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Đặt \(f\left(x\right)=y=\dfrac{1}{3}x^3-\left(2m-1\right)x^2+\left(m^2-m+7\right)x+m-5\)
=>\(y'=\dfrac{1}{3}\cdot3x^2-\left(2m-1\right)\cdot2x^2+\left(m^2-m+7\right)\)
=>\(y'=x^2-\left(4m-2\right)x^2+\left(m^2-m+7\right)\)
Đặt y'=0
\(\text{Δ}=\left(4m-2\right)^2-4\left(m^2-m+7\right)\)
\(=16m^2-16m+4-4m^2+4m-28=12m^2-12m-24\)
Để hàm số f(x) có hai cực trị thì Δ>0
=>\(12\left(m^2-m-2\right)>0\)
=>\(m^2-m-2>0\)
=>(m-2)(m+1)>0
=>\(\left[{}\begin{matrix}m>2\\m< -1\end{matrix}\right.\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=4m-2\\x_1x_2=\dfrac{c}{a}=m^2-m+7\end{matrix}\right.\)
\(x_1^2+x_2^2=\left(\sqrt{74}\right)^2=74\)
=>\(\left(x_1+x_2\right)^2-2x_1x_2=74\)
=>\(\left(4m-2\right)^2-2\left(m^2-m+7\right)=74\)
=>\(16m^2-16m+4-2m^2+2m-14=74\)
=>\(14m^2-14m-84=0\)
=>\(m^2-m-6=0\)
=>(m-3)(m+2)=0
=>\(\left[{}\begin{matrix}m=3\left(nhận\right)\\m=-2\left(nhận\right)\end{matrix}\right.\)
Diện tích hình chữ nhật đó là:
\(42\times24=1008\left(dm^2\right)\)
Diện tích hình tam giác đó là:
\(1008\times3=3024\left(dm^2\right)\)
Chiều cao hình tam giác là:
\(3024\times2:96=63\left(dm\right)\)
Diện tích của hình chữ nhật là:
42 x 24 = 1008 (dm2)
Vì diện tích tam giác gấp 3 lần diện tích hình chữ nhật nên diện của hình tam giác là:
1008 x 3 = 3024 (dm2)
Vậy chiều cao của tam giác đó là:
(3024 : 96) x 2 = 63 (dm)
Đáp số: 63 dm
nhớ tick
1 to know
2 to meet
3 will tell
4 has just gone
5 has already known
6 to do
7 read
8 has just left
9 learn
#\(yGLinh\)
1. They ought ( know ) ..to know.. all the regulations in this school
2. It's very happy ( meet ) ..to meet. them again
3.I am sure they ( tell ) ..will tell. us about themselves soon
4. Nam isn't here. He ( just go ) .has just gone.. out for a walk
5. He (already know ) ..has already known. this regulation
6. Of us is ready ( do ) ..to do. the work
7. Have you ever ( read ) ...read. the story
8. Our teacher ( just leave ) ..has just left. the school
9. Pupils must ( learn) ..learn. their lessons before coming to class
$1+2+3+4+5+6+7+8+9+10$
$=(1+9)+(2+8)+(3+7)+(4+6)+10+5$
$=10+10+10+10+10+5$
$=10\times5+5$
$=50+5=55$
1+2+3+4+5+6+7+8+9+10
=(1+10)+(2+9)+(3+8)+(4+7)+(5+6)
=11+11+11+11+11
=55
2B:
a: \(A=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{4}+\dfrac{5}{8}=\dfrac{3}{8}+\dfrac{5}{8}=\dfrac{8}{8}=1\)
b: \(B=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{3\left(\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}}{\dfrac{3}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}\right)}=\dfrac{1}{3}+\dfrac{2}{3}=1\)
a) \(A=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{4}+\dfrac{5}{8}\)
\(=\dfrac{3}{8}+\dfrac{5}{8}=1\)
b) \(B=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{1}{2}-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{3\left(\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)}\)
\(=\dfrac{1}{3}+\dfrac{1}{\dfrac{3}{2}}=\dfrac{1}{3}+\dfrac{2}{3}=1\)
1 is sleeping
2 eat
3 is running - wants
4 sets
5 is
6 am doing - think - know
7 buys
8 is
9 rains - is raining
10 is cooking - cooks
#\(yGLinh\)
1 is sleeping
2 eat
3 is running - wants
4 sets
5 is
6 am doing - think - know
7 buys
8 is
9 rains - is raining
10 is cooking - cooks
\(B=\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{n}{n+1}\)
\(=\dfrac{1}{n+1}\)
1. chữ n
2. chữ c
C1:Chữ n
C2:Chữ c