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a, \(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)
a, Ta có : \(\left\{{}\begin{matrix}3x+2y=-2\\-x+4y=3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}3\left(4y-3\right)+2y=-2\\x=4y-3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}12y-9+2y=-2\\x=4y-3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}14y=7\\x=4y-3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=\frac{1}{2}\\x=\frac{4.1}{2}-3=-1\end{matrix}\right.\)
Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(-1;\frac{1}{2}\right)\)
b, Ta có : \(\left\{{}\begin{matrix}x+2y=11\\5x-3y=3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=11-2y\\5\left(11-2y\right)-3y=3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=11-2y\\55-10y-3y=3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=11-2y\\-13y=-52\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=11-2.4=3\\y=4\end{matrix}\right.\)
Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;4\right)\)
c, Ta có : \(\left\{{}\begin{matrix}10x-9y=1\\15x+21y=36\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}30x-27y=3\\30x+42y=72\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}10x-9y=1\\-69y=-69\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}10x-9=1\\y=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(1;1\right)\)
d, Ta có : \(\left\{{}\begin{matrix}2x+y=3\\x+y=2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3-2x\\x+2-2x=2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3-2x\\2-x=2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3-2.0=3\\x=0\end{matrix}\right.\)
Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(0;3\right)\)
e, Ta có : \(\left\{{}\begin{matrix}x+y=2\\2x-3y=9\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=2-y\\2\left(2-y\right)-3y=9\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=2-y\\4-2y-3y=9\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=2-y\\-5y=5\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=2+1=3\\y=-1\end{matrix}\right.\)
Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;-1\right)\)
f, Ta có : \(\left\{{}\begin{matrix}x-2y=11\\5x+3y=3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=11+2y\\5\left(11+2y\right)+3y=3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=11+2y\\55+10y+3y=3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=11+2y\\13y=-52\end{matrix}\right.\)
Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;-4\right)\)
g, Ta có : \(\left\{{}\begin{matrix}3x-y=5\\2x+3y=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\2x+3\left(3x-5\right)=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\2x+9x-15=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\11x=33\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=9-5=4\\x=3\end{matrix}\right.\)
Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;4\right)\)
h, Ta có : \(\left\{{}\begin{matrix}5x+3y=-7\\3x-y=-8\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}5x+3\left(3x+8\right)=-7\\y=3x+8\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}5x+9x+24=-7\\y=3x+8\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}14x=-31\\y=3x+8\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=-\frac{31}{14}\\y=3.\left(-\frac{31}{14}\right)+8=\frac{19}{14}\end{matrix}\right.\)
Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(-\frac{31}{14};\frac{19}{14}\right)\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\dfrac{2}{15}\\y\ne-\dfrac{4}{9}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}9y+6+20x-16=0\\\left(5x-4\right)\left(9y+4\right)=\left(3y+2\right)\left(15x-2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}20x+9y=10\\5x+15y=-6\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{5}\\y=-\dfrac{2}{3}\end{matrix}\right.\)
Từ pt dưới:
\(x^2+9y^2=6xy\Leftrightarrow x^2-6xy+9y^2=0\)
\(\Leftrightarrow\left(x-3y\right)^2=0\Leftrightarrow x-3y=0\Leftrightarrow x=3y\)
Thế lên pt trên: \(2.\left(3y\right)^2+y^2=19\)
\(\Leftrightarrow19y^2=19\Leftrightarrow y^2=1\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=3\\y=-1\Rightarrow x=-3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^3-y^3=35\\2x^2+3y^2=4x-9y\left(1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y^3-x^3=-35\\3y^2+9y+2x^2-4x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y^3-x^3=-35\\9y^2+27y+6x^2-12x=0\end{matrix}\right.\)
\(\Rightarrow\left(y^3+9y^2+27y\right)-\left(x^3-6x^2+12x\right)=-35\)
\(\Rightarrow\left(y^3+9y^2+27y+27\right)-\left(x^3-6x^2+12x-8\right)=0\)
\(\Rightarrow\left(y+3\right)^3-\left(x-2\right)^2=0\)
\(\Rightarrow\left(y-x+5\right)\left[\left(y+3\right)^2+\left(y+3\right)\left(x-2\right)+\left(x-2\right)^2\right]=0\)
*Với \(x=y+5\). Thay vào (1) ta được:
\(2\left(y+5\right)^2+3y^2=4\left(y+5\right)-9y\)
\(\Leftrightarrow2y^2+20y+50+3y^2=4y+20-9y\)
\(\Leftrightarrow5y^2+25y+30=0\Leftrightarrow y^2+5y+6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=-2\\y=-3\end{matrix}\right.\)
*\(y=-2\Rightarrow x=3\) ; \(y=-3\Rightarrow x=2\).
*Với \(\left(y+3\right)^2+\left(y+3\right)\left(x-2\right)+\left(x-2\right)^2=0\). Ta có:
\(\left(y+3\right)^2+\left(y+3\right)\left(x-2\right)+\left(x-2\right)^2\)
\(=\left[\left(y+3\right)+\dfrac{\left(x-2\right)}{2}\right]^2+\dfrac{3}{4}\left(x-2\right)^2\ge0\)
Dấu "=" xảy ra khi \(x=2;y=-3\)
Vậy \(x=2;y=-3\)
Thử lại ta có nghiệm (x;y) của hệ đã cho là \(\left(3;-2\right),\left(2;-3\right)\)
Nhận thấy \(x=0\) không phải nghiệm, hệ tương đương:
\(\left\{{}\begin{matrix}21y-20=\dfrac{1}{x^3}\\y^3+20=\dfrac{21}{x}\end{matrix}\right.\)
Cộng vế với vế:
\(y^3+21y=\dfrac{1}{x^3}+\dfrac{21}{x}\)
\(\Leftrightarrow y^3-\dfrac{1}{x^3}+21\left(y-\dfrac{1}{x}\right)=0\)
\(\Leftrightarrow\left(y-\dfrac{1}{x}\right)\left(y^2+\dfrac{y}{x}+\dfrac{1}{x^2}\right)+21\left(y-\dfrac{1}{x}\right)=0\)
\(\Leftrightarrow\left(y-\dfrac{1}{x}\right)\left(y^2+\dfrac{y}{x}+\dfrac{1}{x^2}+21\right)=0\)
\(\Leftrightarrow y=\dfrac{1}{x}\)
\(\Leftrightarrow...\)
`{([10x+y]/[x+y]=6),(xy+25=10y+x):}` `ĐK: x \ne -y`
`<=>{(10x+y=6x+6y),(xy+25=10y+x):}`
`<=>{(y=4/5x),(x. 4/5x+25=10. 4/5x+x):}`
`<=>{(y=4/5x),(4/5x^2-9x+25=0):}`
`<=>{(y=4/5x),([(x=25/4),(x=5):}):}`
`<=>[({(x=25/4),(y=4/5 . 25/4=5):}),({(x=5),(y=4/5 .5=4):}):}` (t/m)
\(\left\{{}\begin{matrix}\dfrac{10x+y}{x+y}=6\\xy+25=10y+x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10x+y=6\left(x+y\right)\\xy-10y-x=-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x-5y=0\\xy-10y-x=-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5y}{4}\\xy-10y-x=-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5y}{4}\\\dfrac{5y}{4}y-10y-\dfrac{5y}{4}=-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5y}{4}\\\dfrac{5y^2-45y}{4}=-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5y}{4}\\5y^2-45y+100=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5y}{4}\\y_1=5\\y_2=4\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm \(\left\{{}\begin{matrix}x=\dfrac{25}{4}\\y=5\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=5\\y=4\end{matrix}\right.\)
a: \(\left\{{}\begin{matrix}3x-2y=11\\4x-5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=11+2y\\4x-5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\4\left(\dfrac{2}{3}y+\dfrac{11}{3}\right)-5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\\dfrac{8}{3}y+\dfrac{44}{3}-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\-\dfrac{7}{3}y=3-\dfrac{44}{3}=-\dfrac{35}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=5\\x=\dfrac{2}{3}\cdot5+\dfrac{11}{3}=\dfrac{10}{3}+\dfrac{11}{3}=\dfrac{21}{3}=7\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}+1\\5x-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\5\left(\dfrac{2}{3}y+2\right)-8y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\\dfrac{10}{3}y+10-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{14}{3}y=3-10=-7\\x=\dfrac{2}{3}y+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=7:\dfrac{14}{3}=7\cdot\dfrac{3}{14}=\dfrac{3}{2}\\x=\dfrac{2}{3}\cdot\dfrac{3}{2}+2=3\end{matrix}\right.\)
c: \(\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x+8\\3x+5\left(2x+8\right)=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2x+8\\3x+10x+40=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=2x+8\\13x=-39\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-3\\y=2\cdot\left(-3\right)+8=8-6=2\end{matrix}\right.\)
d: \(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y\\x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3}y+y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{5}{3}y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=\dfrac{2}{3}\cdot6=4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}150x-135y=15\\150x+210y=360\end{matrix}\right.\)
\(\Leftrightarrow-345y=-345\)
\(\Rightarrow y=1\left(1\right)\)
Thay (1) vào ptr đầu: \(10x-9\cdot1=1\)
\(\Rightarrow y=1\)
Uhm, sorry bạn, bạn sử lại giúp mình dòng suy ra cuối cùng là \(\Rightarrow x=1\) nhé!