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a)
5x2+ 12x- 30= 0
x( 5x +12- 30)= 0
\(\orbr{\begin{cases}x=0\\5x+12-30=0\end{cases}}\)
\(\orbr{\begin{cases}x=0\\5x+12=30\end{cases}}\)
\(\orbr{\begin{cases}x=0\\5x=30-12\end{cases}}\)
\(\orbr{\begin{cases}x=0\\5x=18\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=18:5\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=\frac{18}{5}\end{cases}}\)
Vậy PT có tập nghiệm là T={18/5;0}
P/s: chị nhớ thêm dấu tương đương vào PT nhé :)
a, m=2
=> \(x^2-6x+8=0\)=> \(\orbr{\begin{cases}x=2\\x=4\end{cases}}\)
b, Để phương trình có 2 nghiệm
thì \(\Delta'=\left(m+1\right)^2-m^2-4=2m-3\ge0\)=> \(m\ge\frac{3}{2}\)
Theo viet ta có
\(\hept{\begin{cases}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2+4\end{cases}}\)
Vì x2 là nghiệm của phương trình
nên \(2\left(m+1\right)x_2=x^2_2+m^2+4\)
Khi đó
\(\left(x_1^2+x^2_2\right)+m^2+4\le3m^2+16\)
=> \(\left(x_1+x_2\right)^2-2x_1x_2\le2m^2+12\)
=> \(4\left(m+1\right)^2-2\left(m^2+4\right)\le2m^2+12\)
=.>\(8m\le16\)=>\(m\le2\)
Vậy \(m\le2\)
ờ,,,,hắn là lớp 9,,,
tui quên ccoong thức bậc 3 r,,,,đợi xem lại tý
cứ xem đi bn ...công thức bậc 3 là biến đổi mà