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![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\left\{{}\begin{matrix}2x+2y=4\\2x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5y=-5\\x=2-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=3\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\x+y=10\end{matrix}\right.\)Theo tc dãy tỉ số bằng nhau
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{10}{5}=2\Rightarrow x=4;y=6\)
a.\(\Leftrightarrow\left\{{}\begin{matrix}3x+3y=6\\2x-3y=9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=15\\2x-3y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\2.3-3y=9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
b.\(\Leftrightarrow\left\{{}\begin{matrix}3x=2y\\x+y-10=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x-2y=0\\x+y-10=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-2y=0\\2x+2y=20\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=20\\3x-2y=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=4\\3.4-2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=6\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x^2 - 3x + 2 = 0
\(\Delta=b^2-4ac=\left(-3\right)^2-4.1.2=1\)
=> pt có 2 nghiệm pb
\(x_1=\frac{-\left(-3\right)+1}{2}=2\)
\(x_2=\frac{-\left(-3\right)-1}{2}=1\)
a) Dễ thấy phương trình có a + b + c = 0
nên pt đã cho có hai nghiệm phân biệt x1 = 1 ; x2 = c/a = 2
b) \(\hept{\begin{cases}x+3y=3\left(I\right)\\4x-3y=-18\left(II\right)\end{cases}}\)
Lấy (I) + (II) theo vế => 5x = -15 <=> x = -3
Thay x = -3 vào (I) => -3 + 3y = 3 => y = 2
Vậy pt có nghiệm ( x ; y ) = ( -3 ; 2 )
![](https://rs.olm.vn/images/avt/0.png?1311)
????
xin lỗi nha !
mình mới học lớp 3
mà bài này khó nắm
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\left\{{}\begin{matrix}3x+y=3\\2x-y=7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x=10\\2x-y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2x-7=2\cdot2-7=-3\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là (x,y)=(2;-3)
b) Ta có: \(7x^2-2x+3=0\)
a=7; b=-2; c=3
\(\Delta=\left(-2\right)^2-4\cdot7\cdot3=4-84=-80< 0\)
Suy ra: Phương trình vô nghiệm
Vậy: \(S=\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(\left\{{}\begin{matrix}2x+y=10\\5x-3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10x+5y=50\\10x-6y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}11y=44\\2x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=3\end{matrix}\right.\)
Vậy hpt có nghiệm (x;y) = (3;4)
2)
a) 3x2 - 2x - 1 = 0
\(\Leftrightarrow3x^2-3x+x-1=0\)
\(\Leftrightarrow3x\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=1\end{matrix}\right.\)
Vậy pt có nghiệm x = 1 hoặc x = 3
b) Đặt x2 = t (t \(\ge\) 0)
Pt trở thành: t2 - 20t + 4 = 0
\(\Delta\) = (-20)2 - 4.1.4 = 400 - 16 = 384
=> pt có 2 nghiệm phân biệt t1 = \(\dfrac{20+8\sqrt{6}}{2}=10+4\sqrt{6}\)
t2 = \(\dfrac{20-8\sqrt{6}}{2}=10-4\sqrt{6}\)
=> x1 = \(\sqrt{10+4\sqrt{6}}=\sqrt{\left(2+\sqrt{6}\right)^2}=2+\sqrt{6}\)
x2 = \(2-\sqrt{6}\)
\(a,\left\{{}\begin{matrix}x+y=3\\2x-3y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+2y=6\\2x-3y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5y=5\\2x-3y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=1\\2x-3.1=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(2;1\right)\)
b, \(x^2-7x+10=0\\ \Leftrightarrow x^2-5x-2x+10=0\\ \Leftrightarrow x\left(x-5\right)-2\left(x-5\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
\(a,\)\(\left\{{}\begin{matrix}x+y=3\\2x-3y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+3y=9\\2x-3y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x=10\\2x-3y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2.2-3y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
Vậy hệ pt có nghiệm duy nhất \(\left(x;y\right)=\left(2;1\right)\)
\(b,x^2-7x+10=0\)
\(\Delta=b^2-4ac=\left(-7\right)^2-4.10=9>0\)
\(\Rightarrow\) Pt có 2 nghiệm \(x_1,x_2\)
Ta có :
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{7+3}{2}=5\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{7-3}{2}=2\end{matrix}\right.\)
Vậy \(S=\left\{5;2\right\}\)