1+1+1+1x2+2=......=..........
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ĐKXĐ: x ≠ 1 hoặc x = -1.
Ta có:
⇔ 8x = - 10 ⇔ x = - 5/4.
Vậy phương trình đã cho có nghiệm là x = - 5/4.
ĐKXĐ: x ≠ 1 hoặc x = -1.
Ta có:
⇔ 8x = - 10 ⇔ x = - 5/4.
Vậy phương trình đã cho có nghiệm là x = - 5/4.
a) Ta có P = 4 x 2 ( x − 2 y ) 2 ( x + 2 y ) 2 . ( x + 2 y ) 2 16 x = x 4 ( x − 2 y ) 2
Với x ≠ 0 , x ≠ ± 2 y
b) Ta có Q = 16 x ( x 2 − 16 ) 2 . x 2 − 16 2 x = 8 16 − x 2 với x ≠ 0 , x ≠ ± 4
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{24\times25}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(=1-\dfrac{1}{25}\)
\(=\dfrac{24}{25}\)
\(\dfrac{1}{1\times2}\) + \(\dfrac{1}{2\times3}\) + ...+ \(\dfrac{1}{x\times\left(x+1\right)}\) = \(\dfrac{1}{2}\)
\(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) +...+ \(\dfrac{1}{x}\) - \(\dfrac{1}{x+1}\) = \(\dfrac{1}{2}\)
1 - \(\dfrac{1}{x+1}\) = \(\dfrac{1}{2}\)
\(\dfrac{1}{x+1}\) = 1 - \(\dfrac{1}{2}\)
\(\dfrac{1}{x+1}\) = \(\dfrac{1}{2}\)
\(x+1\) = 1 : \(\dfrac{1}{2}\)
\(x\) + 1 = 2
\(x\) = 2 - 1
\(x\) = 1
Đáp án là B
Ta có:
C = lim x → 1 x − 1 x + 1 − m x − 1 x − 1 x + 1 = lim x → 1 x + 1 − m x + 1 = 2 − m 2
mà C = 2 ⇒ m = − 2.
a) (2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x(2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x
=4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x
=8x.5(2x+1)(2x−1)(2x+1).4x=102x−18x.5(2x+1)(2x−1)(2x+1).4x=102x−1
b) (1x2+x−2−xx+1):(1x+x−2)(1x2+x−2−xx+1):(1x+x−2)
=(1x(x+1)+x−2x+1):1+x2−2xx(1x(x+1)+x−2x+1):1+x2−2xx
=1+x(x−2)x(x+1).xx2−2x+11+x(x−2)x(x+1).xx2−2x+1
=(x2−2x+1)xx(x+1)(x2−2x+1)=1x+1(x2−2x+1)xx(x+1)(x2−2x+1)=1x+1
c) 1x−1−x3−xx2+1.(1x2−2x+1+11−x2)1x−1−x3−xx2+1.(1x2−2x+1+11−x2)
=1x−1−x3−xx2+1.[1(x−1)2−1(x−1)(x+1)]
a) (2x+12x−1−2x−12x+1):4x10x−5(2x+12x−1−2x−12x+1):4x10x−5
= 0 - 0
= 0
b) (1x2+x−2−xx+1):(1x+x−2);(1x2+x−2−xx+1):(1x+x−2)
= (x-xx+1) : (2x-2) : (x-xx+1) : (2x-2)
c) 1x−1−x3−xx2+1.(1x2−2x+1+11−x2)
= -2x-1-xx2+1. (14 - 4x)
= -x2-1-xx2+14-4x
= -6x-xx2+13
1/1.2 + 1/2.3 + .... + 1/x.(x+1) = 1/2
1 - 1/2 + 1/2 - 1/3 + ... + 1/x + 1/x+1 = 1/2
1 - 1/x+1 = 1/2
1/x+1 = 1 - 1/2
1/x+1 = 1/2
=> x + 1 = 2
x = 2 - 1
x = 1
là =8 nha bạn 1000000% lun đó k mình nha bạn
1 + 1 + 1 + 1 x 2 + 2 = 1 + 1 + 1 + 2 + 2 = 7
Duyệt nhé