khoanh tròn :giúp với
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\(c,\) Để PT có 2 nghiệm \(x_1;x_2\Leftrightarrow\Delta=\left(m-4\right)^2+8\left(m-2\right)\ge0\)
\(\Leftrightarrow m^2-8m+16+8m-16\ge0\\ \Leftrightarrow m^2\ge0\left(\text{luôn đúng}\right)\)
Do đó PT có 2 nghiệm với mọi m
\(\text{Viét: }\left\{{}\begin{matrix}x_1+x_2=\dfrac{m-4}{m-2}\left(1\right)\\x_1x_2=\dfrac{2}{2-m}\left(2\right)\end{matrix}\right.\)
Kết hợp \(x_1-x_2=3\text{ với }\left(1\right)\text{ ta được}\)
\(\left\{{}\begin{matrix}x_1=\left(\dfrac{m-4}{m-2}+3\right):2=\dfrac{4m-10}{m-2}\cdot\dfrac{1}{2}=\dfrac{2m-5}{m-2}\\x_2=\dfrac{m-4}{m-2}-\dfrac{2m-5}{m-2}=\dfrac{1-m}{m-2}\end{matrix}\right.\)
Thay vào \(\left(2\right)\Leftrightarrow\dfrac{\left(2m-5\right)\left(1-m\right)}{\left(2-m\right)^2}=\dfrac{2}{2-m}\)
\(\Leftrightarrow\left(2m-5\right)\left(1-m\right)=2\left(2-m\right)\\ \Leftrightarrow7m-2m^2-5=4-2m\\ \Leftrightarrow2m^2-9m+9=0\\ \Leftrightarrow\left[{}\begin{matrix}m=3\\m=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(m=3\text{ và }m=\dfrac{3}{2}\) thỏa đề bài
VII:
1. older than
2. more expensive than
3. more difficult than
4. longer than
5. more modern than
6. younger than
7. more expensive
8. the coldest
9. larger
VI.
1. d → strongest
2. b → shorter
3. b → higher
4. d → as
5. d → youngest
6. d → than
VII.
1. older than
2. more expensive than
3. more difficult than
4. longer than
5. more modern
6. the youngest
7. more expensive
8. the coldest
9. larger than
VIII.
1. c
2. c
3. a
4. c
5. b
`#3107.101107`
Câu 1:
a.
`(1)` Vô cùng nhỏ
`(2)` Trung hòa về điện
`(3)` hạt nhân
`(4)` điện tích dương
`(5)` vỏ nguyên tử
`(6)` các electron
`(7)` điện tích âm
b.
`(8)` chuyển động
`(9)` sắp xếp
c.
`(10)` electron
`(11)` hạt nhân
Bài `3`
Cậu tách cho các câu sau nx nhé^^
\(a,x+\dfrac{1}{2}=\dfrac{7}{3}\\ \Rightarrow x=\dfrac{7}{3}-\dfrac{1}{2}\\ \Rightarrow x=\dfrac{14}{6}-\dfrac{3}{6}\\ \Rightarrow x=\dfrac{11}{6}\\ b,\dfrac{2}{5}x-\dfrac{1}{5}=-0,6\\ \Rightarrow\dfrac{2}{5}x=-\dfrac{3}{5}+\dfrac{1}{5}\\ \Rightarrow\dfrac{2}{5}x=-\dfrac{2}{5}\\ \Rightarrow x=-\dfrac{2}{5}:\dfrac{2}{5}\\ \Rightarrow x=-1\\ c,\left(0,5x-\dfrac{3}{7}\right):\dfrac{1}{2}=1\dfrac{1}{7}\\ \Rightarrow\dfrac{1}{2}x-\dfrac{3}{7}=\dfrac{8}{7}\cdot\dfrac{1}{2}\\ \Rightarrow\dfrac{1}{2}x-\dfrac{3}{7}=\dfrac{8}{14}\\ \Rightarrow\dfrac{1}{2}x=\dfrac{4}{7}+\dfrac{3}{7}\\ \Rightarrow\dfrac{1}{2}x=1\\ \Rightarrow x=1:\dfrac{1}{2}\\ \Rightarrow x=2\)
\(d,\dfrac{2}{3}x-\dfrac{2}{5}=\dfrac{1}{2}x-\dfrac{1}{3}\\ \Rightarrow\dfrac{2}{3}x-\dfrac{1}{2}x=-\dfrac{1}{3}+\dfrac{2}{5}\\ \Rightarrow\left(\dfrac{2}{3}-\dfrac{1}{2}\right)x=\dfrac{1}{15}\\ \Rightarrow\dfrac{1}{6}x=\dfrac{1}{15}\\ \Rightarrow x=\dfrac{1}{15}:\dfrac{1}{6}\\ \Rightarrow x=\dfrac{2}{5}\)
`e,1/2 x+2 1/2=3 1/2 x-3/4`
`=> 1/2 x+ 5/2= 7/2x - 3/4`
`=> 1/2x - 7/2x = -3/4 -5/2`
`=> -3x=-13/4`
`=>x=13/12`
\(f,2x\left(x-\dfrac{1}{7}\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\\ g,\left(\dfrac{2x}{5}-1\right):\left(-5\right)=\dfrac{1}{4}\\ \Rightarrow2x:5-1=\dfrac{1}{4}\cdot\left(-5\right)\\ \Rightarrow2x:5-1=-\dfrac{5}{4}\\ \Rightarrow2x:5=-\dfrac{5}{4}+1\\ \Rightarrow2x:5=-\dfrac{1}{14}\\ \Rightarrow2x=-\dfrac{1}{14}\cdot5\\ \Rightarrow2x=-\dfrac{5}{14}\\ \Rightarrow x=-\dfrac{5}{14}:2\\ \Rightarrow x=-\dfrac{5}{28}\)
\(\left(x-1\right)^3=\dfrac{1}{8}\\ \Rightarrow\left(x-1\right)^3=\left(\dfrac{1}{2}\right)^3\\ \Rightarrow x-1=\dfrac{1}{2}\\ \Rightarrow x=\dfrac{1}{2}+1\\ \Rightarrow x=\dfrac{1}{2}+\dfrac{2}{2}\\ \Rightarrow x=\dfrac{3}{2}\)
đúng k