cho don thuc 3(a+1/a)x^4y^2 voi a la hang so ,a khac 0
a)tim a de da thuc luon khong am voi moi x,y
b)tim a de da thuc luon khong duong voi moi x,y
cho cac da thuc F(x)=4x2+3x-2 G(x)=3x2-2x+5 H(x)=x(5x-2)+3
a. tim x de F(x)+G(x)-H(x)=0
b. chung to F(x)-3x+5 luon duong voi moi x
Giải:
a) \(F\left(x\right)+G\left(x\right)-H\left(x\right)\)
\(=4x^2+3x-2+3x^2-2x+5-\left[x\left(5x-2\right)+3\right]\)
\(=4x^2+3x-2+3x^2-2x+5-\left(5x^2-2x+3\right)\)
\(=4x^2+3x-2+3x^2-2x+5-5x^2+2x-3\)
\(=2x^2+3x\)
Để \(F\left(x\right)+G\left(x\right)-H\left(x\right)=0\)
\(\Leftrightarrow2x^2+3x=0\)
\(\Leftrightarrow x\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(F\left(x\right)-3x+5\)
\(=4x^2+3x-2-3x+5\)
\(=4x^2+3\)
Vì \(x^2\ge0;\forall x\)
\(\Leftrightarrow4x^2\ge0;\forall x\)
\(\Leftrightarrow4x^2+3\ge3>0;\forall x\)
Vậy ...
a)Cho vi du ve 1 don thuc ,1 da thuc nhung khong phai la don thuc ,2 don thuc dong dang voi nhau ,da thuc 1 bien
b)Hay viet 1 don thuc dong dang voi don thuc don thuc xy\(^3\) sao cho tai x=-1 y=2 gia tri cua don thuc do la so nguyen duong nho hon 10
a) 5 ; x+y ;2xy đồng dạng với 4xy ;a (x) = 25x + 2x2
b) -2xy
a) Cho vidu ve 1 don thuc ,moi da thuc nhung k phai la don thuc 2 don thuc dong dang voi nhau,da thuc 1 bien
b) Hay viet 1 don thuc dong dang voi don thuc xy\(^3\) sao cho tai x=-1 y=2 gia tri cua don thuc do la so nguyen duong nho hon 10
CAC BAN GIUP MINH NHA MINH GAP LAM
cho cac da thuc F(x) = 4x^2 + 3x - 2
G(x) = 3x^2 - 2x + 5 H(x) = x(5x-2) +3
a) tim x de F(x) + G(x) - H(x) = 0
b) chung to F(x) - 3x + 5 luon duong voi moi x
chung minh rang bieu thuc 4x(x+y)(x+y+z)(x+y) y^2x^2 luon luon khong am voi moi gia tri cua x,y va z
Đặt \(A=4x\left(x+y\right)\left(x+y+z\right)\left(x+z\right)+y^2z^2\)
\(=4\left(x+y\right)\left(x+z\right)x\left(x+y+z\right)+y^2z^2=4\left(x^2+xz+xy+yz\right)\left(x^2+xy+xz\right)+y^2z^2\)
Đặt x2+xy+xz=t, ta có:
\(A=4\left(t+yz\right)t+y^2z^2=4t^2+4tyz+y^2z^2=\left(2t+yz\right)^2=\left(2x^2+2xy+2xz+yz\right)^2\ge0\)
chung minh rang bieu thuc 4x(x+y)(x+y+z)(x+y) y^2x^2 luon luon khong am voi moi gia tri cua x,y va z
ta có : \(4x\left(x+y\right)\left(x+y+z\right)\left(x+y\right)y^2x^2=4x\left(x+y+z\right)\left(x+y\right)^2y^2x^2\)
không thể khẳng định đc \(\Rightarrow\) bn xem lại đề .
cho da thuc :A(x)=x^2-x+1
a,cmr: a(x)>0 voi moi gtri cua x
b, da thuc A(x) co nghiem hay khong ? vi sao?
a) = x(x-1) +1
x(x-1) = 0 khi x = 0; x=1
còn lại x(x - 1) luôn >0
vậy A(x) >0 với mọi x
b) A(x) vô nghiệm vì A(x) luôn .> 0 (cmt)
cho m la so nguyen duong nho hon 30.tim m de da thuc x2+mx+72 viet thanh tich 2 da thuc bac nhat voi he so nguyen
a,CMR:Bieu thuc n(2n-3)-2n(n+1) luon chia het cho 5 voi moi n la so nguyen
b,CMR:Bieu thuc (n-1)(n+4)-(n-4)(n+10) luon chia het cho 6 voi moi so nguyen n