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a: \(P\left(x\right)=-5x^4+2x^2-8x+\dfrac{1}{2}\)
\(Q\left(x\right)=4x^4+2x^3-5x^2-6x+\dfrac{3}{2}\)
b: \(A\left(x\right)=-5x^4+2x^2-8x+\dfrac{1}{2}+4x^4+2x^3-5x^2-6x+\dfrac{3}{2}=-x^4+2x^3-3x^2-14x+2\)
\(B\left(x\right)=-5x^4+2x^2-8x+\dfrac{1}{2}-4x^4-2x^3+5x^2+6x-\dfrac{3}{2}=-9x^4-2x^3+7x^2-2x-1\)
a, f(y)=4y6−6y2−3y4−3+4y4−4y6+5y
=\(^{y^4-6y^2+5y-3}\)
b, f(0)=\(^{0^4-6.0^2+5.0-3}\)
=-3
f(\(\dfrac{1}{2}\))=(\(\left(\dfrac{1}{2}\right)^4-6.\left(\dfrac{1}{2}\right)^2+5.\dfrac{1}{2}-3\)
=\(\dfrac{1}{16}-\dfrac{3}{2}+\dfrac{5}{2}-\dfrac{6}{2}\)
=\(\dfrac{1}{16}-\dfrac{24}{16}+\dfrac{40}{16}-\dfrac{48}{16}\)
=\(\dfrac{-31}{16}\)
c, A(y)=f(y)+k(y)
=(\(^{y^4-6y^2+5y-3}\))+(\(4y^2-y^4\)
=\(2y^2+5y-3\)
Xin lỗi ad nhìu nha :(( ý d tui hơm nhớ cách làm nên hông dám chỉ bậy:)
\(A=\dfrac{5}{4}\cdot\dfrac{15-4}{3}\cdot\dfrac{-1}{11}=\dfrac{5}{4}\cdot\dfrac{11}{3}\cdot\dfrac{-1}{11}=\dfrac{-5}{12}\)=-50/120
\(B=\dfrac{3}{4}\cdot\dfrac{-1}{12}\cdot\dfrac{-2}{3}=\dfrac{3\cdot2}{4\cdot12\cdot3}=\dfrac{2}{4\cdot12}=\dfrac{1}{24}\)=5/120
\(C=\dfrac{5}{4}\cdot\dfrac{-1}{15}\cdot\dfrac{-2}{5}=\dfrac{2}{4\cdot15}=\dfrac{1}{30}\)=4/120
\(D=3\cdot\dfrac{8-15}{12}\cdot\dfrac{-1}{7}=\dfrac{1}{4}\)=30/120
Vì -50<4<5<30
nên A<C<B<D
A=\(\dfrac{5}{4}\).(5-\(\dfrac{4}{3}\)).(\(-\dfrac{1}{11}\))
= \(\dfrac{5}{4}\).\(\dfrac{11}{3}\).(\(-\dfrac{1}{11}\))
=\(\dfrac{5}{4}\).[\(\dfrac{11}{3}.\left(-\dfrac{1}{11}\right)\text{]}\)
=\(\dfrac{5}{4}.\dfrac{1}{3}\)
=\(\dfrac{5}{12}\) (1)
B=\(\dfrac{3}{4}:\left(-12\right).\left(-\dfrac{2}{3}\right)\) =\(\dfrac{3}{4}:\text{[}\left(-12\right).\left(-\dfrac{2}{3}\right)\text{]}\)
=\(\dfrac{3}{4}:8\) =\(\dfrac{3}{4}.\dfrac{1}{8}\)=\(\dfrac{3}{32}\)(2)
C=\(\dfrac{5}{4}:\left(-15\right).\left(-\dfrac{2}{5}\right)\) =\(\dfrac{5}{4}:\text{[}\left(-15\right).\left(-\dfrac{2}{5}\right)\text{]}\)
=\(\dfrac{5}{4}:6=\dfrac{5}{4}.\dfrac{1}{6}=\dfrac{5}{24}\left(3\right)\)
D=(-3).\(\left(\dfrac{2}{3}-\dfrac{5}{4}\right):\left(-7\right)\) =(-3).\(\left(-\dfrac{7}{12}\right)\):(-7)=\(\dfrac{7}{4}:\left(-7\right)\)=\(\dfrac{7}{4}\).\(\left(\dfrac{-1}{7}\right)\)=\(\dfrac{-1}{4}\) (4)
Từ (1),(2),(3)và(4)=>Ta có thể sắp xếp các kết quả trên theo thứ tự tăng dần là:
(Bạn tự làm nhé! mình bận đi học rồi
)
a) \(x^2y^2-x^2+\left(\dfrac{1}{2}\right)^6x=x^2y^2-x^2+\dfrac{1}{64}x\)
\(\Rightarrow\) đa thức bậc 4
b) \(\left(-9x^2\right)\dfrac{1}{3}y+y\left(-x^2\right)+24x\left(\dfrac{-1}{4}xy\right)\)
\(=-3x^2y-x^2y-6x^2y\)
\(=-10x^2y\)
Thay \(x=1;y=-1\) vào đa thức ta có:
\(-10x^2y=-10.1^2.\left(-1\right)=10\)
\(A=\dfrac{5}{4}\left(5-\dfrac{4}{3}\right)\left(-\dfrac{1}{11}\right)\)
\(A=\dfrac{5}{4}.\dfrac{11}{3}.\left(-\dfrac{1}{11}\right)\)
\(A=-\dfrac{5}{12}\)
\(B=\dfrac{3}{4}:\left(-12\right).\left(-\dfrac{2}{3}\right)\)
\(B=\dfrac{3}{4}.\left(-\dfrac{1}{12}\right).\left(-\dfrac{2}{3}\right)\)
\(B=\dfrac{1}{24}\)
\(C=\dfrac{5}{4}:\left(-15\right).\left(-\dfrac{2}{5}\right)\)
\(C=\dfrac{5}{4}.\left(-\dfrac{1}{15}\right).\left(-\dfrac{2}{5}\right)\)
\(C=\dfrac{1}{30}\)
\(D=\left(-3\right)\left(\dfrac{2}{3}-\dfrac{5}{4}\right):\left(-7\right)\)
\(D=\left(-3\right)\left(-\dfrac{7}{12}\right)\left(-\dfrac{1}{7}\right)\)
\(D=-\dfrac{1}{4}\)
Sắp xếp theo thứ tự tăng dần:
\(A,D,C,B\)
a: \(=-\dfrac{1}{15}x^6y\)
b: \(=\dfrac{4}{5}ab^5\cdot2x^3y\cdot\left(-y\right)=-\dfrac{8}{5}ab^5\cdot x^3y^2\)
c: \(=-16\cdot\dfrac{3}{4}v^3\cdot\dfrac{-2}{5}uv=\dfrac{24}{5}v^4u\)
d: \(=8\cdot\left(-64\right)\cdot5\cdot u^2v^2\cdot\left(-27\right)v^3=69120u^2v^5\)
e: \(=-10y\cdot8y^3z^3\cdot25z^2=-2000y^4z^5\)
Đơn giản
\(C=\dfrac{2}{3}y^4-\dfrac{3}{4}y^2-\dfrac{9}{2}y-\dfrac{1}{3}y^3+y-\dfrac{2}{5}=-\dfrac{2}{5}-\dfrac{7}{2}y-\dfrac{3}{4}y^2-\dfrac{1}{3}y^3+\dfrac{2}{3}y^4\)\(D=\dfrac{1}{5}y^2-y-3y^4+2=2-y+\dfrac{1}{5}y^2-3y^4\)
Bây giờ tìm hiệu của C - D
Ta có hiệu của C - D bằng: ( xin lỗi vì không biết cách đặt tính )
\(C=-\dfrac{2}{5}-\dfrac{7}{2}y-\dfrac{3}{4}y^2-\dfrac{1}{3}y^3+\dfrac{2}{3}y^4-D=2-y+\dfrac{1}{5}y^2-3y^4\)
C - D = \(-\dfrac{12}{5}-\dfrac{5}{2}y-\dfrac{19}{20}y^2-\dfrac{1}{3}y^3+\dfrac{11}{3}y^4\)
Easy