Giải hệ phương trình sau bằng phương pháp thế
1) \(\left\{{}\begin{matrix}x-2y=4\\-2x+5y=-3\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}2x+y=10\\5x-3y=3\end{matrix}\right.\)
3) \(\left\{{}\begin{matrix}x+2y=4\\-3x+y=7\end{matrix}\right.\)
a) \(\left\{{}\begin{matrix}2x+4=0\\4x+2y=-3\end{matrix}\right.\) c)\(\left\{{}\begin{matrix}\left(x-15\right).\left(y+2\right)=x.y\\\left(x+15\right).\left(y-1\right)=x.y\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}2x+4=y\\x+2y=-3\end{matrix}\right.\) d) \(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}=5\\\frac{2}{x}+\frac{5}{y}=7\end{matrix}\right.\) tính bằng phương pháp cộng dại số
Bµi 1: A)\(\left\{{}\begin{matrix}X=35.\left(Y+2\right)\\X=50.\left(Y-1\right)\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}Y=2X-3\\Y=X-1\end{matrix}\right.\)
C) \(\left\{{}\begin{matrix}\left(X+14\right).\left(Y-2\right)=X.Y\\\left(X-4\right).\left(Y+1\right)=X.Y\end{matrix}\right.\)
D)\(\left\{{}\begin{matrix}Y=\frac{6-X}{4}\\Y=\frac{4X-5}{3}\end{matrix}\right.\)GIẢI BÀI 1 BẰNG PHƯƠNG PHAP THẾ
Giải hệ phương trình
\(\left\{{}\begin{matrix}4\left(2x-y+3\right)-3\left(x-2y+3\right)=48\\3\left(3x-4y+3\right)+4\left(4x-2y-9\right)=48\end{matrix}\right.\)
\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\)
\(\left\{{}\begin{matrix}-2\left(2x+1\right)+1,5=3\left(y-2\right)-6x\\11,5-4\left(3-x\right)=2y-\left(5-x\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{8x-5y-3}{7}+\dfrac{11y-4x-7}{5}=12\\\dfrac{9x+4y-13}{5}-\dfrac{3\left(x-2\right)}{4}=15\end{matrix}\right.\)
\(\left\{{}\begin{matrix}2\sqrt{3}x-\sqrt{5}y=2\sqrt{6}-\sqrt{15}\\3x-y=3\sqrt{2}-\sqrt{3}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}6.\left(x+y\right)=8+2x-3y\\5.\left(y-x\right)=5+3x+2y\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x-1\right).\left(y-2\right)=\left(x+1\right).\left(y-3\right)\\\left(x-5\right).\left(y+4\right)=\left(x-4\right).\left(y+1\right)\end{matrix}\right.\)
Mọi Ng giúp em với
Ai làm hết em tick đúng nha ( trước 19:00 hôm nay)
Bài 1: Giải hệ phương trình sau theo m
a, \(\left\{{}\begin{matrix}x-my=m^2+1\\mx+y=m^2+1\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}x+y=m-2\\\left(m+2\right)x-4y=m^2-4\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}2x+my=m+2\\\left(m+1\right)x+2my=2m+4\end{matrix}\right.\)
d, \(\left\{{}\begin{matrix}mx+2y=3\\m^2x-4y=-6\end{matrix}\right.\)
cho he phuong trinh:
\(\left\{{}\begin{matrix}x+2y=m+1\\2x+3y=m-2\end{matrix}\right.\)
a. Giai he pt vs m=1
b. Tim m de he pt co nghiem (x;y) thoa man \(\left\{{}\begin{matrix}x>3\\y< 5\end{matrix}\right.\)
1)\(\left\{{}\begin{matrix}x^2-y^2-2x+2y=0\\x^2-3xy+5y^2-3=0\end{matrix}\right.\)
2)\(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{1-y}=1\\\frac{1}{x-1}-\frac{1}{y}=2\end{matrix}\right.\)
3)\(\left\{{}\begin{matrix}x^2-4x+3=0\\x^2+xy+y^2=1\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}x^2+y^2+x+y=2\\\left(x+1\right)^2-\left(y+2\right)^2=0\end{matrix}\right.\)
giải hệ pt bằng phương pháp thế:
a,\(\left\{{}\begin{matrix}3x+y=-2\\-9x-39=6\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}x+y=101\\-x+y=-1\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}x+y=2\\\dfrac{1}{2}x+y=\dfrac{5}{4}\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}x-5y=16\\10y-2x=-32\end{matrix}\right.\)