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a) \(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\left(5x+1\right)^2=\left(\pm\dfrac{6}{9}\right)\)\(^2\)
\(5x+1=\pm\dfrac{6}{9}\)
+) \(5x+1=\dfrac{6}{9}\)
\(5x=\dfrac{6}{9}-1=\dfrac{6}{9}-\dfrac{9}{9}\)
\(5x=\dfrac{-5}{9}\)
\(x=\dfrac{-5}{9}:5=\dfrac{-1}{45}\)
+) \(5x+1=\dfrac{-6}{9}\)
\(5x=\dfrac{-6}{9}-1=\dfrac{-6}{9}-\dfrac{9}{9}\)
\(5x=\dfrac{-5}{3}\)
\(x=\dfrac{-5}{3}:5=\dfrac{-5}{15}\)
vậy \(x\in\left\{\dfrac{-5}{15};\dfrac{-1}{45}\right\}\)
a,
- Theo đề bài ta có:
(8x-1)2n-1 = 52n-1
=> 8x-1 = 5
8x = 6
x = \(\dfrac{6}{8}\)= \(\dfrac{3}{4}\)
- Vậy x = \(\dfrac{3}{4}\)
b,
- Ta có:
(x - 7)x+1 - (x - 7)x+11 = 0
(x - 7)x . (x - 7) - (x - 7)x . (x - 7)11 = 0
(x - 7)x . [(x - 7) - (x - 7)11] = 0
=> (x - 7)x = 0 hoặc [(x - 7) - (x - 7)11] = 0
- TH1: (x - 7)x = 0
=> x - 7 = 0
=> x = 7
- TH2:
[(x - 7) - (x - 7)11] = 0
=> x - 7 = (x -7)11
=> x - 7 = 1 hoặc x - 7 = 0
+ Nếu x - 7 = 1
x = 8
+ Nếu x - 7 = 0 (TH1)
- Vậy x = 7 hoặc x = 8
c, - Theo đề bài ta có:
\(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
- Thấy \(\left(\dfrac{2}{3}\right)^6=\left(\dfrac{2}{3}\right)^{2\cdot3}\)= \(\left(\dfrac{4}{9}\right)^3\)
=> \(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{4}{9}\right)^3\)
=> \(x-\dfrac{2}{9}=\dfrac{4}{9}\)
=> \(x=\dfrac{4}{9}-\dfrac{2}{9}\)
\(x=\dfrac{2}{9}\)
- Vậy \(x=\dfrac{2}{9}\)
\(f\left(x\right)-g\left(x\right)\)
\(=x^{2n}-x^{2n-1}+...+x^2-x+1+x^{2n+1}-x^{2n}+x^{2n-1}-...-x^2+x-1\)
\(=x^{2n+1}\)
\(=\left(\dfrac{1}{10}\right)^{2n+1}=\dfrac{1}{10^{2n+1}}\)
Ta có:
\(f\left(x\right)-g\left(x\right)=\left(x^{2n}-x^{2n-1}+...+x^2-x+1\right)-\left(-x^{2n+1}+x^{2n}-x^{2n-1}+...+x^2-x+1\right)\)
\(=x^{2n}-x^{2n-1}+...+x^2-x+1+x^{2n+1}-x^{2n}+x^{2n-1}-...-x^2+x-1=x^{2n+1}\)
\(\Rightarrow f\left(\dfrac{1}{10}\right)-g\left(\dfrac{1}{10}\right)=\left(\dfrac{1}{10}\right)^{2n+1}\)
Vậy \(f\left(\dfrac{1}{10}\right)-g\left(\dfrac{1}{10}\right)=\left(\dfrac{1}{10}\right)^{2n+1}\)
a. \(\left(\frac{-1}{5}\right)^n=\frac{-1}{125}\)
<=> \(\left(\frac{-1}{5}\right)^n=\left(\frac{-1}{5}\right)^3\)
<=> n = 3
b. \(\left(\frac{-2}{11}\right)^m=\frac{4}{121}\)
<=> \(\left(\frac{-2}{11}\right)^m=\left(\frac{2}{11}\right)^2\)
<=> m = 2
c. 72n + 72n+2 = 2450
<=> 72n + 72n . 72 = 2450
<=> 72n.(1+72) = 2450
<=> 72n = 72
<=> 2n = 2
<=> n = 1
a: =>5x+1=6/7 hoặc 5x+1=-6/7
=>5x=-1/7 hoặc 5x=-13/7
=>x=-1/35 hoặc x=-13/35
b: =>x-2/9=4/9
=>x=6/9=2/3
c: =>8x+1=5
=>8x=4
hay x=1/2
a.\(2n^2-3n+1=2n\times\left(n-1\right)-\left(n-1\right)=\left(2n-1\right)\times\left(n-1\right)\Rightarrow2n-1⋮n-1\)
\(\Rightarrow2\left(n-1\right)+1⋮n-1\Rightarrow1⋮n-1\Rightarrow n-1\inƯ\left(1\right)=\left\{1\right\}\Rightarrow n=2\)
b.Tách tương tự nha
\(2n^2-3n+1=\left(2n^2-2n\right)-n+1=2n\left(n-1\right)-n+1\)\(\Rightarrow-n+1⋮n-1\Rightarrow-\left(n-1\right)⋮n-1\)
vậy với mọi x thuộc N đều t/m
b) tương tự nha