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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 \(x\in\left\{\dfrac{-5}{15};\dfrac{-1}{45}\right\}\)
\(\left(3x^2-51\right)^{2n}=\left(-24\right)^{2n}\)
\(\Rightarrow3x^2-51\in\left\{-24;24\right\}\)
+) \(3x^2-51=-24\)
=> 3x2 = -24 + 51
=> 3x2 = 27
=> x2 = 27 : 3
=> x2 = 9 = 32 = (-3)2
=> x \(\in\){-3; 3}.
+) \(3x^2-51=24\)
=> 3x2 = 24 + 51
=> 3x2 = 75
=> x2 = 75 : 3
=> x2 = 25 = 52 = (-5)2
=> x \(\in\){-5; 5}.
Vậy có 4 giá trị của x thỏa mãn.
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 \(x=\dfrac{2}{9}\)
\(\left(3x^2-51\right)^{2n}=\left(-24\right)^{2n}\)
\(3x^2-51=-24\)
\(3x^2=27\)
\(x^2=9\)
\(x^2=3^2=\left(-3\right)^2\)
TH1: x=3
TH2: x=-3
=.= hok tốt!!
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: \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)
\(=n^3+2n^2+3n^2+6n-n-2+n^3+2\)
\(=5n^2+5n=5\left(n^2+n\right)⋮5\)
b: \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)
\(=6n^2+30n+n+5-6n^2+3n-10n+5\)
\(=24n+10⋮2\)
d: \(=\left(n+1\right)\left(n^2+2n\right)\)
\(=n\left(n+1\right)\left(n+2\right)⋮6\)
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
a: =>|3x-5|=|x+2|
=>3x-5=x+2 hoặc 3x-5=-x-2
=>2x=7 hoặc 4x=3
=>x=7/2 hoặc x=3/4
b: \(\Leftrightarrow\left\{{}\begin{matrix}3x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
c: \(\Leftrightarrow\left|3x-5\right|=x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(3x-5-x+2\right)\left(3x-5+x-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(2x-3\right)\left(4x-7\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
d: \(\dfrac{11}{2}\le\left|x\right|< \dfrac{17}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{11}{2}< =x< \dfrac{17}{2}\\-\dfrac{17}{2}< x< =-\dfrac{11}{2}\end{matrix}\right.\)
a: \(\left(3x^2-51\right)^{2n}=\left(-24\right)^{2n}=24^{2n}\)
\(\Leftrightarrow3x^2-51=24\) hoặc 3x2-51=-24
=>3x2=75 hoặc 3x2=27
=>x2=25 hoặc x2=9
hay \(x\in\left\{5;-5;3;-3\right\}\)
b: =>x-3>=0 và x-8<=0
=>3<=x<=8