}\begin{matrix}y=2x-3\\y=x-1\end{matrix}\right.\)
suy ra: 2x-3=x-1
=>x=2
=>y=1
vậy \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
c.\(\left\{{}\begin{matrix}\left(x+14\right).\left(y-2\right)=xy\\\left(x-4\right).\left(y-1\right)=xy\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2x+14=0\\-x-y=0\end{matrix}\right.\)
vậy:\(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}y=\frac{6-x}{4}\\y=\frac{4x-5}{3}\end{matrix}\right.\)
x=2
y=1
vậy...
a) Ta có: \(\left\{{}\begin{matrix}3x-2\left|y\right|=9\\2x+3\left|y\right|=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-4\left|y\right|=18\\6x+9\left|y\right|=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-13\left|y\right|=15\\3x-2\left|y\right|=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|y\right|=\dfrac{-15}{13}\\3x-2\left|y\right|=9\end{matrix}\right.\Leftrightarrow\)Phương trình vô nghiệmVậy: \(S=\varnothing\)
$\begin{cases}3x-2|y|=9\\2x+3|y|=1\\\end{cases}$
`<=>` $\begin{cases}6x-4|y|=18\\6x+9|y|=3\\\end{cases}$
`<=>` $\begin{cases}13|y|=-15(loại)\\|3x|-2|y|=9\\\end{cases}$
Vậy HPT vô nghiệm
a)
HPT \(\Leftrightarrow \left\{\begin{matrix} 4x+8y=0(1)\\ 4x+2y=-3(2)\end{matrix}\right.\)
Lấy $(1)-(2)$ ta thu được: $8y-2y=3$
$\Leftrightarrow 6y=3\Leftrightarrow y=\frac{1}{2}$
Khi đó: $x=\frac{-4y}{2}=-2y=-1$
Vậy..........
b)
HPT \(\Leftrightarrow \left\{\begin{matrix} 2x-y=-4(1)\\ 2x+4y=-6(2)\end{matrix}\right.\)
Lấy $(1)-(2)$ suy ra: $-y-4y=-4-(-6)$
$\Leftrightarrow -5y=2\Rightarrow y=\frac{-2}{5}$
$\Rightarrow x=-3-2y=\frac{-11}{5}$
c)
HPT \(\Leftrightarrow \left\{\begin{matrix} xy+2x-15y-30=xy\\ xy-x+15y-15=xy\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} 2x-15y=30\\ -x+15y=15\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} 2x-15y=30(1)\\ -2x+30y=30(2)\end{matrix}\right.\)
Lấy $(1)+(2)$ suy ra $-15y+30y=60$
$\Leftrightarrow 15y=60\Leftrightarrow y=4$
$\Rightarrow x=15y-15=45$
Vậy.......
d)
HPT \(\Leftrightarrow \left\{\begin{matrix} \frac{2}{x}+\frac{2}{y}=10(1)\\ \frac{2}{x}+\frac{5}{y}=7(2)\end{matrix}\right.\)
Lấy \((2)-(1)\Rightarrow \frac{3}{y}=7-10=-3\Rightarrow y=-1\)
\(\Rightarrow \frac{1}{x}=5-\frac{1}{y}=5-\frac{1}{-1}=6\Rightarrow x=\frac{1}{6}\)
Vậy........
\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+6y=8+2x-3y\\5y-5x=5+3x+2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2x+6y+3y=8\\-5x-3x+5y-2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-8x+3y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-24x+9y=15\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}28x=-7\\4x+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{7}{28}=-\dfrac{1}{4}\\4.\left(-\dfrac{1}{4}\right)+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\y=1\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(-\dfrac{1}{4};1\right)\)
a) Ta có: \(\left\{{}\begin{matrix}2\left(x+1\right)-3\left(y-2\right)=5\\-4\left(x-2\right)+5\left(y-3\right)=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+2-3y+6=5\\-4x+8+5y-15=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-3\\-4x+5y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-6y=-6\\-4x+5y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-y=0\\2x-3y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\2x-3\cdot0=-3\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=0\end{matrix}\right.\)
Vậy: hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=0\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}8\left(x-3\right)-3\left(y+1\right)=-2\\3\left(x+2\right)-2\left(1-y\right)=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}8x-24-3y-3=-2\\3x+6-2+2y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}8x-3y=25\\3x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}24x-9y=75\\24x+16y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-25y=67\\3x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-67}{25}\\3x=1-2y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=1-2\cdot\dfrac{-67}{25}=\dfrac{159}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)
a) HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-3\\-4x+5y=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x-6y=-6\\-4x+5y=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-y=0\\x=\dfrac{3y-3}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(-\dfrac{3}{2};0\right)\)
b) HPT \(\Leftrightarrow\left\{{}\begin{matrix}8x-3y=25\\3x+2y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}16x-6y=50\\9x+6y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}25x=53\\y=\dfrac{1-3x}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(\dfrac{53}{25};-\dfrac{67}{25}\right)\)
\(\begin{cases} xy=20\\ x+y=9 \end{cases} \)
\(\Leftrightarrow\)\(\begin{cases} xy=20 (1)\\ x=9-y (2) \end{cases} \)
Thế (2) vào (1) ta được:
\((9-y)y=20\)
\(\Leftrightarrow\)\(9y-y^2-20=0\)
\(\Leftrightarrow\)\(\begin{cases} y=4\\ y=5 \end{cases} \)
Với y = 4 thay vào (2) ta được x = 5
Với y = 5 thay vào (2) ta được x = 4
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
Lời giải:
a.
Từ $x+y=2\Rightarrow y=2-x$. Thay vào PT(2):
$(m+1)x+m(2-x)=7$
$\Leftrightarrow x+2m=7$
$\Leftrightarrow x=7-2m$
$y=2-x=2-(7-2m)=2m-5$
Vậy hpt có nghiệm $(x,y)=(7-2m, 2m-5)(*)$
Nếu $x,y$ có 1 số $\geq 0$, một số $\leq 0$ thì $xy\leq 0< 1$
Nếu $x,y$ cùng $\geq 0$ thì áp dụng BĐT Cô-si:
$2=x+y\geq 2\sqrt{xy}\Rightarrow xy\leq 1$
Vậy tóm lại $xy\leq 1(**)$
Từ $(*); (**)$ suy ra với mọi $m$ thì hpt luôn có nghiệm $x,y$ thỏa mãn $xy\leq 1$
b.
$xy>0$
$\Leftrightarrow (7-2m)(2m-5)>0$
$\Leftrightarrow 7> 2m> 5$
$\Leftrightarrow \frac{7}{2}> m> \frac{5}{2}$
Do $m$ nguyên nên $m=3$
Thử lại thấy đúng.
Câu 1 \(\left\{{}\begin{matrix}2x+2y+2xy=10\left(1\right)\\x^2+y^2=5\left(2\right)\end{matrix}\right.\)
=>2.(2) - (1)=\(\left(x-1\right)^2+\left(y-1\right)^2+\left(x-y\right)^2=0\)
<=>\(\left\{{}\begin{matrix}x-1=0\\y-1=0\\x-y=0\end{matrix}\right.\) =>x=y=1
Câu 2 dùng vi-et đảo
Câu 3 rút x=y+1 từ pt trên rồi thế xuống dưới
Câu 4 lấy pt trên cộng pt dưới rồi xét dấu GTTĐ
\(\left\{{}\begin{matrix}\left(x-2\right)\left(y+1\right)=xy\\\left(x+8\right)\left(y-2\right)=xy\end{matrix}\right.\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\\\ \) \(\left\{{}\begin{matrix}xy+x-2y-2-xy=0\\xy-2x+8y-16-xy=0\end{matrix}\right.\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \)\(\left\{{}\begin{matrix}x-2y=2\\-2x+8y=16\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x-2y=2\\-x+4y=8\end{matrix}\right.\)\(\left\{{}\begin{matrix}2y=10\\x-2y=2\end{matrix}\right.\) \(\left\{{}\begin{matrix}y=5\\x-10=2\end{matrix}\right.\)\(\left\{{}\begin{matrix}y=5\\x=12\end{matrix}\right.\)
Vậy hpt có nghiệm duy nhất là (x;y) = (12;5)
Ta có: \(\left\{{}\begin{matrix}\left(x-2\right)\left(y+1\right)=xy\\\left(x+8\right)\left(y-2\right)=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy+x-2y-2-xy=0\\xy-2x+8y-16-xy=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2y-2=0\\-2x+8y-16=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=2\\-2x+8y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=4\\-2x+8y=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4y=20\\x-2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=2+2y=2+2\cdot5=12\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=12\\y=5\end{matrix}\right.\)