Đáp án:
Giải thích các bước giải:
4) 5x²-17x+12=0
⇔5x²-5x-12x+12=0
⇔(5x²-5x)-(12x+12)=0
⇔5x(x-1)-12(x-1)=0
⇔(5x-12)(x-1)=0
⇔\(\left[ \begin{array}{l}5x-12=0\\x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}5x=12\\x=1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=12/5\\x=1\end{array} \right.\)
5) 3x²-19x-22 =0
⇔ 3x²-3x+22x-22=0
⇔(3x²-3x)+(22x-22)=0
⇔3x(x-1)+22(x-1)=0
⇔(3x+22)(x-1)=0
⇔\(\left[ \begin{array}{l}3x-22=0\\x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}3x=22\\x=1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=22/3\\x=1\end{array} \right.\)
6) x²-(1+√2)x+√2=0
⇔x²-x-x√2+√2=0
⇔(x²-x)-(x√2-√2)=0
⇔x(x-1)-√2(x-1)=0
⇔(x-√2)(x-1)=0
⇔\(\left[ \begin{array}{l}x-√2=0\\x-1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=√2\\x=1\end{array} \right.\)
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