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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:
\(\left\{{}\begin{matrix}x^3-y^3=9\\3x^2+6y^2=3x-12y\end{matrix}\right.\)
\(\Leftrightarrow x^3-y^3-3x^2-6y^2=9-3x+12y\)
\(\Leftrightarrow x^3-3x^2+3x-1=y^3+6y^2+12y+8\)
\(\Leftrightarrow\left(x-1\right)^3=\left(y+2\right)^3\)
\(\Leftrightarrow x-1=y+2\Rightarrow x=y+3\)
\(\Rightarrow\left(y+3\right)^2+2y^2=y+3-4y\)
\(\Leftrightarrow y^2+3y+2=0\Rightarrow\left[{}\begin{matrix}y=-1\Rightarrow x=2\\y=-2\Rightarrow x=1\end{matrix}\right.\)
c/ \(y=0\) không phải nghiệm
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+1+y\left(x+y\right)=4y\\y\left(x+y\right)^2-2\left(x^2+1\right)=7y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x^2+1}{y}+x+y=4\\\left(x+y\right)^2-2\left(\frac{x^2+1}{y}\right)=7\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\\frac{x^2+1}{y}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=4\\a^2-2b=7\end{matrix}\right.\) \(\Rightarrow a^2-2\left(4-a\right)=7\)
\(\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}a=3\Rightarrow b=1\\a=-5\Rightarrow b=9\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+y=3\\\frac{x^2+1}{y}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=3-x\\x^2+1-y=0\end{matrix}\right.\)
\(\Rightarrow x^2+1-\left(3-x\right)=0\Rightarrow...\)
TH2: làm tương tự
a/ \(\left\{{}\begin{matrix}\left(x-y\right)\left(x^2+y^2\right)=13\\\left(x-y\right)\left(x+y\right)^2=25\end{matrix}\right.\)
Do \(x=y;x=-y\) đều ko phải nghiệm
\(\Rightarrow\frac{x^2+y^2}{\left(x+y\right)^2}=\frac{13}{25}\Leftrightarrow25\left(x^2+y^2\right)=13\left(x+y\right)^2\)
\(\Leftrightarrow12x^2-26xy+12y^2=0\)
\(\Leftrightarrow\left(2x-3y\right)\left(3x-2y\right)=0\Rightarrow\left[{}\begin{matrix}y=\frac{2}{3}x\\y=\frac{3}{2}x\end{matrix}\right.\)
Thay vào 1 trong 2 pt ban đầu là xong
b/ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\y\ge0\end{matrix}\right.\) \(\Rightarrow x+y>0\)
\(xy+x+y+y^2=x^2-y^2\)
\(\Leftrightarrow x\left(y+1\right)+y\left(y+1\right)=\left(x-y\right)\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)\left(y+1\right)=\left(x+y\right)\left(x-y\right)\)
\(\Leftrightarrow y+1=x-y\Rightarrow x=2y+1\)
Thay vào pt dưới:
\(\left(2y+1\right)\sqrt{2y}+y\sqrt{2y}=2\left(y+1\right)\)
\(\Leftrightarrow\sqrt{2y}\left(3y+1\right)=2\left(y+1\right)\)
\(\Leftrightarrow y\left(9y^2+6y+1\right)=2\left(y^2+2y+1\right)\)
\(\Leftrightarrow9y^3+2y^2-3y-2=0\)
Nghiệm quá xấu, bạn coi lại đề
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
1,ĐK: \(x,y\ne-2\)
HPT<=> \(\left\{{}\begin{matrix}x\left(x+2\right)+y\left(y+2\right)=\left(x+2\right)\left(y+2\right)\left(1\right)\\x^2\left(x+2\right)^2+y^2\left(y+2\right)^2=\left(x+2\right)^2\left(y+2\right)^2\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x^2\left(x+2\right)^2+2xy\left(x+2\right)\left(y+2\right)+y^2\left(y+2\right)^2=\left(x+2\right)^2\left(y+2\right)^2\\x^2\left(x+2\right)^2+y^2\left(y+2\right)^2=\left(x+2\right)^2\left(y+2\right)^2\end{matrix}\right.\)
=> \(2xy\left(x+2\right)\left(y+2\right)=0\)
<=>\(2xy=0\) (do x+2 và y+2 \(\ne0\))
<=> \(\left[{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Tại x=0 thay vào (1) có: \(y\left(y+2\right)=2\left(y+2\right)\) <=> y= \(\pm2\) => y=2 (vì y khác -2)
Tại y=0 thay vào (1) có: \(x\left(x+2\right)=2\left(x+2\right)\) => x=2
Vậy HPT có 2 nghiệm duy nhất (2,0),(0,2)
2, ĐK: \(y\ne-1\)
HPT <=> \(\left\{{}\begin{matrix}x^2=2\left(x+3\right)\left(y+1\right)\left(1\right)\\\frac{3x^2}{y+1}=4-x\end{matrix}\right.\)
=> \(\frac{6\left(3+x\right)\left(y+1\right)}{y+1}=4-x\)
<=> 6(x+3)=4-x
<=> \(14=-7x\)
<=> \(x=-2\) thay vào (1) có \(4=2\left(y+1\right)\)
<=>y=1\(\)( tm)
Vậy hpt có một nghiệm duy nhất (-2,1)
3,\(\left\{{}\begin{matrix}x^2-y=y^2-x\left(1\right)\\x^2-x=y+3\left(2\right)\end{matrix}\right.\)
PT (1) <=> \(\left(x-y\right)\left(x+y\right)+\left(x-y\right)=0\)
<=> (x-y)(x+y+1)=0
<=>\(\left[{}\begin{matrix}x=y\\y=-x-1\end{matrix}\right.\)
Tại x=y thay vào (2) có \(y^2-y=y+3\) <=> \(y^2-2y-3=0\) <=> (y-3)(y+1)=0 <=> \(\left[{}\begin{matrix}y=3\\y=-1\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Tại y=-1-x thay vào (2) có: \(x^2-x=-1-x+3\) <=> \(x^2=2\) <=> \(\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}y=-1-\sqrt{2}\\y=-1+\sqrt{2}\end{matrix}\right.\)
Vậy hpt có 4 nghiệm (3,3),(-1,-1), ( \(\sqrt{2},-1-\sqrt{2}\)),( \(-\sqrt{2},-1+\sqrt{2}\))
4,\(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=\frac{9}{2}\left(1\right)\\xy+\frac{1}{xy}+\frac{x}{y}+\frac{y}{x}=5\left(2\right)\end{matrix}\right.\)(đk:\(x\ne0,y\ne0\))
<=> \(\left\{{}\begin{matrix}\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)=\frac{9}{2}\\\left(y+\frac{1}{y}\right)\left(x+\frac{1}{x}\right)=5\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+\frac{1}{x}=u\\y+\frac{1}{y}=v\end{matrix}\right.\)
Có \(\left\{{}\begin{matrix}u+v=\frac{9}{2}\\uv=5\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=\frac{9}{2}-v\\v\left(\frac{9}{2}-v\right)=5\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=\frac{9}{2}-v\\\left(v-\frac{5}{2}\right)\left(v-2\right)=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=\frac{9}{2}-v\\\left[{}\begin{matrix}v=\frac{5}{2}\\v=2\end{matrix}\right.\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}v=\frac{5}{2}\\u=2\end{matrix}\right.\\\left[{}\begin{matrix}v=2\\u=\frac{5}{2}\end{matrix}\right.\end{matrix}\right.\)
Tại \(\left\{{}\begin{matrix}v=\frac{5}{2}\\u=2\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x+\frac{1}{x}=2\\y+\frac{1}{y}=\frac{5}{2}\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-2\right)\left(y-\frac{1}{2}\right)=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=1\\\left[{}\begin{matrix}y=2\\y=\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left[{}\begin{matrix}x=1\\y=\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
Tại \(\left\{{}\begin{matrix}v=2\\u=\frac{5}{2}\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x+\frac{1}{x}=\frac{5}{2}\\y+\frac{1}{y}=2\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}\left(x-2\right)\left(x-\frac{1}{2}\right)=0\\\left(y-1\right)^2=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\end{matrix}\right.\\y=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left[{}\begin{matrix}x=\frac{1}{2}\\y=1\end{matrix}\right.\end{matrix}\right.\)
Vậy hpt có 4 nghiệm (1,2),( \(1,\frac{1}{2}\)) ,( 2,1),(\(\frac{1}{2},1\)).
10.
\(\left\{{}\begin{matrix}2x^2-3xy+y^2+x-y=0\\x^2+x+1=y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-2xy-xy+y^2+x-y=0\\x^2+x+1=y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(2x-y+1\right)=0\\x^2+x+1=y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\y=2x+1\end{matrix}\right.\\x^2+x+1=y^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x^2+x+1=y^2\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\x^2+x+1=y^2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x^2+x+1=x^2\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\x^2+x+1=\left(2x+1\right)^2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\3x\left(x+1\right)=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=1\\\left[{}\begin{matrix}\left\{{}\begin{matrix}y=2x+1\\x=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\x=-1\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y=-1\\\left\{{}\begin{matrix}x=0\\y=-\frac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=-1\\\left\{{}\begin{matrix}x=0\\y=-\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
d: =>6y+2-4x+4=5 và 15y+5-8x+8=9
=>-4x+6y=-1 và -8x+15y=-4
=>x=-3/4; y=-2/3
c: \(\Leftrightarrow\left\{{}\begin{matrix}x+1=-1\\y+1=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-3\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}3y-15+2x-6=0\\7x-28+3y+3y-3=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=21\\7x+6y=45\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{19}{3}\end{matrix}\right.\)
\(e,\left\{{}\begin{matrix}\left(\frac{x}{y}\right)^3+\left(\frac{x}{y}\right)^2=12\\\left(xy\right)^2+xy=6\end{matrix}\right.\left(x;y\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy\in\left\{2;-3\right\}\end{matrix}\right.\)
Vì \(\frac{x}{y}=2>0\Rightarrow xy>0\Rightarrow xy=2\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y\\2y^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\left(h\right)\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)
\(a,\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+\frac{x}{y}=3\\x+\frac{1}{y}+\frac{x}{y}=3\end{matrix}\right.\left(x;y\ne0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\frac{1}{y}\right)^2-\frac{x}{y}=3\\\left(x+\frac{1}{y}\right)+\frac{x}{y}=3\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+\frac{1}{y}=a\\\frac{x}{y}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^2-b=3\\a+b=3\end{matrix}\right.\)
Làm nốt nha
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