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\(A=\dfrac{1}{5}\left(\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{496}-\dfrac{1}{501}\right)\)
\(=\dfrac{1}{5}\cdot\dfrac{55}{334}=\dfrac{11}{334}\)
\(B=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}=\dfrac{20}{21}\)
Bài 1
a, -25 . 63 - 25 . 3
= 25 . (-63) - 25 . 3
= 25 . [(-63) - 3]
= 25 . (-66) = -1650
Bài 2
c, (x + 1)2 = 16
=> (x + 1)2 = 42
=> x + 1 = 4
=> x = 3
d, (-38) - (x - 2) = -16
=> (x - 2) = -38 - (-16)
=> x - 2 = -38 + 16 = -22
=> x = 2 + (-22)
=> x = -20
a) 2^x . 16^2 = 1024 b) 64 . 4^x = 16^8 c) 2^x = 16
=> 2^x . 256 = 1024 => 64 . 4^x = (4^2) ^ 8 => 2^x = 2^4
=> 2^x = 1024 : 256 => 4^3 . 4^x = 4^16 => x = 4
=> 2^x = 4 => 4^x = 4^16 : 4^3
=> 2^x = 2^2 => 4^x = 4^13
=> x = 13
=> x = 2
a) \(2^x.16^2=1024\Rightarrow2^x=1024:16^2=2^{10}:\left(2^4\right)^2=2^{10}:2^8=2^2\)\(\Rightarrow x=2\)
b) \(64.4^x=16^8\Rightarrow4^x=16^8:64=\left(4^2\right)^8:4^3=4^{16}:4^3=4^{13}\Rightarrow x=13\)
c)\(2^x=16\Rightarrow2^x=2^4\Rightarrow x=4\)
\(a,\frac{1}{2}x+\frac{2}{3}\left(x-2\right)=\frac{1}{3}\)
\(\frac{1}{2}x+\frac{2}{3}x-\frac{4}{3}=\frac{1}{3}\)
\(\frac{7}{6}x=\frac{5}{3}\)
\(x=\frac{10}{7}\)
\(b,16^{x-1}:16=4^4\)
\(16^{x-1}=4096\)
\(16^{x-1}=16^3\)
\(\Rightarrow x-1=3\)
\(x=4\)
=.= hk tốt!!
a) \(\frac{1}{2}x+\frac{2}{3}\left(x-2\right)=\frac{1}{3}\)
<=>\(\frac{x}{2}+\frac{2x}{3}-\frac{4}{3}-\frac{1}{3}=0\)<=>\(\frac{7x}{6}-\frac{5}{3}=0\)=>x=\(\frac{10}{7}\)
b)16x-1:16=256 => 16x-1=4096=163
T thấy x-1=3 =>x=2
Chúc bạn học tốt
a. ( x+1)^4 =16
=> x+1 = 2
=> x=1
b. ( x-1) ^4=16
=> x-1=2
=> x= 3
a. ( x+1) ^4=16
=> x+1=-2
=> x= -3
b. ( x-1)^4= 16
=> x-1=-2
=> x=-1
cái này dễ nà!
ta có:
5x + 2 ⋮ x + 1
=> (5x+5) - 5 + 2 ⋮ x + 1
=> (5x+5.1) - 3 ⋮ x + 1
=> 5(x+1) - 3 ⋮ x + 1
có x+1 ⋮ x+1 => 5 (x+1) ⋮ x + 1
=> - 3 ⋮ x + 1
=> x + 1 ∈ Ư(-3)
x ∈ Z => x + 1 ∈ Z
=> x + 1 ∈ {-1;-3;1;3}
=> x ∈ {-2;-4;0;2}
vậy____
\(5x+2\)\(⋮\)\(x+1\)
\(\Leftrightarrow\)\(5\left(x+1\right)-3\)\(⋮\)\(x+1\)
Ta thấy \(5\left(x+1\right)\)\(⋮\)\(x+1\)
\(\Leftrightarrow\)\(3\)\(⋮\)\(x+1\)
\(\Rightarrow\)\(x+1\)\(\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\)\(x=\left\{-4;-2;0;2\right\}\)
(x + 1)2 = 16
x+1 = + -4
x = 3; -5
x=3