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:
\(\left\{{}\begin{matrix}3x-2\left|y\right|=9\\2x+3\left|y\right|=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left|x-2\right|+2\left|y-1\right|=9\\x+\left|y-1\right|=-1\end{matrix}\right.\)
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
giải hệ
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,\(\left\{{}\begin{matrix}x^3-y^3=9\\x^2+2y^2=x-4y\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}\left(x-y\right)\left(2x+3y\right)=12\\6\left(x-y\right)+xy\left(x-y\right)=12\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}x^2+y^2+1=2\left(x+y\right)\\y\left(2x-y\right)=\left(2y+1\right)\end{matrix}\right.\)
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:
\(\left\{{}\begin{matrix}x^3-y^3=9\\3x^2+6y^2=3x-12y\end{matrix}\right.\)
\(\Leftrightarrow x^3-y^3-3x^2-6y^2=9-3x+12y\)
\(\Leftrightarrow x^3-3x^2+3x-1=y^3+6y^2+12y+8\)
\(\Leftrightarrow\left(x-1\right)^3=\left(y+2\right)^3\)
\(\Leftrightarrow x-1=y+2\Rightarrow x=y+3\)
\(\Rightarrow\left(y+3\right)^2+2y^2=y+3-4y\)
\(\Leftrightarrow y^2+3y+2=0\Rightarrow\left[{}\begin{matrix}y=-1\Rightarrow x=2\\y=-2\Rightarrow x=1\end{matrix}\right.\)
Câu 3:
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(2x+3y\right)=12\\\left(x-y\right)\left(xy+6\right)=12\end{matrix}\right.\)
\(\Leftrightarrow\left(x-y\right)\left(2x+3y\right)=\left(x-y\right)\left(xy+6\right)\)
\(\Leftrightarrow2x+3y=xy+6\)
\(\Leftrightarrow x\left(y-2\right)-3\left(y-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(y-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)
TH1: \(x=3\Rightarrow\left(3-y\right)\left(3y+6\right)=12\)
\(\Leftrightarrow y^2-y-2=0\Rightarrow\left[{}\begin{matrix}y=-1\\y=2\end{matrix}\right.\)
TH2: \(y=2\Rightarrow\left(x-2\right)\left(2x+6\right)=12\)
\(\Leftrightarrow x^2+x-12=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)
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 các hệ phương trình sau :
a, \(\left\{{}\begin{matrix}x^2+xy=y^2+1\\3x+y=y^2+3\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}x^2-y^2=4x-2y-3\\x^2+y^2=5\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}x^2+x-xy-2y^2-2y=0\\x^2+y^2=1\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}2\left(y+z\right)=yz\\xy+yz+zx=108\\xyz=180\end{matrix}\right.\)
Giải các hệ phương trình:
a) \(\left\{{}\begin{matrix}\left|x-2\right|+2\left|y-1\right|=9\\x+\left|y-1\right|=-1\end{matrix}\right.\);
b) \(\left\{{}\begin{matrix}x+y+\left|x\right|=25\\x-y+\left|y\right|=30\end{matrix}\right..\)
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
Giải hệ phương trình
\(\left\{{}\begin{matrix}x^2+y^2+xy=13\\x^4+y^4+x^2y^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=13\\\left(x^2+y^2\right)^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=13+xy\\\left[\left(x+y\right)^2-2xy\right]^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=13\\\left(13-xy\right)^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy=3\\\left(x+y\right)^2=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\) hoặc x+y = -4
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=-4\\xy=3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)
Mọi người có thể giải thích từ dấu tương đương thứ 3 xuống 4. tại sao lại như vậy k?
Giải hệ phương trình
a)\(\left\{{}\begin{matrix}6x^2-3xy+x=1-y\\x^2+y^2=1\end{matrix}\right.\) c)\(\left\{{}\begin{matrix}\left|x+1\right|+\left|y-1\right|=5\\\left|x+1\right|-4y+4=0\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}2x^2-2x+xy-y=0\\x^2-3xy+4=0\end{matrix}\right.\)
a \(\Leftrightarrow\left\{{}\begin{matrix}6x^2-3xy+x=1-y\left(1\right)\\x^2+y^2=1\left(2\right)\end{matrix}\right.\) Từ (1) \(\Rightarrow6x^2-3xy+x-1+y=0\)
\(\Leftrightarrow\left(6x^2+x-1\right)-\left(3xy-y\right)=0\) \(\Leftrightarrow\left(6x^2+3x-2x-1\right)+y\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(2x+1\right)+y\left(3x-1\right)=0\) \(\Leftrightarrow\left(3x-1\right)\left(2x+1+y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\2x+y=-1\end{matrix}\right.\)
*Nếu 3x-1=0⇔x=\(\dfrac{1}{3}\) Thay vào (2) ta được:
\(\dfrac{1}{9}+y^2=1\Leftrightarrow y^2=\dfrac{8}{9}\Leftrightarrow y=\dfrac{\pm2\sqrt{2}}{3}\)
*Nếu 2x+y=-1\(\Leftrightarrow y=-1-2x\) Thay vào (2) ta được :
\(\Rightarrow x^2+\left(-2x-1\right)^2=1\Leftrightarrow x^2+4x^2+4x+1=1\Leftrightarrow5x^2+4x=0\Leftrightarrow x\left(5x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-4}{5}\end{matrix}\right.\)
.Nếu x=0⇒y=0
.Nếu x=\(\dfrac{-4}{5}\) \(\Rightarrow y=-1+\dfrac{4}{5}=-\dfrac{1}{5}\) Vậy...
Câu b)
\(\left\{{}\begin{matrix}2x^2-2x+xy-y=0\\x^2-3xy+4=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}2x\left(x-1\right)+y\left(x-1\right)\\x^2-3xy+4=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}\left(x-1\right)\left(2x+y\right)=0\\x^2-3xy+4=0\left(2\right)\end{matrix}\right.\)
Để (x-1)(2x+y) = 0 thì: \(\left[{}\begin{matrix}x-1=0\\2x+y=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=1\\2x+y=0\end{matrix}\right.\)
Thay x=1 vào PT (2) ta có:
(2) ⇔12-3.1.y+4=0
⇔1-3y +4=0
⇔-3y+5=0
⇔y=\(\dfrac{5}{3}\)
Vậy HPT có nghiệm (x:y) = (1;\(\dfrac{5}{3}\))
b\(\left\{{}\begin{matrix}2x^2-2x+xy-y=0\left(1\right)\\x^2-3xy+4=0\left(2\right)\end{matrix}\right.\)
Từ (1) \(\Rightarrow2x\left(x-1\right)+y\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x+y\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+y=0\end{matrix}\right.\)
*Nếu x-1=0⇔x=1 Thay vào (2) ta được: \(1-3y+4=0\Leftrightarrow3y=5\Leftrightarrow y=\dfrac{5}{3}\)
*Nếu 2x+y=0\(\Leftrightarrow y=-2x\) Thay vào (2) ta được:
\(\Rightarrow x^2+6x^2+4=0\Leftrightarrow7x^2=-4\) Vô lí ⇒ Trường hợp này ko có x,y (L)
Vậy...
