giải hệ phương trình:
\(\left\{{}\begin{matrix}\left(x-y\right)\left(x^2-y^2\right)=3\\\left(x+y\right)\left(x^2+y^2\right)=15\end{matrix}\right.\)
Giải hệ phương trình sau:
\(\left\{{}\begin{matrix}3x-2\left|y\right|=9\\2x+3\left|y\right|=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left|x-2\right|+2\left|y-1\right|=9\\x+\left|y-1\right|=-1\end{matrix}\right.\)
a) Ta có: \(\left\{{}\begin{matrix}3x-2\left|y\right|=9\\2x+3\left|y\right|=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-4\left|y\right|=18\\6x+9\left|y\right|=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-13\left|y\right|=15\\3x-2\left|y\right|=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|y\right|=\dfrac{-15}{13}\\3x-2\left|y\right|=9\end{matrix}\right.\Leftrightarrow\)Phương trình vô nghiệmVậy: \(S=\varnothing\)
$\begin{cases}3x-2|y|=9\\2x+3|y|=1\\\end{cases}$
`<=>` $\begin{cases}6x-4|y|=18\\6x+9|y|=3\\\end{cases}$
`<=>` $\begin{cases}13|y|=-15(loại)\\|3x|-2|y|=9\\\end{cases}$
Vậy HPT vô nghiệm
$\begin{cases}|x-2|+2|y-1|=9\\x+|y-1|=-1\\\end{cases}$
`<=>` $\begin{cases}|x-2|+2|y-1|=9\\2x+2|y-1|=-2\\\end{cases}$
`<=>` $\begin{cases}|x-2|-2x=11\\x+|y-1|=-1\\\end{cases}$
`<=>` $\begin{cases}|x-2|=2x+11\\x+|y-1|=-1\\\end{cases}$
Đến đây dễ rồi bạn tự giải :D
Bài 3: Giải các hệ phương trình sau:
a)\(\left\{{}\begin{matrix}2\left(x-2\right)+3\left(1+y\right)=-2\\3\left(x-2\right)-2\left(1+y\right)=-3\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)
a: =>2x-4+3+3y=-2 và 3x-6-2-2y=-3
=>2x+3y=-2+4-3=2-3=-1 và 3x-2y=-3+6+2=5
=>x=1; y=-1
b: =>x^2-x+xy-y=x^2+x-xy-y+2xy
=>-x-y=x-y và y^2+y-yx-x=y^2-2y+xy-2x-2xy
=>x=0 và y-x=-2y-2x
=>x=0 và y=0
giải hệ phương trình a)\(\left\{{}\begin{matrix}2\left(x+1\right)-3\left(y-2\right)=5\\-4\left(x-2\right)+5\left(y-3\right)=-1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}8\left(x-3\right)-3\left(y+1\right)=-2\\3\left(x+2\right)-2\left(1-y\right)=5\end{matrix}\right.\)
Help me ~~~
a) Ta có: \(\left\{{}\begin{matrix}2\left(x+1\right)-3\left(y-2\right)=5\\-4\left(x-2\right)+5\left(y-3\right)=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+2-3y+6=5\\-4x+8+5y-15=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-3\\-4x+5y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-6y=-6\\-4x+5y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-y=0\\2x-3y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\2x-3\cdot0=-3\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=0\end{matrix}\right.\)
Vậy: hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=0\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}8\left(x-3\right)-3\left(y+1\right)=-2\\3\left(x+2\right)-2\left(1-y\right)=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}8x-24-3y-3=-2\\3x+6-2+2y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}8x-3y=25\\3x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}24x-9y=75\\24x+16y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-25y=67\\3x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-67}{25}\\3x=1-2y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=1-2\cdot\dfrac{-67}{25}=\dfrac{159}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)
a) HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=-3\\-4x+5y=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x-6y=-6\\-4x+5y=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-y=0\\x=\dfrac{3y-3}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(-\dfrac{3}{2};0\right)\)
b) HPT \(\Leftrightarrow\left\{{}\begin{matrix}8x-3y=25\\3x+2y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}16x-6y=50\\9x+6y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}25x=53\\y=\dfrac{1-3x}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{53}{25}\\y=-\dfrac{67}{25}\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(\dfrac{53}{25};-\dfrac{67}{25}\right)\)
Giải hệ phương trình sau: \(\left\{{}\begin{matrix}2x^3-y^3=15\\\left(x-y\right)\left(x^2+2y^2\right)=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x^3-2y^3=30\\5\left(x-y\right)\left(x^2+2y^2\right)=30\end{matrix}\right.\)
Trừ vế cho vế:
\(5\left(x-y\right)\left(x^2+2y^2\right)-\left(4x^3-2y^3\right)=0\)
\(\Leftrightarrow x^3-5x^2y+10xy^2-8y^3=0\)
\(\Leftrightarrow\left(x-2y\right)\left(x^2-3xy+4y^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2y\\x=y=0\left(ktm\right)\end{matrix}\right.\)
Thay vào pt đầu:
\(\Rightarrow2\left(2y\right)^3-y^3=15\)
\(\Rightarrow y^3=1\Rightarrow y=1\Rightarrow x=2\)
giải hệ phương trình:\(\left\{{}\begin{matrix}\left(x-3\right)\left(y+6\right)=xy\\\left(x+2\right)\left(y-2\right)=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy+6x-3y-18=xy\\xy-2x+2y-4=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-3y=18\\-2x+2y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=6\\-x+y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=10\end{matrix}\right.\)
Giải các hệ phương trình sau
a,\(\left\{{}\begin{matrix}\sqrt{3}x-y=\sqrt{2}\\x-\sqrt{2}y=\sqrt{3}\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}5\left(x-y\right)-3\left(2x+3y\right)=12\\3\left(x+2y\right)-4\left(x+2y\right)=5\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}\dfrac{x+2}{y-1}=\dfrac{x-4}{y+2}\\\dfrac{2x+3}{y-1}=\dfrac{4x+1}{2y+1}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{x+2}{y-1}=\dfrac{x-4}{y+2}\\\dfrac{2x+3}{y-1}=\dfrac{4x+1}{2y+1}\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\left(x+2\right)\left(y+2\right)=\left(y-1\right)\left(x-\text{4}\right)\\\left(2x+3\right)\left(2y+1\right)=\left(y-1\right)\left(4x+1\right)\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}xy+2x+2y+4=xy-4y-x+4\\4xy+2x+6y+3=4xy-4x+y-1\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}3x+6y=0\\6x+5y=-4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=-\dfrac{8}{7}\\y=\dfrac{4}{7}\end{matrix}\right.\)(TM)
\(\left\{{}\begin{matrix}5\left(x-y\right)-3\left(2x+3y\right)=12\\3\left(x+2y\right)-4\left(x+2y\right)=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}5x-5y-6x-9y=12\\3x+6y-4x-8y=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}-x-14y=12\\-x-2y=5\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=-\dfrac{26}{3}\\y=-\dfrac{7}{12}\end{matrix}\right.\)
Vậy HPT có nghiệm (x;y) = (\(-\dfrac{26}{3};-\dfrac{7}{12}\))
Giải hệ phương trình \(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=\left(x+1\right)\left(y-3\right)\\\left(x-5\right)\left(y+4\right)=\left(x-4\right)\left(y+1\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x-2\right)\left(y+1\right)=xy
\\\left(x+8\right)\left(y-2\right)=xy\end{matrix}\right.\) GIÚP MÌNH VỚI Ạ MÌNH CẢM ƠN
\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+6y=8+2x-3y\\5y-5x=5+3x+2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2x+6y+3y=8\\-5x-3x+5y-2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-8x+3y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-24x+9y=15\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}28x=-7\\4x+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{7}{28}=-\dfrac{1}{4}\\4.\left(-\dfrac{1}{4}\right)+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\y=1\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(-\dfrac{1}{4};1\right)\)
Bài 2
Giải hệ phương trình sau \(\left\{{}\begin{matrix}x\left(y-2\right)=\left(x+2\right)\left(y-4\right)\\\left(x-3\right)\left(2y+7\right)=\left(2x-7\right)\left(y+3\right)\end{matrix}\right.\)
=>xy-2x=xy-4x+2y-8 và 2xy+7x-6y-21=2xy+6x-7y-21
=>2x-2y=-8 và x+y=0
=>x-y=-4 và x+y=0
=>2x=-4 và x+y=0
=>x=-2 và y=2
Giải hệ phương trình: \(\left\{{}\begin{matrix}\sqrt{x+2}.\left(x-y+3\right)=\sqrt{y}\\x^2+\left(x+3\right)\left(2x-y+5\right)=x+16\end{matrix}\right.\)
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\) thì pt đầu trở thành:
\(a\left(a^2-b^2+1\right)=b\)
\(\Leftrightarrow a\left(a-b\right)\left(a+b\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+1\right)=0\)
\(\Leftrightarrow a=b\Rightarrow\sqrt{x+2}=\sqrt{y}\Rightarrow y=x+2\)
Thay xuống pt dưới:
\(x^2+\left(x+3\right)\left(x+3\right)=x+16\)
\(\Leftrightarrow2x^2+5x-7=0\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=3\\x=-\dfrac{2}{7}\left(loại\right)\end{matrix}\right.\)
Giải các hệ phương trình
a) \(\left\{{}\begin{matrix}\left(x+y\right)\left(x^2+y^2\right)=15\\\left(x-y\right)\left(x^2-y^2\right)=3\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}x^2+y^2+1=2\left(x+y\right)\\y\left(2x-y\right)=2y+1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x+y\right).\left(x^2+y^2\right)=15\\\left(x-y\right).\left(x^2-y^2\right)=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right).\left(x^2+y^2\right)=15\\\left(x+y\right).\left(x-y\right)^2=3\end{matrix}\right.\)
\(\Rightarrow\frac{x^2+y^2}{\left(x-y\right)^2}=\frac{15}{3}=5\left(\text{vì }x+y\ne0\right)\)
\(\Rightarrow5.\left(x-y\right)^2-x^2-y^2=0\)
\(\Rightarrow\left[\left(x-y\right)^2-x^2\right]+\left[\left(2x-2y\right)^2-y^2\right]=0\)
\(\Rightarrow\left(x-y-x\right).\left(x-y+x\right)+\left(2x-2y-y\right).\left(2x-2y+y\right)=0\)
\(\Rightarrow-y.\left(2x-y\right)+\left(2x-3y\right).\left(2x-y\right)=0\)
\(\Rightarrow\left(2x-y\right).\left(2x-3y-y\right)=0\)
\(\Rightarrow\left(2x-y\right).\left(2x-4y\right)=0\Rightarrow\left[{}\begin{matrix}2x=y\\x=2y\end{matrix}\right.\)
Thay vào giải tiếp
Lười nghĩ quá :v cứ nhân tung tóe ra vậy
a) \(pt\Leftrightarrow\left(x+y\right)\left(x^2+y^2\right)=5\left(x-y\right)\left(x^2-y^2\right)\)
\(\Leftrightarrow4x^3+4y^3+6x^2y+6xy^2=0\)
\(\Leftrightarrow2x^3+2y^3+3x^2y+3xy^2=0\)
\(\Leftrightarrow2\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(2x^2+xy+2y^2\right)=0\)
Vì \(2x^2+xy+2y^2>0\forall x;y\)
\(\Rightarrow x=-y\)
Thế x vào tìm y
a) \(\left\{{}\begin{matrix}x^3+y^3+x^2y+xy^2=15\\x^3+y^3-x^2y-xy^2=3\end{matrix}\right.\)⇒ \(\left\{{}\begin{matrix}x^3+y^3=9\\x^2y+xy^2=6\end{matrix}\right.\)
⇒ \(x^3+y^3+3x^2y+3xy^2=27\)
⇒ \(x+y=3\Rightarrow x^2+y^2=5\)(1)
Thay \(y=3-x\) vào (1) , ta được:
\(2x^2-6x+4=0\Rightarrow\left(x-1\right)\left(x-2\right)=0\)
⇒ \(\left[{}\begin{matrix}x=1;y=2\\x=2;y=1\end{matrix}\right.\)