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Bài 2:
a: \(\Leftrightarrow\left\{{}\begin{matrix}2-x+y-3x-3y=5\\3x-3y+5x+5y=-2\end{matrix}\right.\)
=>-4x-2y=3 và 8x+2y=-2
=>x=1/4; y=-2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)
=>y=6 và x-2=5/4
=>x=13/4; y=6
c: =>x+y=24 và 3x+y=78
=>-2x=-54 và x+y=24
=>x=27; y=-3
d: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}=2+3\cdot1=5\end{matrix}\right.\)
=>y+2=1 và x-1=25
=>x=26; y=-1
Ta có hpt \(\left\{{}\begin{matrix}xy+3y-5x-15=xy\\2xy+30x-y^2-15y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}5x=3y-15\\6\left(3y-15\right)-y^2-15y=0\end{matrix}\right.\)
Ta có pt (2) \(\Leftrightarrow3y-y^2-80=0\Leftrightarrow y^2-3y+80=0\left(VN\right)\)
=> hpy vô nghiệm
c) Ta có hpt \(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left(xy+x+y\right)=30\\xy\left(x+y\right)+xy+x+y=11\end{matrix}\right.\)
Đặt j\(xy\left(x+y\right)=a;xy+x+y=b\), ta có hpt
\(\left\{{}\begin{matrix}ab=30\\a+b=11\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=5;b=6\\a=6;b=5\end{matrix}\right.\)
với a=5;b=6, ta có \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}xy=1;x+y=5\\xy=5;x+y=1\end{matrix}\right.\)
đến đây thì thế y hoặc x ra pt bậc 2, còn TH còn lại bn tự giải nhé !
a)
\(\left\{{}\begin{matrix}x+y+xy=7\\x^2+y^2+xy=13\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+xy=7\\\left(x+y\right)^2-xy=13\end{matrix}\right.\)
Đặt x+y = S, xy = P,ta có hệ
\(\left\{{}\begin{matrix}S+P=17\\S^2-P=13\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}P=S-17\\S^2-S+4=0\end{matrix}\right.\)
\(S^2-S+4>0\)
=> Hệ phương trình vô nghiệm
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.
\(\left\{{}\begin{matrix}2x^2-3xy+y^2+x-y=0\\x^2+x+1=y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-2xy-xy+y^2+x-y=0\\x^2+x+1=y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(2x-y+1\right)=0\\x^2+x+1=y^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\\y=2x+1\end{matrix}\right.\\x^2+x+1=y^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x^2+x+1=y^2\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\x^2+x+1=y^2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x^2+x+1=x^2\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\x^2+x+1=\left(2x+1\right)^2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=y\\x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\3x\left(x+1\right)=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=1\\\left[{}\begin{matrix}\left\{{}\begin{matrix}y=2x+1\\x=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=2x+1\\x=-1\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y=-1\\\left\{{}\begin{matrix}x=0\\y=-\frac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=-1\\\left\{{}\begin{matrix}x=0\\y=-\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^3+y^3=1+y-x+xy\left(1\right)\\7xy+y-x=7\left(2\right)\end{matrix}\right.\)
Từ(2)\(\Rightarrow x-y=7xy-7\)
\(\left(1\right)\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)=1+y-x+xy\)
\(\Leftrightarrow\left[\sqrt{\left(x-y\right)^2+4xy}\right]\left[\left(x-y\right)^2+xy\right]=1+7-7xy+xy\)
\(\Leftrightarrow7\left[\sqrt{\left(7xy-7\right)^2+4xy}\right]\left(7xy-7+xy\right)=-6xy+8\)
Đặt xy=a
\(\Rightarrow7\left[\sqrt{\left(7a-7\right)^2+4a}\right]\left(8a-7\right)=-6a+8\)
\(\Leftrightarrow49\left(\sqrt{\left(a-1\right)^2}\right)\left(8a-7\right)+6a-8=0\)
Với \(a-1\ge0\Leftrightarrow a\ge1\)
\(\Rightarrow49\left(8a^2-15a+7\right)+6a-8=0\)
\(\Leftrightarrow392a^2-729a+335=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{729+\sqrt{6161}}{784}\left(TM\right)\\a=\dfrac{729-\sqrt{6161}}{784}\left(KTM\right)\end{matrix}\right.\)\(\Rightarrow xy=\dfrac{729+\sqrt{6161}}{784}\)\(\Rightarrow y=\dfrac{\dfrac{729+\sqrt{6161}}{784}}{x}\)
Thay vào (2)\(\Rightarrow\)\(x\approx1,125;y\approx0,915\)
Với \(a-1< 0\Leftrightarrow a< 1\)
\(\Rightarrow49\left(-a+1\right)\left(8a-7\right)=-6a+8\)
\(\Leftrightarrow-49\left(8a^2-15a+7\right)+6a-8=0\)
\(\Leftrightarrow-392a^2+741a-351=0\)(vô nghiệm).
Vậy hpt có nghiệm (x;y)=(1,125;0,915).
\(\left\{{}\begin{matrix}x^3+y^3=1-x+y+xy\left(1\right)\\7xy+y-x=7\left(2\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^3+y^3=1-x+y+xy\\x-y=7xy-7\end{matrix}\right.\)
Từ pt (1) suy ra: \(x^3+y^3=1+xy-\left(x-y\right)\)
\(\Leftrightarrow x^3+y^3=1+xy-7xy+7\)
\(\Leftrightarrow x^3+y^3=-6xy+8\)
\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)=-6xy+8\)
\(\Leftrightarrow\left(x+y\right)^3-8=-6xy+3xy\left(x+y\right)\)
\(\Leftrightarrow\left(x+y-2\right)\left[\left(x+y\right)^2+2\left(x+y\right)+4\right]=3xy\left(x+y-2\right)\)
\(\Leftrightarrow\left(x+y-2\right)\left[\left(x+y\right)^2+2\left(x+y\right)+4-3xy\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y-2=0\\\left(x+y\right)^2+2\left(x+y\right)+4-3xy=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=2\left(3\right)\\\left(x+y\right)^2+2\left(x+y\right)+4-3xy=0\left(4\right)\end{matrix}\right.\)
TH1: Từ (2) và (4) suy ra: \(\Leftrightarrow\left[{}\begin{matrix}x+y=2\\7xy+y-x=7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2-y\\7\left(2-y\right)y+y-2+y=7\end{matrix}\right.\)
Suy ra: 14y - 7y2 + y - 2 + y = 7
<=> 7y2 - 16y +9 = 0
\(\Leftrightarrow\left[{}\begin{matrix}y=1\rightarrow x=1\\y=\frac{9}{7}\rightarrow x=\frac{5}{9}\end{matrix}\right.\)
TH2:Thay vào tính cho kết quả ko thỏa mãn
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