1,tim nghiem cua cac da thuc sau
a,f(x)=(x-1).(1-3x)
b,g(x)=(2x+1).(x^2+5)
c,h(x)=x^3-4x
d,k(x)= can bac 2.x+1
Tim nghiem cua cac da thuc sau:
a,f(x)=(x-1).(1-3x)
b,g(x)=(2x+1).(x^2+5)
c,h(x)=x^3-4x
d,k(x)=can bac 2.x+1
Bai2: Cho da thuc f(x)=x\(^2\)+4x-5
a, So -5 co phai la nghiem cua f(x) ko?
bViet tap hop S tat ca cac nghiem cua f(x)
Bai 3; Thu gon roi tim nghiem cua cac da thuc sau
a, f(x)=x(1-2x)+(2x\(^2\)-x+4)
b,g(x)=x(x-5)-x(x+2)+7x
c, h(x)=x(x-1)+1
moi nguoi xin hay giup to
Bài 2 mk giải luôn nhé
f(x)=x^2+4x-5=x^2-x+5x-5
=x(x-1)+5(x-1)
=(x+5)(x-1)
Vậy x=-5 hoặc x=1 là nghiệm của đa thức f(x)
tim nghiem cua cac da thuc
a,x^2+x
b,x^2+2x+1
c,2x^2+3x-5
d,x^2-4x+3
e,x^2+6x+5
f,3x(12x-4)-9x(4x-3)=30
g,2x(x-1)+x(5-2x)=15
Cho 2 da thuc: P(x)=-3x^2+2x+1
Q(x)=-3x^2+x-2
a, Tinh Q(1/2)
b, x=1 co phai la nghiem cua P(x) ko?
c, Tim da thuc H(x)=P(x)-Q(x)
d, Tim tat ca cac gia tri cua x sao cho P(x)=Q(x)
GIUP MIK VOI MIK DANG CAN GAP
THANK YOU
\(a,Q\left(\dfrac{1}{2}\right)=-3.\left(\dfrac{1}{2}\right)^2+\dfrac{1}{2}-2\)
\(Q\left(\dfrac{1}{2}\right)=-3.\dfrac{1}{4}+\dfrac{1}{2}-2\)
\(Q\left(\dfrac{1}{2}\right)=-\dfrac{3}{4}+\left(-\dfrac{3}{2}\right)\)
\(Q\left(\dfrac{1}{2}\right)=-\dfrac{9}{4}\)
\(b,P\left(1\right)=-3.1^2+2.1+1\)
\(P\left(1\right)=-3.1+2+1\)
\(P\left(1\right)=-3+2+1\)
\(P\left(1\right)=0\)
Vậy x = 1 là nghiệm của đa thức P(x)
\(c,H\left(x\right)=\left(-3x^2+2x+1\right)-\left(-3x^2+x-2\right)\)
Câu c thì dễ rồi bn tự làm đi nha còn câu d thì mik chịu
Bai 4: Xac dinh he so m de cac da thuc sau nhan 1 lam nghiem
a,mx\(^2\)+2x+8 b, 7x\(^2\)+mx-1 c, x\(^5\)-3x\(^2\)+m
Bai5: Cho da thuc f(x)=x\(^2\)+mx +2
a, Xac dinh m de f(x) nhan -2 lam 1 nghiem
b, Tim tap hop cac nghiem cua f(x) ung voi gia tri vua tim duoc cua m
Bai 6: Tim da thuc f(X) roi tim nghien cua f(x) biet
\(x^3+2x^2\left(4y-1\right)-4xy^2-9y^3-f\left(x\right)=-5x^3+8x^2y-4xy^2-9y^3\)
bai 1: cho cac da thuc
f(x)= x^5-3x^2+7x^4-x^5+2x^2-9x^3+x^2-1/4x+2x-3
g(x)=5x^4-x^5+1/2x^4+x^5+x^2-4x^4-2x^3+3x^2+x^3-1/4
a, thu gon va sap xep cac da thuc tren theo luy thua giam dancua ien
b,tinh f(1);f(-1); g(1); g(-1)
c,tinh f(x)+g(x);f(x)-g(x)
bai 1: cho cac da thuc
f(x)= x^5-3x^2+7x^4-x^5+2x^2-9x^3+x^2-1/4x+2x-3
g(x)=5x^4-x^5+1/2x^4+x^5+x^2-4x^4-2x^3+3x^2+x^3-1/4
a, thu gon va sap xep cac da thuc tren theo luy thua giam dancua ien
b,tinh f(1);f(-1); g(1); g(-1)
c,tinh f(x)+g(x);f(x)-g(x)
a)\(f\left(x\right)=x^5-3x^2+7x^4-x^5+2x^2-9x^3+x^2-\frac{1}{4}x+2x-3\)
\(=x^5-x^5+7x^4-9x^3-3x^2+2x^2+x^2-\frac{1}{4}x+2x-3\)
\(=7x^4-9x^3+\frac{7}{4}x-3\)
\(g\left(x\right)=5x^4-x^5+\frac{1}{2}x^2+x^5+x^2-4x^4-2x^3+3x^2+x^3-\frac{1}{4}\)
\(=-x^5+x^5+5x^4-4x^4-2x^3+x^3+\frac{1}{2}x^2+x^2+3x^2-\frac{1}{4}\)
\(=x^4-x^3+\frac{9}{2}x^2-\frac{1}{4}\)
b)\(f\left(1\right)=7.1^4-9.1^3+\frac{7}{4}.1-3=7-9+\frac{7}{4}-3=-\frac{13}{4}\)
\(f\left(-1\right)=7.\left(-1\right)^4-9.\left(-1\right)^3+\frac{7}{4}.\left(-1\right)-3=7+9-\frac{7}{4}-3=\frac{45}{4}\)
\(g\left(1\right)=1^4-1^3+\frac{9}{2}.1^2-\frac{1}{4}=1-1+\frac{9}{2}-\frac{1}{4}=\frac{17}{4}\)
\(g\left(-1\right)=\left(-1\right)^4-\left(-1\right)^3+\frac{9}{2}.\left(-1\right)^2-\frac{1}{4}=1+1+\frac{9}{2}-\frac{1}{4}=\frac{25}{4}\)
c) Ta có: f(x)+g(x)=\(7x^4-9x^3+\frac{7}{4}x-3+x^4-x^3+\frac{9}{2}x^2-\frac{1}{4}=7x^4+x^4-9x^3-x^3+\frac{9}{2}x^2+\frac{7}{4}x-3-\frac{1}{4}\)
\(=8x^4-10x^3+\frac{9}{2}x^2+\frac{7}{4}x-\frac{13}{4}\)
f(x)-g(x) =\(7x^4-9x^3+\frac{7}{4}x-3-x^4+x^3-\frac{9}{2}x^2+\frac{1}{4}=7x^4-x^4-9x^3+x^3-\frac{9}{2}x^2+\frac{7}{4}x-3+\frac{1}{4}\)
\(=6x^4-8x^3-\frac{9}{2}x^2+\frac{7}{4}x-\frac{11}{4}\)
Tim nghiem cua cac da thuc sau:
a)\(f\left(x\right)=3x^2-4x-7\)
b)\(g\left(x\right)=2x^2+3x+1\)
c)\(h\left(x\right)=x^2+9x+14\)
d)\(k\left(x\right)=x^3+4x^2+5x+2\)
e)\(Q\left(x\right)=x^2-11x-102\)
cho da thuc f(x)+3x^2+2x-5va g(x)=-3x^2-2x+2 tinh k=f+g va tim bac cua k
`K(x)=F(x)+G(x)`
`K(x)=(3x^2+2x-5)+(-3x^2-2x+2)`
`= 3x^2+2x-5-3x^2-2x+2`
`= (3x^2-3x^2)+(2x-2x)+(-5+2)`
`= -3`
Bậc của đa thức: `0`
`@` `\text {dnammv}`