Tìm \(x\in N\) :
a) \(3^x.81^{2x+1}=81\)
b) \(4^x-25=89\)
c) \(9^x+3^{2x+4}=7290\)
d) \(2^{x-2}+2^x+2^{x+2}=48\)
GIÚP MK VS NHA !
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) ( x+1)2 =25
Ta có :
52=25
=> ( x+1)2)=52
=> x+1=5
=> x=4
b) 32x=210
=> 32x=(22.5)
=> 32x=(25)2
=> 32x=322
=> x=2
c) 9x=81
Ta có :
92=81
=> 9x=92
=> x=2
d) 3. (2x +1)2=48
=> (2x +1)2=42:3
=> (2x +1)2=16
Mà :
42=16
=> (2x +1)2=42
=> 2x+1=4
=> 2x=3
=> x=1,5
e) (x-1)4 = (x-1)2
TH1: (x-1)4 = (x-1)2=0
=> x=1
TH2: (x-1)4 = (x-1)2=1
=> x=2
Xong :)
a) \(\left(x+1\right)^2=25\)
\(\left(x+1\right)^2=5^2=\left(-5\right)^2\)
\(\Rightarrow x=5-1=4\)
\(x=\left(-5\right)-1=-6\)
\(\left(x+4\right)^2-81=0\Leftrightarrow\left(x+4\right)^2-9^2=0\)
\(\Leftrightarrow\left(x+4+9\right)\times\left(x+4-9\right)=0\)
\(\Leftrightarrow\left(x+13\right)\times\left(x-5\right)=0\)
\(\left[{}\begin{matrix}x+13=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=5\end{matrix}\right.\)
mk chi lam duoc cau c thoi
suy ra 2x-1=5 hoac 2x-1=-5
2x=6 2x=-4
x=3 x=-2
vay x=3 hoac x=-2
a. 3/4.x+1/3=-1/2
=>3/4.x=-1/2-1/3=-5/6
=>x=-5/6 chia 3/4=-10/9
b. -x/4=-9/x =>-x*x=4*-9
=>-2x=-36 =>x=18
c./2x-1/=5
=> 2x-1=5 =>2x=5+1=6 =>x=3
hoặc 2x-1=-5 =>2x=-5+1=-4 =>x=-2
d,e: Sai đề rồi
a, (x-2)^2 - (x-3)(x+3)=6
x^2-4x+4-(x^2-9)=6
x^2-4x+4-x^2+9=6
(x^2-x^2)-4x+13=6
-4x=-7
x=1,75
b, 4(x-3)^2 - (2x-1)(2x+1)=10
4(x^2-6x+9)-(4x^2-1)=10
4x^2-24x+36-4x^2+1=10
-24x+37=10
x=9/8
c,(x-4)^2 - (x+2)(x-2)=6
x^2-8x+16-(x^2-4)=6
x^2-8x+16-x^2+4=6
-8x+20=6
x=7/4
d, 9(x+1)^2 - (3x-2)(3x+2)=10
9(x^2+2x+1)-(9x^2-4)=10
9x^2+18x+9-9x^2+4=10
18x+13=10
x=-1/6
\(a,\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(-4x+13=6\)
\(-4x=6-13\)
\(-4x=-7\)
\(x=\frac{-7}{-4}\)
\(x=\frac{7}{4}\)
Vậy \(x=\frac{7}{4}\)
\(b,4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(4\left(x^2-6x+9\right)-\left(4x^2-1\right)=10\)
\(4x^2-24x+36-4x^2+1=10\)
\(-24x+37=10\)
\(x=\frac{9}{8}\)
Vậy \(x=\frac{9}{8}\)
\(c,\left(x-4\right)^2-\left(x+2\right)\left(x-2\right)=6\)
\(x^2-8x+16-\left(x^2-4\right)=6\)
\(x^2-8x+16-x^2+4=6\)
\(-8x+20=6\)
\(x=\frac{7}{4}\)
Vậy \(x=\frac{7}{4}\)
\(d,9\left(x+1\right)^2-\left(3x-2\right)\left(3x+2\right)=10\)
\(9\left(x^2+2x+1\right)-\left(9x^2-4\right)=10\)
\(9x^2+18x+9-9x^2+4=10\)
\(18x+13=10\)
\(x=\frac{-1}{6}\)
Vậy \(x=\frac{-1}{6}\)
\(a)3^x\cdot81^{2x+1}=81\\ 3^x\cdot\left(3^4\right)^{2x+1}=81\\ 3^x\cdot3^{8x+4}=3^4\\ 3^{9x+4}=3^4\\ \Leftrightarrow9x+4=4\\ \Leftrightarrow9x=0\\ \Rightarrow x=0\)
\(b)4^x-25=89\\ \Leftrightarrow4^x=64\\ \Leftrightarrow4^x=4^3\\ \Rightarrow x=3\)