cho truoc n diem trong do k co 3 diem nao thang hang ve duong thang di qua cac cap diem. Tim n biet rang neu co them 1 diem (k thang hang voi bat cu 2 diem nao trong tong so n diem da cho) thi so duong thang ve dc tang them 8 duong
xác định hàm số y = | a |x biet rang do thi ham so di qua diem A ( -3,1 )
Tim m de duong thang d di qua I(-1;2) co he so goc -m cat (C) tai 3 diem phan biet A, B,I sao cho I la trung diem cua AB biet (C): y=x3 + 3x2 +mx+m
cho hai duong thang : (d1) : y = 4mx -(m+5) (d2): y =(3m^2 +1)x + (m^2 +9) a,voi gia tri naocua m thi (d1)//(d2) , b,voi gia tri nao cua m thi (d1) cat (d2) tim toa do giao diem khi m =2 , c, C /m rang khi m thay doi thi duong thang (d1) luon di qua diem co dinh A ,(d2) d qua diem co di B .Tinh AB ?
Cho f(x)=ax2+bx+c voi a,b,c la cac so huu ti.Chung minh rang:f(-2).f)3)≤0.Biet rang:13a+b+2c=0.
f(-2).f(3) = (4a-2b+c).(9a+3b+c)
= (4a-2b+c).(13a+b+2c-(4a-2b+c))
Mà 13a+b+2c = 0 theo giả thiết
=> f(-2).f(3) = -[(4a-2b+c)^2]
Có (4a-2b+c)^2 luôn >= 0 => f(-2).f(3) luôn nhỏ hơn hoặc bằng 0
tại sao (4a-2b+c)^2 luôn >=2 mà lại suy ra được f(-2)f(3) luôn nhỏ hơn = 0
Cho truoc n diem trong do ko co 3 diem nao thang hang.Ve cac duong thang di qua cac cap diem,Tim n biet rang neu co them 1 diem(ko thang hang voi bat ki 2 diem nao trong so n diem da cho) thi so duong thang ve duoc tang them la 8.
Minh dang can gap!Cac ban giup minh voi nhe!
1) Biet rang do thi ham so y = ax+b di qua diem E(2;-1) va song song voi duong thang ON voi O la goc toa do va N( 1; 3). Tinh gia tri bieu thuc S = a2 + b2
A. S= -40 B. S = 58 C. S = -4 D. S = -58
Phương trình đường thẳng ON có dạng \(y=a'x+b'\left(d'\right)\)
\(\left\{{}\begin{matrix}b'=0\\a'+b'=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}b'=0\\a'=3\end{matrix}\right.\Rightarrow y=3x\left(d'\right)\)
\(y=ax+b\left(d\right)\) đi qua \(E\left(2;-1\right)\Rightarrow2a+b=-1\left(1\right)\)
\(\left(d\right)//\left(d'\right)\Leftrightarrow\left\{{}\begin{matrix}a=3\\b\ne0\end{matrix}\right.\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow b=-7\)
\(\Rightarrow S=a^2+b^2=58\)
cho n diem trong do khong co 3 diem nao thang hang . Cu qua 2 diem ta ve 1 duong thang . Biet rang co tat ca 105 duong thang . tim n ?
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105 = n.(n-1):2
=n.(n-1)=105.2
=n.(n-1)=210
=n.(n-1)=15.14
vậy n = 15
ta co 105=n[n-1]chia 2
210=n[n-1]
14.15=n[n-1]
vay n=15
Cho parabol (p) y=1/2×x2 va 2 diem A va B thuoc (p) co hoanh do lan luot la -1; 2. Duong thang (d) co phuong trinh y=mx+n
a) tim toa do hai diem A va B. Tim m va n biet (d) di qua 2 diem A va B
b) tim do dai duong cao OH cua tam giac OAB. biet O la goc toa do
a) xa =-1 =>ya =1/2.(-1)^2 =1/2=> A(-1;1/2)
xb=2 =>yb =1/2.2^2 =2=> B(2;2)
\(\left\{{}\begin{matrix}\dfrac{1}{2}=-m+n\\2=2m+n\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-2m+2n=1\\2m+n=2\end{matrix}\right.\)=> n=1; m =1/2
b) \(AB=\sqrt{\left(x_b-x_a\right)^2+\left(y_b-y_a\right)^2}=\sqrt{3^2+\left(\dfrac{3}{2}\right)^2}=\sqrt{\dfrac{3^2\left(4^2+1\right)}{4^2}}=\dfrac{3\sqrt{17}}{4}\)\(S\Delta_{AOB}=\dfrac{1}{2}\left(\left|x_a\right|+\left|x_b\right|\right)\left(y_b-y_a\right)=\dfrac{1}{2}\left(1+2\right).\left(2-\dfrac{1}{2}\right)=\dfrac{1}{2}.3.\dfrac{3}{2}=\left(\dfrac{3}{2}\right)^2\)\(S_{\Delta AOC}=\dfrac{1}{2}OH.AB\)
\(OH=2.\dfrac{\dfrac{9}{4}}{\dfrac{3\sqrt{17}}{4}}=\dfrac{6}{\sqrt{17}}=\dfrac{6\sqrt{17}}{17}\)
