Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a,
\(\left(4x-\dfrac{1}{3}\right)^6=1\\ \Rightarrow\left[{}\begin{matrix}4x-\dfrac{1}{3}=1\\4x-\dfrac{1}{3}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x=\dfrac{4}{3}\\4x=\dfrac{-2}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{-1}{6}\end{matrix}\right.\)
b,
\(\left(5x-\dfrac{2}{3}\right)^2=0\\ \Rightarrow5x-\dfrac{2}{3}=0\\ 5x=\dfrac{2}{3}\\ x=\dfrac{2}{15}\)
c,
\(\left(\dfrac{1}{3}x-\dfrac{1}{2}\right)^3=-8\\ \Rightarrow\dfrac{1}{3}x-\dfrac{1}{2}=-2\\ \dfrac{1}{3}x=\dfrac{-3}{2}\\ x=\dfrac{-9}{2}\)
d,
\(\dfrac{81}{3^n}=3\\ \Leftrightarrow3^4:3^n=3^1\\\Leftrightarrow3^{4-n}=3^1 \\ \Rightarrow n=3\)
e,
\(\dfrac{\left(-2\right)^x}{64}=-2\\ \Leftrightarrow\left(-2\right)^x:\left(-2\right)^6=\left(-2\right)^1\\ \Leftrightarrow\left(-2\right)^{x-6}=\left(-2\right)^1\\ \Rightarrow x=7\)
f,
\(\left(-20\right)^n:10^n=16\\ \left[\left(-20\right):10\right]^n=16\\ \left(-2\right)^n=\left(-2\right)^4\\ \Rightarrow n=4\)
Bài 1:
a) \(\left(4x-\dfrac{1}{3}\right)^6=1\)
\(\Rightarrow4x-\dfrac{1}{3}=1\)
\(4x=1+\dfrac{1}{3}\)
\(4x=\dfrac{4}{3}\)
\(x=\dfrac{4}{3}:4\)
\(x=\dfrac{1}{3}\)
b) \(\left(5x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow5x-\dfrac{2}{3}=0\)
\(5x=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}:5\)
\(x=\dfrac{2}{15}\)
c) \(\left(\dfrac{1}{3}x-\dfrac{1}{2}\right)^3=-8\)
\(\Rightarrow\left(\dfrac{1}{3}x-\dfrac{1}{2}\right)^3=\left(-2\right)^3\)
\(\dfrac{1}{3}x-\dfrac{1}{2}=-2\)
\(\dfrac{1}{3}x=-2+\dfrac{1}{2}\)
\(\dfrac{1}{3}x=\dfrac{-3}{2}\)
\(x=\dfrac{-3}{2}:\dfrac{1}{3}\)
\(x=\dfrac{-9}{2}\)
d) \(\dfrac{81}{3^n}=3\)
\(\Rightarrow\dfrac{3^4}{3^n}=3\)
\(\Rightarrow3^n.3=3^4\)
\(3^{n+1}=3^4\)
n + 1 = 4
n = 4 - 1
n = 3
e) \(\dfrac{\left(-2\right)^x}{64}=-2\)
\(\Rightarrow\dfrac{\left(-2\right)^x}{\left(-2\right)^6}=-2\)
\(\Rightarrow\left(-2\right)^x=\left(-2\right)^6.\left(-2\right)\)
\(\left(-2\right)^x=\left(-2\right)^7\)
x = 7
f) (-20)n : 10n = 16
(-20 : 10)n = 16
(-2)n = 16
(-2)n = (-2)4
n = 4.
1. A = \(\dfrac{3n-7}{n-1}=\dfrac{3n-3}{n-1}+\dfrac{-7}{n-1}=3+\dfrac{-7}{n-1}\)
Tại giá trị \(A\notin Z,3\in Z\)\(\Rightarrow\dfrac{-7}{n-1}\in Z\)\(\Rightarrow n-1\inƯ\left(-7\right)\) với \(x\ne1\) (mẫu sẽ có giá trị là 0 nếu x = 1)
Tại \(n-1=7\)\(\Leftrightarrow n=7+1=8\)
Tại \(n-1=-7\Leftrightarrow n=-7+1=-6\)
Tại \(n-1=1\Leftrightarrow n=1+1=2\)
Tại \(n-1=-1\Leftrightarrow n=-1+1=0\)
2. B = \(\dfrac{4n+1}{2n-3}=\dfrac{4n+6}{2n-3}+\dfrac{-5}{2n-3}=2+\dfrac{-5}{2n-3}\)
Tại giá trị \(B\in Z,2\in Z\)\(\Rightarrow\dfrac{-5}{2n-3}\in Z\)\(\Rightarrow2n-3\inƯ\left(-5\right)\) với \(x\ne\dfrac{3}{2}\)
Tại \(2n-3=5\Leftrightarrow2n=8\Leftrightarrow n=4\)
Tại \(2n-3=-5\Leftrightarrow2n=-2\Leftrightarrow n=-1\)
Tại \(2n-3=1\Leftrightarrow2n=4\Leftrightarrow n=2\)
Tại \(2n-3=-1\Leftrightarrow2n=2\Leftrightarrow n=1\)
Lời giải:
a) \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{n-1}-1\right)\left(\frac{1}{n}-1\right)\)
\(=\frac{1-2}{2}.\frac{1-3}{3}.\frac{1-4}{4}...\frac{-(n-2)}{n-1}.\frac{-(n-1)}{n}\)
\(=\frac{(-1)(-2)(-3)...[-(n-2)][-(n-1)]}{2.3.4...(n-1)n}\)
\(=\frac{(-1)^{n-1}(1.2.3....(n-2)(n-1))}{2.3.4...(n-1)n}=(-1)^{n-1}.\frac{1}{n}\)
b) \(B=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{n^2}-1\right)\)
\(=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.....\frac{1-n^2}{n^2}\)
\(=\frac{(-1)(2^2-1)}{2^2}.\frac{(-1)(3^2-1)}{3^2}....\frac{(-1)(n^2-1)}{n^2}\)
\(=(-1)^{n-1}.\frac{(2^2-1)(3^2-1)...(n^2-1)}{2^2.3^2....n^2}\)
\(=(-1)^{n-1}.\frac{(2-1)(2+1)(3-1)(3+1)...(n-1)(n+1)}{2^2.3^2....n^2}\)
\(=(-1)^{n-1}.\frac{(2-1)(3-1)...(n-1)}{2.3...n}.\frac{(2+1)(3+1)...(n+1)}{2.3...n}\)
\(=(-1)^{n-1}.\frac{1.2.3...(n-1)}{2.3...n}.\frac{3.4...(n+1)}{2.3.4...n}\)
\(=(-1)^{n-1}.\frac{1}{n}.\frac{n+1}{2}=(-1)^{n-1}.\frac{n+1}{2n}\)
\(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}\)
\(A=\left(-1\right)^{2n+n+n+1}\)
\(A=\left(-1\right)^{4n+1}\)
\(B=\left(10000-1^2\right).\left(10000-2^2\right)...\left(10000-1000^2\right)\)
\(B=\left(10000-1^2\right)\left(10000-2^2\right)...\left(10000-100^2\right)...\left(10000-1000^2\right)\)
\(B=\left(10000-1^2\right)\left(10000-2^2\right)...\left(10000-10000\right)...\left(10000-1000^2\right)\)
\(B=\left(10000-1^2\right)\left(10000-2^2\right)...0\left(10000-1000^2\right)\)
\(B=0\)
\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{5^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...0....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(C=0\)
\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-10^3\right)}\)
\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-1000\right)}\)
\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...0}\)
\(D=1999^0\)
\(D=1\)
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+....+\dfrac{1}{18.19.20}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{4}-\dfrac{1}{2.19.20}< \dfrac{1}{4}\)
Cái B TT nhé
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\\ =1-\dfrac{1}{n}< 1\)
D TT
E mk thấy nó ss ớ
a) ( x + 5 )3 = -64
x + 5 = - 4
x = - 4 - 5
x = -9
b) (2x - 3)2=9
2x - 3 = 3
2x = 3+3
2x = 6
x = 6 : 2
x = 3
e) \(\dfrac{8}{2x}=4\)
=> 4 . 2x = 8
8x =8
x = 8 : 8
x = 1
g) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\)
\(\left(\dfrac{1}{2}\right)^{2x}:\left(\dfrac{1}{2}\right)^1=\dfrac{1}{8}\)
\(\left(\dfrac{1}{2}\right)^{2x}:\dfrac{1}{2}=\dfrac{1}{8}\)
\(\left(\dfrac{1}{2}\right)^{2x}=\dfrac{1}{8}.\dfrac{1}{2}\)
\(\left(\dfrac{1}{2}\right)^{2x}=\dfrac{1}{16}\)
\(\left(\dfrac{1}{2}\right)^{2x}=\left(\dfrac{1}{2}\right)^{2.2}\)
=> x = 2
h) \(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\)
\(\dfrac{1}{4}.x=\dfrac{1}{32}\)
x = \(\dfrac{1}{32}:\dfrac{1}{4}\)
x = \(\dfrac{1}{8}\)
i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\)
\(x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)\)
\(x=\dfrac{-1}{27}\)
a) (x + 5)3 = -64
=> (x + 5)3 = (-4)3
x + 5 = -4
x = -4 - 5
x = -9
b) (2x - 3)2 = 9
=> (2x - 3)2 = (\(\pm\)3)2
=> 2x - 3 = 3 hoặc 2x - 3 = -3
*2x - 3 = 3
2x = 3 + 3
2x = 9
x = \(\dfrac{9}{2}\)
*2x - 3 = -3
2x = -3 + 3
2x = 0
x = 0 : 2
x = 0
Vậy x \(\in\left\{\dfrac{9}{2};0\right\}\)
c) \(\dfrac{x}{\dfrac{4}{2}}=\dfrac{4}{\dfrac{x}{2}}\)
=> \(x.\dfrac{x}{2}=4.\dfrac{4}{2}\)
\(\dfrac{x}{2}=8\)
x = 8 : 2
x = 4
d) \(\dfrac{-32}{\left(-2\right)^n}=4\)
\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)
=> (-2)n . (-2)2= (-2)5
(-2)n = (-2)5 : (-2)2
(-2)n = (-2)3
Vậy n = 3
e) \(\dfrac{8}{2x}=4\)
=> 2x . 4 = 8
2x = 8 : 4
2x = 2
x = 1
g) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^3\)
2x - 1 = 3
2x = 3 + 1
2x = 4
x = 4 : 2
x = 2
h) \(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\)
\(x=\left(\dfrac{1}{2}\right)^5:\left(\dfrac{1}{2}\right)^2\)
\(x=\left(\dfrac{1}{2}\right)^3\)
\(x=\dfrac{1}{8}\)
i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\)
\(x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)\)
\(x=\left(\dfrac{-1}{3}\right)^4:\left(\dfrac{-1}{3}\right)\)
\(x=\left(\dfrac{-1}{3}\right)^3\)
\(x=\dfrac{-1}{27}\).