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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 !
y = f(x) = 2x - x2
f(1) = 2.1 - 12 = 2 - 1 = 1
y = g(x) = x - 7 + 3
g(2) = 2 - 7 + 3 = -2
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,f(x)-g(x)+h(x)=2x-`1
b,đặt S(x)=f(x)-g(x)+h(x)
S(x)=0<=>2x+1=0=>x=\(\dfrac{-1}{2}\)
câu 4: b, đề bài là tính giá trị của A tại x =-1/2;y=-1
Tk
Bài 2
a) F(x)-G(x)+H(x)= \(x^3-2x^2+3x+1-\left(x^3+x-1\right)+\left(2x^2-1\right)\)
= \(x^3-2x^2+3x+1-x^3-x+1+2x^2-1\)
= \(x^3-x^3-2x^2+2x^2+3x-x+1+1-1\)
= 2x + 1
b) 2x + 1 = 0
2x = -1
x=\(\dfrac{-1}{2}\)
