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a, Đặt \(\hept{\begin{cases}\frac{1}{x}=u\\\frac{1}{y}=v\end{cases}}\left(u;v\ne0\right)\)
\(\Leftrightarrow\hept{\begin{cases}u+v=\frac{5}{6}\\\frac{1}{6}u+\frac{1}{5}v=\frac{3}{20}\end{cases}}\Leftrightarrow\hept{\begin{cases}u=\frac{5}{6}-v\left(1\right)\\\frac{1}{6}u+\frac{1}{5}v=\frac{3}{20}\left(2\right)\end{cases}}\)
Thay (1) vào (2) ta được : \(\frac{1}{6}\left(\frac{5}{6}-v\right)+\frac{1}{5}v=\frac{3}{20}\)
\(\Leftrightarrow\frac{5}{36}-\frac{v}{6}+\frac{v}{5}=\frac{3}{20}\)
\(\Leftrightarrow\frac{-v}{6}+\frac{v}{5}=\frac{3}{20}-\frac{5}{36}\Leftrightarrow\frac{v}{30}=\frac{1}{90}\Leftrightarrow v=\frac{1}{3}\)(*)
hay \(v=\frac{1}{3}=\frac{1}{y}\Rightarrow y=3\)
Thay (*) vào (1) ta được : \(u=\frac{5}{6}-\frac{1}{3}=\frac{1}{2}\)hay \(u=\frac{1}{2}=\frac{1}{x}\Rightarrow x=2\)
Vậy x = 2 ; y = 3
b, \(\hept{\begin{cases}4\left(x+y\right)=5\left(x-y\right)\\\frac{40}{x+y}+\frac{40}{x-y}=9\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{4}{x-y}=\frac{5}{x+y}\left(1\right)\\\frac{40}{x+y}+\frac{40}{x-y}=9\left(2\right)\end{cases}}\)
Xét phương trình 1 ta có : \(\frac{4}{x-y}-\frac{5}{x+y}=0\)
\(\Leftrightarrow\frac{4\left(x+y\right)-5\left(x-y\right)}{\left(x+y\right)\left(x-y\right)}=0\Leftrightarrow4x+4y-5x+5y=0\)
\(\Leftrightarrow-x+9y=0\Leftrightarrow x=9y\)(*)
Thay vào 2 ta có : \(\frac{40}{9y+y}+\frac{40}{9y-y}=9\)
\(\Leftrightarrow\frac{4}{y}+\frac{5}{y}=9\Leftrightarrow\frac{9}{y}=9\Leftrightarrow y=1\)
Thay y = 1 vào (*) ta có : \(x=9.1=9\)
Vậy x = 9 ; y = 1
Câu 1:
\(\left\{{}\begin{matrix}\left(x+y\right)\left(x^2+y^2\right)=15\\\left(x+y\right)\left(x-y\right)^2=3\end{matrix}\right.\)
\(\Leftrightarrow\left(x+y\right)\left(x^2+y^2\right)=5\left(x+y\right)\left(x-y\right)^2\)
\(\Leftrightarrow x^2+y^2=5\left(x-y\right)^2\)
\(\Leftrightarrow2x^2-5xy+2y^2=0\)
\(\Leftrightarrow\left(2x-y\right)\left(x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2x\\x=2y\end{matrix}\right.\)
TH1: \(y=2x\Rightarrow3x\left(x^2+4x^2\right)=15\Leftrightarrow x^3=1\Rightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
TH2: \(x=2y\Rightarrow3y\left(4y^2+y^2\right)=15\Rightarrow y^3=1\Rightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Câu 2:
\(\left\{{}\begin{matrix}x^3-y^3=9\\3x^2+6y^2=3x-12y\end{matrix}\right.\)
\(\Leftrightarrow x^3-y^3-3x^2-6y^2=9-3x+12y\)
\(\Leftrightarrow x^3-3x^2+3x-1=y^3+6y^2+12y+8\)
\(\Leftrightarrow\left(x-1\right)^3=\left(y+2\right)^3\)
\(\Leftrightarrow x-1=y+2\Rightarrow x=y+3\)
\(\Rightarrow\left(y+3\right)^2+2y^2=y+3-4y\)
\(\Leftrightarrow y^2+3y+2=0\Rightarrow\left[{}\begin{matrix}y=-1\Rightarrow x=2\\y=-2\Rightarrow x=1\end{matrix}\right.\)
3a)\(\left\{{}\begin{matrix}\dfrac{1}{x-2}+\dfrac{1}{2y-1}=2\\\dfrac{2}{x-2}-\dfrac{3}{2y-1}=1\end{matrix}\right.\) (ĐK: x≠2;y≠\(\dfrac{1}{2}\))
Đặt \(\dfrac{1}{x-2}=a;\dfrac{1}{2y-1}=b\) (ĐK: a>0; b>0)
Hệ phương trình đã cho trở thành
\(\left\{{}\begin{matrix}a+b=2\\2a-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\2\left(2-b\right)-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\4-2b-3b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\b=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{7}{5}\left(TM\text{Đ}K\right)\\b=\dfrac{3}{5}\left(TM\text{Đ}K\right)\end{matrix}\right.\) Khi đó \(\left\{{}\begin{matrix}\dfrac{1}{x-2}=\dfrac{7}{5}\\\dfrac{1}{2y-1}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\left(x-2\right)=5\\3\left(2y-1\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7x-14=5\\6y-3=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{7}\left(TM\text{Đ}K\right)\\y=\dfrac{4}{3}\left(TM\text{Đ}K\right)\end{matrix}\right.\) Vậy hệ phương trình đã cho có nghiệm duy nhất (x;y)=\(\left(\dfrac{19}{7};\dfrac{4}{3}\right)\)
b) Bạn làm tương tự như câu a kết quả là (x;y)=\(\left(\dfrac{12}{5};\dfrac{-14}{5}\right)\)
c)\(\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=13\\2\sqrt{x-1}-\sqrt{y}=4\end{matrix}\right.\)(ĐK: x≥1;y≥0)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}+2\sqrt{y}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}+4\sqrt{x-1}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7\sqrt{x-1}=13\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}49\left(x-1\right)=169\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}49x-49=169\\\sqrt{y}=2\sqrt{x-1}-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{218}{49}\\y=\dfrac{4}{49}\end{matrix}\right.\left(TM\text{Đ}K\right)\)
Bài 4:
Theo đề, ta có hệ:
\(\left\{{}\begin{matrix}3\left(3a-2\right)-2\left(2b+1\right)=30\\3\left(a+2\right)+2\left(3b-1\right)=-20\end{matrix}\right.\)
=>9a-6-4b-2=30 và 3a+6+6b-2=-20
=>9a-4b=38 và 3a+6b=-20+2-6=-24
=>a=2; b=-5
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...
b) Lấy pt đầu trừ pt dưới thu được:
\(x^3-y^3+2\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2+2\right)=0\)
Do \(x^2+xy+y^2=\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+2>0\)
Do đó x = y. Thay vào pt đầu thu được:
\(x^3-2x-1=0\Leftrightarrow\left(x+1\right)\left(x^2-x-1\right)=0\)
c) Lấy pt trên trừ pt dưới:
\(2\left(x^2-y^2\right)-3\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(2x+2y-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\2x+2y-3=0\end{matrix}\right.\)
Auto làm nốt:D
P/s: Is that true?
`{(x+1)(y-1)=2),((x-3)(y+1)=-6):}`
`<=>{(xy-x+y-1=2),(xy+x-3y-3=-6):}`
`<=>{(x(y-1)=3-y),(xy+x-3y-3=-6):}`
`<=>{(x=[3-y]/[y-1]\text{ (1)}),(xy+x-3y=-3\text{ (2)}):}`
Thay `(1)` vào `(2)` có:
`[3-y]/[y-1] .y+[3-y]/[y-1]-3y=-3`
`=>3y-y^2+3-y-3y^2+3y=-3y+3`
`<=>4y^2-8y=0`
`<=>[(y=0),(y=2):}`
`=>[(x=[3-0]/[0-1]=-3),(x=[3-2]/[2-1]=1):}`
Vậy `S={(-3;0),(1;2)}`