5x2 – 20x
Tìm GTLN của biểu thức:
K= -5x2+20x-2021
Lời giải:
$K=-5x^2+20x-2021=-2001-5(x^2-4x+4)=-2001-5(x-2)^2$
Vì $(x-2)^2\geq 0, \forall x\in\mathbb{R}$
$\Rightarrow K=-2001-5(x-2)^2\leq -2001$
Vậy $K_{\max}=-2001$ khi $(x-2)^2=0\Leftrightarrow x=2$
Ta có: \(K=-5x^2+20x-2021\)
\(=-5\left(x^2-4x+\dfrac{2021}{5}\right)\)
\(=-5\left(x^2-4x+4+\dfrac{2001}{5}\right)\)
\(=-5\left(x-2\right)^2-2001\le-2001\forall x\)
Dấu '=' xảy ra khi x=2
Đa thức 5x2 – 20x = 5x.(x – …). Chỗ trống cần điền đơn thức thích hợp là
A. 4
B. 5
C. 10
D. 15
Cho f(x) là đa thức thỏa mãn lim x → 2 f ( x ) - 20 x - 2 = 10 . Tính lim x → 2 6 f ( x ) + 5 3 - 5 x 2 + x + 6
A. T = 12 25
B. T = 4 25
C. T = 5 25
D. T = 6 25
Đặt
Vì lim x → 2 f ( x ) - 20 x - 2 = 10 nên f( x) -20 =0 hay f( x) = 20 nên P =5
Khi đó
Suy ra
T= lim x → 2 f ( x ) - 20 x - 2 . lim x → 2 6 ( x - 3 ) ( P 2 + 5 P + 25 ) = 10 . 6 5 . 75 = 4 25
Chọn B.
M= -5x2+20x+17
mọi ng giải giùm em
em thanks ạ
bài 3 phân tích đa thức sau thành nhân tử
a 4x2 -16 + (3x +12) (4-2x)
b x3 + X2Y -15x -15y
c 3(x+8) -x2 -8x
d x3 -3x2 + 1 -3x
e 5x2 -5y2 -20x + 20y
kkk =0)
a) \(4x^2-16+\left(3x+12\right)\left(4-2x\right)\)
\(=\left(2x-4\right)\left(2x+4\right)-3\left(x+4\right)\left(2x-4\right)\)
\(=\left(2x-4\right)\left(2x+4-3x-12\right)\)
\(=-\left(2x-4\right)\left(x+8\right)\)
b) \(x^3+x^2y-15x-15y\)
\(=x^2\left(x+y\right)-15\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-15\right)\)
c) \(3\left(x+8\right)-x^2-8x\)
\(=3\left(x+8\right)-x\left(x+8\right)\)
\(=\left(x+8\right)\left(3-x\right)\)
d) \(x^3-3x^2+1-3x\)
\(=x^3+1-3x^2-3x\)
\(=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
d) \(5x^2-5y^2-20x+20y\)
\(=5\left(x^2-y^2\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y-4\right)\)
(x−1)(5x2−3x+2)=x(5x2−3x+2)−1(5x2−3x+2)
(x−1)(5x2−3x+2)=x(5x2−3x+2)−1(5x2−3x+2)
=x.5x2+x.(−3x)+x.2+(−1).5x2+(−1)(−3x)+(−1).2=x.5x^2+x.\left(-3x\right)+x.2+\left(-1\right).5x^2+\left(-1\right)\left(-3x\right)+\left(-1\right).2=x.5x2+x.(−3x)+x.2+(−1).5x2+(−1)(−3x)+(−1).2
=5x3−3x2+2x−5x2+3x−2=5x^3-3x^2+2x-5x^2+3x-2=5x3−3x2+2x−5x2+3x−2
=5x3−8x2+5x−2=5x^3-8x^2+5x-2=5x3−8x2+5x−2.
(x−1)(5x2−3x+2)=x(5x2−3x+2)−1(5x2−3x+2)
=x.5x2+x.(−3x)+x.2+(−1).5x2+(−1)(−3x)
=5x3−3x2+2x−5x2+3x−2=5x^3-3x^2+2x-5x^2+3x-2=5x3−3x2+2x−5x2+3x−2
=5x3−8x2+5x−2=5x^3-8x^2+5x-2=5x3−8x2+5x−2.
(5+5x2)+5+5x2 = ?
( 5+5 x 2 ) +5 +5 x2= 30
Học tốt !
( 5 + 5 x 2 ) + 5 + 5 x 2
= 15 + 5 + 5 x 2
= 15 + 5 + 10
= 30
\(\left(5+5\times2\right)+5+5\times\)2
\(=15+5+5\times2\)
\(=15+5+10\)
\(=20+10\)
\(=30\)
#z
B=x^6-20x^5-20x^4-20x^3-20x^2-20x+3 tại x=21
Thay20=x-1 vào B
Ta có \(x=21\Rightarrow x-1=20\)
biểu thức B có dạng :
\(B=x^6-\left(x-1\right)x^5-\left(x-1\right)x^4-\left(x-1\right)x^3-\left(x-1\right)x^2-\left(x-1\right)x+3\)
\(=x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x+3=x+3\)
Vậy \(B=21+3=24\)
Rút gọn
A= \(x^{10}+20x^9+20x^8+20x^7+...+20x^3+20x^2+20x\)
Với \(x=-24\)