Tính l i m x → - 3 + 2 x 2 + 5 x - 3 x + 3 2
A. + ∞
B. - ∞
C. 0
D. – 7
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bài 2
A = 3+3^2 +3^3+ ...+3^100
3.A = 3^2+3^3+3^4+...+3^101
3.A-A=(3^2+3^3+3^4+...+3^101)-(3+3^2+3^3+...+3^100)
2.A=3^101-3
Ta có: 2A+3=3^ x
\(\Rightarrow\)(3^101-3)+3=3^x
\(\Rightarrow\)3^101-(3+3)=3^x
\(\Rightarrow\)3^101=3^x
\(\Rightarrow\)x=101
Vậy x=101
h: =>7x-21-15+5x=11x-5
=>12x-36=11x-5
hay x=31
l: (x+3)(x-4)=0
=>x+3=0 hoặc x-4=0
=>x=-3 hoặc x=4
i: =>5x-30-2x-6=12
=>3x-36=12
hay x=16
m: =>119+27x=8x52
=>27x=297
hay x=11
Nhầm 1 chút nhé, Bài 1 câu a) ( x^2 - 5x ) . ( x^2 - x +3) - ( x^3 - x^2 +1) . ( 2x - 1 )
Bài 1:
a) \(\left(m+2\right).3-5=4\)
\(\Leftrightarrow3m+6-5=4\)
\(\Leftrightarrow3m+1=4\)
\(\Leftrightarrow3m=4-1\)
\(\Leftrightarrow3m=3\)
\(\Leftrightarrow m=1\)
Vậy: m = 1
b) \(\left(m-3\right).\left(-2\right)+8=-10\)
\(\Leftrightarrow-2m+6+8=-10\)
\(\Leftrightarrow-2m+14=-10\)
\(\Leftrightarrow-2m=-10-14\)
\(\Leftrightarrow-2m=-24\)
\(\Leftrightarrow m=12\)
Vậy: m = 12
Bài 2:
a) \(\left(x-2\right)^2=9\)
\(\Leftrightarrow\left(x-2\right)^2=3^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
b) \(\left(x+3\right)^2-0,16=0\)
\(\Leftrightarrow\left(x+3\right)^2=0,16\)
\(\Leftrightarrow\left(x+3\right)^2=\left(0,4\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0,4\\x+3=-0,4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2,6\\x=-3,4\end{matrix}\right.\)
c) \(x^3=25x\)
\(\Leftrightarrow x^3-25x=0\)
\(\Leftrightarrow x\left(x^2-25\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-25=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm5\end{matrix}\right.\)
bài 1:
\(5^x+5^{x+1}+5^{x+2}+5^{x+3}+3900=0\)
=> \(5^x+5^{x+1}+5^{x+2}+5^{x+3}=-3900\)
=> \(5^x(5^1+5^2+5^3)=-3900\)
=> \(5^x.155=-3900\)
=> \(5^x=-3900:155\)
=> \(5^x\approx-25\)
=> \(5^x=-\left(5^2\right)\)
=> x=2
bài 2:
A= \(2+2^2+2^3+.....+2^{2018}\)
=> 2A= \(\left(1+2+2^2+....+2^{2019}\right)\)
=> 2A-A= \(\left(1+2+2^2+....+2^{2018}\right)\) -( \(2+2^2+2^3+.....+2^{2019}\))
=> 2A-A= \(\left(1+2+2^2+....+2^{2018}\right)\)+ \(2-2^2-2^3-.....-2^{2019}\)
=> A= 1- \(2^{2019}\)
Thay A= 1- \(2^{2019}\) vào ta được
1-\(2^{2019}\) +2 =\(2^x\)
=> 1-(1+1).\(2^{2020}\) =\(2^x\)
=> -1. \(2^{2020}\) = \(2^x\)
=> -(\(2^{2020}\)) =\(2^x\)
=> x= 2010
a) Ta có: \(\frac{x^3-3x^2+x-3}{x-3}\)
\(=\frac{x^2\left(x-3\right)+\left(x-3\right)}{\left(x-3\right)}=\frac{\left(x-3\right)\left(x^2+1\right)}{x-3}=x^2+1\)
b) Ta có: \(\frac{x^2+2x+x^2-4}{x+2}\)
\(=\frac{x\left(x+2\right)+\left(x+2\right)\left(x-2\right)}{x+2}=\frac{\left(x+2\right)\left(x+x-2\right)}{x+2}=2x-2\)
c) Ta có: \(\frac{2x^3-5x^2+6x-15}{2x-5}\)
\(=\frac{x^2\left(2x-5\right)+3\left(2x-5\right)}{2x-5}=\frac{\left(2x-5\right)\left(x^2+3\right)}{2x-5}=x^2+3\)
Tính nhanh mỗi biểu thức sau:
a, 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20
= (0 + 20) + (1 + 19) + (2 + 18) + (3 + 17) + (4 + 16) + (5 + 15) + (6 + 14) + (7 + 13) + (8 + 12) + (9 + 11) + 10
= 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 10
= 20 x 10 + 10
= 200 + 10
= 210
b, 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x (4 x 9 - 36)
= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x (36 - 36)
= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 0
= A x 0
= 0
c, (81 - 7 x 9 - 18) : (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
= (81 - 63 - 18) : (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
= (18 - 18) : (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
= 0 :(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
= 0 : A
= 0
d, (6 x 5 + 7 - 37) x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)
= (30 + 7 - 37) x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)
= (37 - 37) x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)
= 0 x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)
= 0 x A
= 0
e, (11 x 9 - 100 + 1) : (1 x 2 x 3 x 4 x ... x 10)
= (99 - 100 + 1) : (1 x 2 x 3 x 4 x ... x 10)
= (99 + 1 - 100) : (1 x 2 x 3 x 4 x ... x 10)
= (100 - 100) : (1 x 2 x 3 x 4 x ... x 10)
= 0 : (1 x 2 x 3 x 4 x ... x 10)
= 0 : A
= 0
g, (m : 1 - m x 1) : (m x 2008 + m x 2008)
= (m - m) : (m x 2008 + m x 2008)
= 0 : (m x 2008 + m x 2008)
= 0 : A
= 0
h, (2 + 4 + 6 + 8 + m x n) x (324 x 3 - 972)
= (2 + 4 + 6 + 8 + m x n) x (972 - 972)
= (2 + 4 + 6 + 8 + m x n) x 0
= A x 0
= 0
l, (1 + 2 + 3 + ... + 99) x (13 x 15 - 12 x 15 - 15)
= (1 + 2 + 3 + ... + 99) x (15 x (13 - 12 - 1))
= (1 + 2 + 3 + ... + 99) x (15 x 0)
= (1 + 2 + 3 + ... + 99) x 0
= A x 0
= 0
i, (0 x 1 x 2 x...x 99 x 100) : (2 + 4 + 6 +...+ 98)
= 0 x : (2 + 4 + 6 +...+ 98)
= 0 x A
= 0
k, (0 + 1 + 2 +...+ 97 + 99) x (45 x 3 - 45 x 2 - 45)
= (0 + 1 + 2 +...+ 97 + 99) x (45 x (3 - 2 - 4))
= (0 + 1 + 2 +...+ 97 + 99) x (45 x 0)
= (0 + 1 + 2 +...+ 97 + 99) x 0
= A x 0
= 0
L = lim x → − 3 + 2 x 2 + 5 x − 3 x + 3 2 = lim x → − 3 + 2 x − 1 x + 3 x + 3 2 = lim x → − 3 + 2 x − 1 x + 3
Ta có x → − 3 + ⇒ x > − 3 ⇒ x + 3 > 0 ⇒ lim x → − 3 + x + 3 = 0 , và lim x → − 3 + 2 x − 1 = − 7.
Kết luận L = − ∞
Chọn đáp án B.