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Chọn D

18 tháng 4 2018

h(x) + g(x) = f(x)

=> h(x)= f(x) - g(x) = \(3x^4+2x^2-2x^4+x^2-5x-\left(x^4-x^2-2x+6+3x^2\right)=x^2-3x-6\)\(h\left(-\dfrac{1}{3}\right)=\left(-\dfrac{1}{3}\right)^2-3\left(-\dfrac{1}{3}\right)-6=\dfrac{-44}{9}\)

\(h\left(\dfrac{3}{2}\right)=\left(\dfrac{3}{2}\right)^2-3\cdot\dfrac{3}{2}-6=-\dfrac{33}{4}\)

\(x^2-3x-6=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{33}}{6}\\x=\dfrac{3-\sqrt{33}}{6}\end{matrix}\right.\)

a: Ta có: \(x^2+x+1\)

\(=x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)

b: Ta có: \(-x^2+x+2\)

\(=-\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{9}{4}\right)\)

\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

17 tháng 12 2017

\(\left\{{}\begin{matrix}h\left(x\right)+g\left(x\right)=x^2+2x+1\\h\left(x\right)-g\left(x\right)=x^2+2x-3\end{matrix}\right.\)

Lấy pt(1) cộng pt(2) ta được:

\(h\left(x\right)+g\left(x\right)+h\left(x\right)-g\left(x\right)=x^2+2x+1+x^2+2x-3\)

\(2h\left(x\right)=2x^2+4x-2\)

\(h\left(x\right)=x^2+2x-1\)

17 tháng 12 2017

Thanks

1 tháng 7 2017

a)Đặt \(A=x^2+x+1\)

         \(A=x^2+2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}\)

           \(A=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

                      Dấu = xảy ra khi \(x+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)

Vậy Min A =3/4 khi x=-1/2

b)Đặt \(B=2+x-x^2\)

        \(B=-\left(x^2-x-2\right)\)

        \(B=-\left(x^2-2.\frac{1}{2}x+\frac{1}{4}-\frac{9}{4}\right)\)

        \(B=\frac{9}{4}-\left(x-\frac{1}{2}\right)^2\le\frac{9}{4}\)

                        Dấu = xảy ra khi \(x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)

vậy Max B = 9/4 khi x=1/2

1 tháng 7 2017

c)\(h\left(h+1\right)\left(h+2\right)\left(h+3\right)\)

\(=\left(h^2+3h\right)\left(h^2+3h+2\right)\)

\(=t\left(t+2\right)\left(vi`...h^2+3h=t\right)\)

\(=t^2+2t=t^2+2t+1-1\)

\(=\left(t+1\right)^2-1=\left(h^2+3h\right)^2-1\ge1\)

Xảy ra khi \(h^2+3h=0\Rightarrow\orbr{\begin{cases}h=0\\h=-3\end{cases}}\)

3 tháng 8 2015

Gợi ý đc không làm dài lắm 

29 tháng 3 2019

a. f(x)+g(x)=2x5−4x4+3x3−x2+5x−1+(−x5+2x4−3x3−x2−2x+7)

=2x5-x5-4x4+2x4+3x3-3x3-x2-x2+5x-2x-1+7

=x5-2x4-2x2+3x+6

b. f(x)+h(x)=2x5−4x4+3x3−x2+5x−1+x5−2x4−2x2−x−3

=2x5+x5-4x4-2x4+3x3-x2-2x2+5x-x-1-3

=3x5-6x4+3x3-3x2+6x-4

c. g(x)+h(x)=−x5+2x4−3x3−x2−2x+7+x5−2x4−2x2−x−3

=-x5+x5+2x4-2x4-3x3-x2-2x2-2x-x+7-3

=-3x3-3x2-3x+4

d. f(x)-g(x)=2x5−4x4+3x3−x2+5x−1-(−x5+2x4−3x3−x2−2x+7)

=2x5−4x4+3x3−x2+5x−1-x5-2x4+3x3+x2+2x-7

=2x5-x5-4x4-2x4+3x3+3x3-x2+x2+5x+2x-1-7

=x5-6x4+6x3+7x-8

e. f(x)-h(x)=2x5−4x4+3x3−x2+5x−1-(x5−2x4−2x2−x−3)

=2x5−4x4+3x3−x2+5x−1-x5+2x4+2x2+x+3

=2x5-x5-4x4+2x4+3x3-x2+2x2+5x+x-1+3

=x5-2x4+3x3+x2+6x-4

h. g(x)-h(x)=−x5+2x4−3x3−x2−2x+7-(x5−2x4−2x2−x−3)

=−x5+2x4−3x3−x2−2x+7-x5+2x4+2x2+x+3

=-x5-x5+2x4+2x4-3x3-x2+2x2-2x+x+7+3

=-2x5+4x4-3x3+x2-x+10

f. f(x)+g(x)+h(x)=2x5−4x4+3x3−x2+5x−1+(−x5+2x4−3x3−x2−2x+7)+x5−2x4−2x2−x−3

=2x5-x5+x5-4x4+2x4-2x4+3x3-3x3-x2-x2-2x2+5x-2x-x-1+7-3

=2x5-4x4-4x2+2x+3

g. f(x)+g(x)-h(x)=2x5−4x4+3x3−x2+5x−1+(−x5+2x4−3x3−x2−2x+7)-(x5−2x4−2x2−x−3)

=2x5−4x4+3x3−x2+5x−1+(−x5+2x4−3x3−x2−2x+7)-x5+2x4+2x2+x+3

=2x5-x5-x5-4x4+2x4+2x4+3x3-3x3-x2-x2+2x2+5x-2x+x-1+7+3

=4x+9

n. f(x)-g(x)+h(x)=2x5−4x4+3x3−x2+5x−1-(−x5+2x4−3x3−x2−2x+7)+x5−2x4−2x2−x−3

=2x5−4x4+3x3−x2+5x−1-x5-2x4+3x3+x2+2x-7+x5−2x4−2x2−x−3

=2x5-x5+x5-4x4-2x4-2x4+3x3+3x3-x2+x2-2x2+5x+2x-x-1-7-3

=2x5-8x4+6x3-2x2+6x-11

m. f(x)-g(x)-h(x)=2x5−4x4+3x3−x2+5x−1-(−x5+2x4−3x3−x2−2x+7)-(x5−2x4−2x2−x−3)

=2x5−4x4+3x3−x2+5x−1-x5-2x4+3x3+x2+2x-7-x5+2x4+2x2+x+3

=2x5-x5-x5-4x4-2x4+2x4+3x3+3x3-x2+x2+2x2+5x+2x+x-1-7+3

=-4x4+6x3+2x2+8x-5

a: \(P\left(x\right)=x^4+x^3-x^2+2x-5\)

\(Q\left(x\right)=x^4+5x^3-3x^2-2x-5\)

b: \(H\left(x\right)=P\left(x\right)-Q\left(x\right)=-4x^3+2x^2+4x\)

c: Bậc của H(x) là 3

 

20 tháng 5 2022

Còn câu D bạn ơi