Bài 2:
a) Khi m = 2 ta có pt:
\(x^2-4x+m+1=0\)
\(\Leftrightarrow x^2-4x+3=0\)
\(\Delta=\left(-4\right)^2-4\cdot1\cdot3=4>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{4+\sqrt{4}}{2}=3\\x_2=\dfrac{4-\sqrt{4}}{2}=1\end{matrix}\right.\)
b) Để pt có nghiệm thì:
\(\Delta\ge0\)
\(\Leftrightarrow\left(-4\right)^2-4\cdot1\cdot\left(m+1\right)\ge0\)
\(\Leftrightarrow16-4m-4\ge0\)
\(\Leftrightarrow-4m+12\ge0\)
\(\Leftrightarrow4m\le12\)
\(\Leftrightarrow m\le3\)
c) \(x^2-4x+m+1=0\)
\(\Delta=\left(-4\right)^2+4\cdot1\cdot\left(m+1\right)=12-4m\)
Để pt có 2 nghiệm phân biệt thì m < 3
\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{4+\sqrt{12-4m}}{2}=2+\sqrt{3-m}\\x_2=\dfrac{4-\sqrt{12-4m}}{2}=2-\sqrt{3-m}\end{matrix}\right.\)
Mà: \(x^2_1+x^2_2=10\)
\(\Leftrightarrow\left(2+\sqrt{3-m}\right)^2+\left(2-\sqrt{3-m}\right)^2=10\)
\(\Leftrightarrow4+4\sqrt{3-m}+3-m+4-4\sqrt{3-m}+3-m=10\)
\(\Leftrightarrow14-2m=10\)
\(\Leftrightarrow2m=4\)
\(\Leftrightarrow m=2\left(tm\right)\)
d) Ta có: \(x^3_1+x^3_2=34\)
\(\Leftrightarrow\left(2+\sqrt{3-m}\right)^2+\left(2-\sqrt{3-m}\right)^2=34\)
\(\Leftrightarrow8+12\sqrt{3-m}+6\left(3-m\right)+\left(3-m\right)\sqrt{3-m}+8-12\sqrt{3-m}+6\left(3-m\right)-\left(3-m\right)\sqrt{3-m}=34\)
\(\Leftrightarrow16+12\left(3-m\right)=34\)
\(\Leftrightarrow2\left(3-m\right)=18\)
\(\Leftrightarrow3-m=9\)
\(\Leftrightarrow m=-6\left(tm\right)\)
Bài 1:
a: \(x^2-11x+30=0\)
=>\(x^2-5x-6x+30=0\)
=>\(x\left(x-5\right)-6\left(x-5\right)=0\)
=>(x-5)(x-6)=0
=>\(\left[{}\begin{matrix}x-5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
b: \(x^2-10x+21=0\)
=>\(x^2-3x-7x+21=0\)
=>\(x\left(x-3\right)-7\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(x-7\right)=0\)
=>\(\left[{}\begin{matrix}x-3=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)
c: \(x^2-12x+27=0\)
=>\(x^2-3x-9x+27=0\)
=>\(x\left(x-3\right)-9\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(x-9\right)=0\)
=>\(\left[{}\begin{matrix}x-3=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=9\end{matrix}\right.\)
d: \(5x^2-17x+12=0\)
=>\(5x^2-5x-12x+12=0\)
=>\(5x\left(x-1\right)-12\left(x-1\right)=0\)
=>(x-1)(5x-12)=0
=>\(\left[{}\begin{matrix}x-1=0\\5x-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{12}{5}\end{matrix}\right.\)
e: \(3x^2-19x-22=0\)
=>\(3x^2-22x+3x-22=0\)
=>\(x\left(3x-22\right)+\left(3x-22\right)=0\)
=>(3x-22)(x+1)=0
=>\(\left[{}\begin{matrix}3x-22=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{22}{3}\\x=-1\end{matrix}\right.\)
f: \(x^2-14x+33=0\)
=>\(x^2-3x-11x+33=0\)
=>\(x\left(x-3\right)-11\left(x-3\right)=0\)
=>(x-3)(x-11)=0
=>\(\left[{}\begin{matrix}x-3=0\\x-11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=11\end{matrix}\right.\)
g: \(6x^2-13x-48=0\)
\(\text{Δ}=\left(-13\right)^2-4\cdot6\cdot\left(-48\right)=1321\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{13-\sqrt{1321}}{12}\\x_2=\dfrac{13+\sqrt{1321}}{12}\end{matrix}\right.\)
h: \(3x^2+5x+61=0\)
\(\text{Δ}=5^2-4\cdot3\cdot61=25-732=-707< 0\)
=>phương trình vô nghiệm
i: \(x^2-\sqrt{3}x-2-\sqrt{6}=0\)
=>\(\left(x^2-2\right)-\sqrt{3}\left(x+\sqrt{2}\right)=0\)
=>\(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)-\sqrt{3}\left(x+\sqrt{2}\right)=0\)
=>\(\left(x+\sqrt{2}\right)\left(x-\sqrt{2}-\sqrt{3}\right)=0\)
=>\(\left[{}\begin{matrix}x+\sqrt{2}=0\\x-\sqrt{2}-\sqrt{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{2}\\x=\sqrt{2}+\sqrt{3}\end{matrix}\right.\)
