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20 tháng 3 2017
a) f(x)-g(x)+h(x)= (2x^2-3x^3)-(3x-3x^3+2x-2)+(2x^2+1)
=2x^2-3x^3-3x+3x^3-2x+2+2x^2+1
=(2x^2+2x^2)+(-3x^3-3x^3)+(2x+3x)+(-2+1)
=4x^2-6x^3+5x-1
b)g(x)-f(x)+h(x)=3x-3x^3+2x-2-2x^2+3x^3+2x^2+1
=(3x+2x)+(-3x^3+3x^3)+(-2x^2+2x^2)+(-2+1)
=5x-1
bạn ơi, cái chỗ mình bỏ trống là như trên nha
5 tháng 11 2017
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
KN
22 tháng 6 2019
a) \(f\left(x\right)=5x^3-7x^2+2x+5\)
\(\Rightarrow f\left(1\right)=5.1^3-7.1^2+2.1+5\)
\(\Rightarrow f\left(1\right)=5.1-7.1+2+5\)
\(\Rightarrow f\left(1\right)=5-7+7\)
\(\Rightarrow f\left(1\right)=5\)
Vậy f(1) = 5.
\(g\left(x\right)=7x^3-7x^2+2x+5\)
\(\Rightarrow g\left(\frac{1}{2}\right)=7.\left(\frac{1}{2}\right)^3-7.\left(\frac{1}{2}\right)^2+2.\frac{1}{2}+5\)
\(\Leftrightarrow g\left(\frac{1}{2}\right)=7.\frac{1}{8}-7.\frac{1}{4}+1+5\)
\(\Leftrightarrow g\left(\frac{1}{2}\right)=\frac{7}{8}-\frac{14}{8}+6\)
\(\Leftrightarrow g\left(\frac{1}{2}\right)=\frac{-7}{8}+\frac{48}{8}\)
\(\Leftrightarrow g\left(\frac{1}{2}\right)=\frac{41}{8}\)
Vậy \(g\left(\frac{1}{2}\right)=\frac{41}{8}\)
KN
22 tháng 6 2019
\(h\left(x\right)=2x^3+4x+1\)
\(\Rightarrow h\left(0\right)=2.0^3+4.0+1\)
\(\Rightarrow h\left(0\right)=0+0+1\)
\(\Rightarrow h\left(0\right)=1\)
Vậy \(h\left(0\right)=1\)
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
11 tháng 7 2016
a) Bậc của đa thức H(x): 3
b) H(2) = 23 – 2.22 + 5. 2 – 10= 8 – 8 + 10 – 10 = 0
H(-1) = (-1)3 – 2.(-1)2 + 5. (-1) – 10 = -1 – 2.1 – 5 + 10 = 2
c) G(x) + H(x) = (– 2x3 + 3x2 – 8x – 1) + (x3 – 2x2 + 5x – 10)
= -2x3 + 3x2 – 8x – 1 + x3 – 2x2 + 5x – 10
= (-2x3 + x3) + (3x2 – 2x2) + ( – 8x + 5x ) – (10+1)
= -x3 + x2 – 3x – 11
G(x) – H(x) = (– 2x3 + 3x2 – 8x – 1) – (x3 – 2x2 + 5x – 10)
= – 2x3 + 3x2 – 8x – 1 – x3 + 2x2 – 5x + 10
= (-2x3 – x3) + (3x2 + 2x2) – (8x + 5x) + (-1+ 10)
= -3x3 + 5x2 – 13x + 9
SM
4
IM
30 tháng 7 2016
\(f\left(x\right)=4x^4+3x-1-\left(3x^2-2x-3\right)=4x^4-3x^2+5x+2\)
\(\left\{{}\begin{matrix}h\left(x\right)+g\left(x\right)=x^2+2x+1\\h\left(x\right)-g\left(x\right)=x^2+2x-3\end{matrix}\right.\)
Lấy pt(1) cộng pt(2) ta được:
\(h\left(x\right)+g\left(x\right)+h\left(x\right)-g\left(x\right)=x^2+2x+1+x^2+2x-3\)
\(2h\left(x\right)=2x^2+4x-2\)
\(h\left(x\right)=x^2+2x-1\)
Thanks