\(a,\Leftrightarrow\Delta'=\left(m-1\right)^2-\left(m-2\right)\ge0\\ \Leftrightarrow m^2-3m+3\ge0\\ \Leftrightarrow\left(m-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge0\left(\text{luôn đúng}\right)\)

Vậy PT có 2 nghiệm pb với mọi m

\(b,\Leftrightarrow0< x_1< x_2\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2>0\\x_1x_2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\left(m-1\right)>0\\m-2>0\end{matrix}\right.\Leftrightarrow m>2\\ c,\text{Thay }x=2\Leftrightarrow4-4\left(m-1\right)+m-2=0\\ \Leftrightarrow m=2\\ \Leftrightarrow x^2-2x=0\\ \Leftrightarrow x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ d,\text{Viét: }\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=m-2\end{matrix}\right.\\ x_1^2+x_2^2=8\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=8\\ \Leftrightarrow4\left(m-1\right)^2-2\left(m-2\right)=8\\ \Leftrightarrow4m^2-10m=0\\ \Leftrightarrow m\left(2m-5\right)=0\Leftrightarrow\left[{}\begin{matrix}m=0\\m=\dfrac{5}{2}\end{matrix}\right.\)

Đăng tách ra bạn nhé 

Vì AD là pg \(\dfrac{AB}{AC}=\dfrac{BD}{DC}\Rightarrow\dfrac{5}{4}=\dfrac{3}{DC}\Rightarrow DC=\dfrac{12}{5}cm\)

BC = DC + DB = 12/5 + 3 = 27/5 cm 

chọn B 

16,2 nhe . nho k mik ^ - ^

ta có: 30,9-28,2+25,5-22,8+....+3,9-1,2

=(30,9+25,5 +...+ 3,9)-(28,2+22,8+...+1,2)

=[(30,9+3,9)*6 : 2 ]-[(28,2+1,2)*6 : 2]

= 104,4-88,2

=16,2

16,2 bạn nhé

3: \(\left(3x+5\right)\left(2x-7\right)\)

\(=6x^2-21x+10x-35\)

\(=6x^2-11x-35\)

4: \(\left(5x-2\right)\left(3x+4\right)\)

\(=15x^2+20x-6x-8\)

\(=15x^2+14x-8\)

bai tap nay lop may day

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