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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: \(\Leftrightarrow\left\{{}\begin{matrix}3y-15+2x-6=0\\7x-28+3y+3y-3=14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=21\\7x+6y=45\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{19}{3}\end{matrix}\right.\)
a.
\(\left\{{}\begin{matrix}\left|x-1\right|+\left|y-2\right|=2\\\left|x-1\right|+y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|y-2\right|-y=-1\\\left|x-1\right|+y=3\end{matrix}\right.\)
Xét phương trình: \(\left|y-2\right|-y=-1\)
TH1: \(y\ge2\)
\(\Rightarrow y-2-y=-1\Leftrightarrow-2=-1\) (loại)
TH2: \(y\le2\)
\(\Rightarrow2-y-y=-1\Rightarrow y=\dfrac{3}{2}\)
Thế vào \(\left|x-1\right|+y=3\)
\(\Rightarrow\left|x-1\right|+\dfrac{3}{2}=3\Rightarrow\left|x-1\right|=\dfrac{3}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{3}{2}\Rightarrow x=\dfrac{5}{2}\\x-1=-\dfrac{3}{2}\Rightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)
b.
\(\left\{{}\begin{matrix}\left|x+1\right|+\left|y-1\right|=5\\\left|x+1\right|-4y+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|y-1\right|+4y-4=5\\\left|x+1\right|-4y+4=0\end{matrix}\right.\)
Xét phương trình: \(\left|y-1\right|+4y-4=5\)
TH1: \(y\ge1\)
\(\Rightarrow y-1+4y-4=5\Rightarrow y=2\)
Thế vào \(\left|x+1\right|-4y+4=0\)
\(\Rightarrow\left|x+1\right|=4\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
TH2: \(y\le1\)
\(\Rightarrow1-y+4y-4=5\Rightarrow y=\dfrac{8}{3}>1\) (không thỏa mãn)
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:
\(\left\{{}\begin{matrix}x^3-y^3=9\\3x^2+6y^2=3x-12y\end{matrix}\right.\)
\(\Leftrightarrow x^3-y^3-3x^2-6y^2=9-3x+12y\)
\(\Leftrightarrow x^3-3x^2+3x-1=y^3+6y^2+12y+8\)
\(\Leftrightarrow\left(x-1\right)^3=\left(y+2\right)^3\)
\(\Leftrightarrow x-1=y+2\Rightarrow x=y+3\)
\(\Rightarrow\left(y+3\right)^2+2y^2=y+3-4y\)
\(\Leftrightarrow y^2+3y+2=0\Rightarrow\left[{}\begin{matrix}y=-1\Rightarrow x=2\\y=-2\Rightarrow x=1\end{matrix}\right.\)
a) \(\left\{{}\begin{matrix}\left(x+3\right)\left(y+5\right)=\left(x+1\right)\left(y+8\right)\\\left(2x-3\right)\left(5y+7\right)=2\left(5x-6\right)\left(y+1\right)\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}xy+5x+3y+15=xy+8x+y+8\\10xy+14x-15y-21=10xy+10x-12y-12\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}-3x+2y=-7\\4x-3y=9\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}-9x+6y=-21\\8x-6y=18\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}-x=-3\\8x-6y=18\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=3\\8.3-6y=18\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm (x;y)=(3;1)
b) ĐKXĐ:\(\left\{{}\begin{matrix}2y-5\ne0\\3y-4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y\ne\dfrac{5}{2}\\y\ne\dfrac{4}{3}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2x-3}{2y-5}=\dfrac{3x+1}{3y-4}\\2\left(x-3\right)-3\left(y+2\right)=-16\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\left(2x-3\right)\left(3y-4\right)=\left(3x+1\right)\left(2y-5\right)\\2x-6-3y-6=-16\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}6xy-8x-9y+12=6xy-15x+2y-5\\2x-3y=-4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}7x-11y=-17\\2x-3y=-4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}14x-22y=-34\\14x-21y=-28\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}14x-22y=-34\\-y=-6\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}14x-22.6=-34\\y=6\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=7\left(TM\right)\\y=6\left(TM\right)\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm (x;y)=(7;6)
a: \(\Leftrightarrow\left\{{}\begin{matrix}8x-4y+12-3x+6y-9=48\\9x-12y+9+16x-8y-36=48\end{matrix}\right.\)
=>5x+2y=48-12+9=45 và 25x-20y=48+36-9=48+27=75
=>x=7; y=5
b: \(\Leftrightarrow\left\{{}\begin{matrix}6x+6y-2x+3y=8\\-5x+5y-3x-2y=5\end{matrix}\right.\)
=>4x+9y=8 và -8x+3y=5
=>x=-1/4; y=1
c: \(\Leftrightarrow\left\{{}\begin{matrix}-4x-2+1,5=3y-6-6x\\11,5-12+4x=2y-5+x\end{matrix}\right.\)
=>-4x-0,5=-6x+3y-6 và 4x-0,5=x+2y-5
=>2x-3y=-5,5 và 3x-2y=-4,5
=>x=-1/2; y=3/2
e: \(\Leftrightarrow\left\{{}\begin{matrix}x\cdot2\sqrt{3}-y\sqrt{5}=2\sqrt{3}\cdot\sqrt{2}-\sqrt{5}\cdot\sqrt{3}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
=>\(x=\sqrt{2};y=\sqrt{3}\)
b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)
\(\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)
\(A=\left(x-1\right)\left(x-8\right)\left(x-4\right)\left(x-5\right)+2002\)
\(\Leftrightarrow A=\left(x^2-9x+8\right)\left(x^2-9x+20\right)+2002\)
Đặt \(x^2-9x+14=y\)
\(\Rightarrow A=\left(y-6\right)\left(y+6\right)+2002\)
\(\Leftrightarrow A=y^2-36+2002\)
\(\Leftrightarrow A=y^2+1966\ge1966\)
Dấu "=" xảy ra khi
\(x^2-9x+14=0\)
\(\Leftrightarrow x=2,7\)