Cho he phuong trinh (1) 4x + 5y = 14 ; (2) 5x + 4y = 13 hỏi 9x + 9y = ?
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Đo´n nhan so nghiem cua cac he phuong trinh sau bang hinh hoc
a)4x+5y=20
0.8x+y=4
b)4x+5y=20
2x+2.5y=5
\(\left\{{}\begin{matrix}4x+3x=-6\\\dfrac{x+3y}{3}-\dfrac{y-2}{5}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7x=-6\\\dfrac{5\left(x+3y\right)-3\left(y-2\right)}{15}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\5x+15y-3y+6=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\12y=9-5x=9+5\cdot\dfrac{6}{7}=9+\dfrac{30}{7}=\dfrac{93}{7}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\y=\dfrac{93}{7\cdot12}=\dfrac{93}{84}=\dfrac{31}{28}\end{matrix}\right.\)
\(\hept{\begin{cases}3x-y=2m+3\\x+2y=3m+1\end{cases}}\Leftrightarrow\hept{\begin{cases}6x-2y=4m+6\\x+2y=3m+1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=m+1\\y=m\end{cases}}\)khi đó: \(^{x^2+y^2=5\Leftrightarrow2m^2+2m+1=5\Leftrightarrow2m^2+2m-4=0\Leftrightarrow\orbr{\begin{cases}m=1\\m=-2\end{cases}}}\)
a) Thay \(x=1\)vào pt ta được :
\(1+k-4-4=0\)
\(\Leftrightarrow k-7=0\)
\(\Leftrightarrow k=7\)
b) Thay \(k=7\)vào pt ta được :
\(x^3+7x^2-4x-4=0\)
\(\Leftrightarrow\left(x^3-x^2\right)+\left(8x^2-8x\right)+\left(4x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)+8x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+8x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2+8x+4=0\end{cases}}\)
* \(x-1=0\Leftrightarrow x=1\)
* \(x^2+8x+4=0\)
Ta có : \(\Delta=8^2-4\times4=48>0\)
\(\Rightarrow\)pt có 2 nghiệm : \(\orbr{\begin{cases}x_1=\frac{-8-\sqrt{48}}{2}=-4-2\sqrt{3}\\x_2=\frac{-8+\sqrt{48}}{2}=-4+2\sqrt{3}\end{cases}}\)
Vậy ...
\(\left\{{}\begin{matrix}4x+5y=14\\5x+4y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14-5y}{4}\\\dfrac{70-25y}{4}+4y=13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14-5y}{4}\\\dfrac{70-9y}{4}=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14-5y}{4}\\y=\dfrac{70-\left(13.4\right)}{9}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14-5.2}{4}\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
thay x = 1 ; y=2 và 9x + 9y ta đc
\(\Leftrightarrow9+9.2=9+18=27\)