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a.
\(\Leftrightarrow\left\{{}\begin{matrix}4xy+8x-6y-12=4xy-12x+54\\3xy-3x+3y-3=3xy+3y-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}20x-6y=66\\-3x=-9\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
b.
\(\Leftrightarrow\left\{{}\begin{matrix}y=1-x\\x^2+xy+3=0\end{matrix}\right.\)
\(\Leftrightarrow x^2+x\left(1-x\right)+3=0\)
\(\Leftrightarrow x+3=0\Rightarrow x=-3\Rightarrow y=4\)
c.
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{2x-5}{3}\\x^2-y^2=40\end{matrix}\right.\)
\(\Rightarrow x^2-\left(\frac{2x-5}{3}\right)^2-40=0\)
\(\Leftrightarrow9x^2-\left(4x^2-20x+25\right)-360=0\)
\(\Leftrightarrow5x^2+20x-385=0\)
\(\Rightarrow\left[{}\begin{matrix}x=7\Rightarrow y=3\\x=-11\Rightarrow y=-9\end{matrix}\right.\)
d.
\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{36-3x}{2}\\\left(x-2\right)\left(y-3\right)=18\end{matrix}\right.\)
\(\Rightarrow\left(x-2\right)\left(\frac{36-3x}{2}-3\right)=18\)
\(\Leftrightarrow\left(x-2\right)\left(10-x\right)=12\)
\(\Leftrightarrow-x^2+12x-32=0\Rightarrow\left[{}\begin{matrix}x=4\Rightarrow y=12\\x=8\Rightarrow y=6\end{matrix}\right.\)
a/ \(\left\{{}\begin{matrix}\left(x^2+x\right)+\left(y^2+y\right)=18\\\left(x^2+x\right)\left(y^2+y\right)=72\end{matrix}\right.\)
Theo Viet đảo, \(x^2+x\) và \(y^2+y\) là nghiệm của:
\(t^2-18t+72=0\Rightarrow\left[{}\begin{matrix}t=12\\t=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x=6\\y^2+y=12\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x=12\\y^2+y=6\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\left\{2;-3\right\}\\y=\left\{3;-4\right\}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\left\{3;-4\right\}\\y=\left\{2;-3\right\}\end{matrix}\right.\end{matrix}\right.\)
b/ ĐKXĐ: ...
\(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y+1}=1\\x=\frac{3y-1}{y}\end{matrix}\right.\)
Nhận thấy \(y=\frac{1}{3}\) không phải nghiệm
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y+1}=1\\\frac{1}{x}=\frac{y}{3y-1}\end{matrix}\right.\) \(\Rightarrow\frac{y}{3y-1}+\frac{1}{y+1}=1\)
\(\Leftrightarrow y\left(y+1\right)+3y-1=\left(3y-1\right)\left(y+1\right)\)
\(\Leftrightarrow y^2-y=0\Rightarrow\left[{}\begin{matrix}y=0\left(l\right)\\y=1\end{matrix}\right.\) \(\Rightarrow x=2\)
Bài 1:
Đặt 2x+1=a
Theo đề, ta có: \(\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}=3\)
=>3a^2(a+1)^2=a^2+2a+1+a^2
=>3a^2(a^2+2a+1)-2a^2-2a-1=0
=>3a^4+6a^3+a^2-2a-1=0
=>(a^2+a-1)(3a^2+3a+1)=0
=>\(a\in\left\{\dfrac{-1+\sqrt{5}}{2};\dfrac{-1-\sqrt{5}}{2}\right\}\)
=>\(2x+1\in\left\{\dfrac{-1+\sqrt{5}}{2};\dfrac{-1-\sqrt{5}}{2}\right\}\)
=>\(2x\in\left\{\dfrac{-3+\sqrt{5}}{2};\dfrac{-3-\sqrt{5}}{2}\right\}\)
hay \(x\in\left\{\dfrac{-3+\sqrt{5}}{4};\dfrac{-3-\sqrt{5}}{4}\right\}\)
bài 2
ta có \(\left(\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\right)^2\)
\(=\left(\sqrt{a}.\sqrt{\frac{8a^2+1}{a}}+\sqrt{b}.\sqrt{\frac{8b^2+1}{b}}+\sqrt{c}.\sqrt{\frac{8c^2+1}{c}}\right)^2\)\(=\left(A\right)\)
Áp dụng bất đẳng thức Bunhiacopxki ta có;
\(\left(A\right)\le\left(a+b+c\right)\left(8a+\frac{1}{a}+8b+\frac{1}{b}+8c+\frac{8}{c}\right)\)
\(=\left(a+b+c\right)\left(9a+9b+9c\right)=9\left(a+b+c\right)^2\)
\(\Rightarrow3\left(a+b+c\right)\ge\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\)(đpcm)
Dấu \(=\)xảy ra khi \(a=b=c=1\)
Những bài còn lại chỉ cần phân tích ra rồi rút gọn là được nha. Bạn tự làm nha!
Đặt \(\hept{\begin{cases}x+y=a\\x-y=b\end{cases}}\)\(\Rightarrow\)ta có hệ \(\hept{\begin{cases}2a+3b=4\\a+2b=5\end{cases}}\Rightarrow\hept{\begin{cases}a=-7\\b=6\end{cases}}\)Từ đó ta có \(\hept{\begin{cases}x+y=-7\\x-y=6\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=-\frac{13}{2}\end{cases}}\)PS: Cái đề chỗ 3(x+y) phải thành 3(x-y) chứ
\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy\right)^3+7\left(xy+x+y+1\right)=31\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3+\left(xy\right)^3+7\left(xy+x+y\right)=30\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\)
\(\Rightarrow\left\{{}\begin{matrix}uv=2\\u^3+v^3+7\left(u+v\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3-3uv\left(u+v\right)+7\left(u+v\right)=30\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3+\left(u+v\right)-30=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\u+v=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=2\\v=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\) \(\Leftrightarrow\left(x;y\right)=\left(1;1\right)\)
2.
ĐKXĐ: \(0\le x\le\dfrac{3}{2}\)
\(\Leftrightarrow9x\left(3-2x\right)+81+54\sqrt{x\left(3-2x\right)}=49x+25\left(3-2x\right)+70\sqrt{x\left(3-2x\right)}\)
\(\Leftrightarrow9x^2-14x-3+8\sqrt{x\left(3-2x\right)}=0\)
\(\Leftrightarrow9\left(x^2-2x+1\right)-4\left(3-x-2\sqrt{x\left(3-2x\right)}\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2-\dfrac{36\left(x-1\right)^2}{3-x+2\sqrt{x\left(3-2x\right)}}=0\)
\(\Leftrightarrow9\left(x-1\right)^2\left(1-\dfrac{4}{3-x+2\sqrt{x\left(3-2x\right)}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\3-x+2\sqrt{x\left(3-2x\right)}=4\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\sqrt{x\left(3-2x\right)}=x+1\)
\(\Leftrightarrow4x\left(3-2x\right)=x^2+2x+1\)
\(\Leftrightarrow9x^2-10x+1=0\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)
\(a\orbr{x=\frac{\pm\sqrt{5}-3}{4}}\)
\(b\hept{\begin{cases}x=5\\y=4\end{cases}}\)
2)\(\Leftrightarrow\left(x^3-x^2y\right)+\left(y^3-xy^2\right)=5\)
\(\Leftrightarrow x^2\left(x-y\right)+y^2\left(y-x\right)=5\)
\(\Leftrightarrow x^2\left(x-y\right)-y^2\left(x-y\right)=5\)
\(\Leftrightarrow\left(x-y\right)\left(x^2-y^2\right)=5\)
TH1\(\hept{\begin{cases}x-y=1\\x^2-y^2=5\end{cases}\Leftrightarrow\hept{\begin{cases}x=3\\y=2\end{cases}\left(N\right)}}\)
TH2\(\hept{\begin{cases}x-y=5\\x^2-y^2=1\end{cases}\Leftrightarrow\hept{ }x,y\in\varnothing}\)
TH3\(\hept{\begin{cases}x-y=-1\\x^2-y^2=-5\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=3\end{cases}\left(N\right)}}\)
TH4\(\hept{\begin{cases}x-y=-5\\x^2-y^2=-1\end{cases}\Leftrightarrow\hept{ }x,y\in\varnothing}\)
Vậy......