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\(3^{x+2}+3^x=90\Leftrightarrow3^x.3^2+3^x=90\Leftrightarrow3^x\left(3^2+1\right)=90\Leftrightarrow3^x.10=90\Leftrightarrow3^x=9\Leftrightarrow3^x=3^2\Leftrightarrow x=2\)
Vậy ...
\(\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{1}{9}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{9}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{2}{3}\right)^2\Leftrightarrow x+\dfrac{1}{2}=\dfrac{2}{3}\Leftrightarrow x=\dfrac{1}{6}\)
Vậy ...
1.
a) \(3^{x+2}+3^x=90\)
\(\Leftrightarrow3^x\left(3^2+1\right)=90\)
\(\Leftrightarrow3^x.10=90\)
\(\Leftrightarrow3^x=9=3^2\)
\(\Leftrightarrow x=2\)
vậy...
b) \(\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{1}{9}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\pm\dfrac{2}{3}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{3}\\x+\dfrac{1}{2}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=\dfrac{-7}{6}\end{matrix}\right.\)
Vậy...
tik mik nha !!!
a)90-6.x =6.(24-4)
90-6.x=6.(16-4)
90-6.x=6.12
90-6.x=72
6.x=90-72
6.x=18
x=18:6
x=3
Vậy x=3
b)35-3|x|=5.(23-4)
35-3|x|=5.(8-4)
35-3|x|=5.4
35-3|x|=20
3|x|=35-20
3|x|=15
|x|=15:3
|x|=5
suy ra x=5 hoặc x= -5
Vậy x=5;x= -5
c/ 2x - 1 = \(5^{98}:5^{96}\)
2x - 1 = \(5^2\) = 25
2x = 25 + 1 = 26
x = 26 : 2
x = 13
d/ 7x + 3 = \(3^5.2^3.9\)
7x + 3 = \(3^5.3^2.8=3^7.8=2187.8\)
7x + 3 = \(17496\)
7x = 17496 - 3 = 17493
x = 17493 : 7
x = 2499
e/\(2^{2x+6}=1\)
\(2^{2x+6}=2^0\)
2x + 6 = 0
2x = 0 - 6 = - 6
x = - 6 : 2
x = - 3
j/ \(2^x=8\)
\(2^x=2^3\)
x = 3
g/ \(2^x:2^3=16\)
\(2^{x-3}=2^4\)
x - 3 = 4
x = 4 + 3
x = 7
h/ \(2^x+2^{x+1}+2^{x+2}=56\)
\(2^x\left(1+2+2^2\right)\) = 56
\(2^x.7=56\)
\(2^x=56:7\)
\(2^x=8\)
\(2^x=2^3\)
x = 3
Bài a, b thiên phong giải r, mk chỉ làm những bài còn lại thôi. Chúc bạn học tốt!!!
Bài 1 tự làm!
Bài 2:
a, \(\left(3x-4\right)\left(x-1\right)^3=0\Rightarrow\left[{}\begin{matrix}3x-4=0\\\left(x-1\right)^3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=1\end{matrix}\right.\)
b, \(2^{2x-1}:4=8^3\Rightarrow2^{2x-1}:2^2=2^9\)
\(\Rightarrow2x-1-2=9\Rightarrow2x-3=9\Rightarrow2x-12\Rightarrow x=6\)
c, Đề chưa rõ
d, \(\left(x+2\right)^5=2^{10}\Rightarrow\left(x+2\right)^5=4^5\Rightarrow x+2=4\Rightarrow x=2\)
e, \(\left(3x-2^4\right).7^3=2.7^4\Rightarrow3x-2^4=2.7^4:7^3\Rightarrow3x-16=2.7=14\)
\(\Rightarrow3x=14+16=30\Rightarrow x=\dfrac{30}{3}=10\)
f, \(\left(x+1\right)^2=\left(x+1\right)^0\Rightarrow\left(x+1\right)^2=1\) (vì x0 = 1)
\(\Rightarrow x+1=1\Rightarrow x=0\)
6)\(3.\left(x+1\right)-2^3.2=11\)
\(3\left(x+1\right)-2^4=11\)
\(3.\left(x+1\right)=11+16\)
3.(x+1)=27
x+1=27:3
x+1=9
x=9-1
x=8
Vậy x=8
7) \(180-2.\left(x+5\right)^2=130\)
\(2.\left(x+5\right)^2=180-130\)
2.(x+5)2=50
(x+5)2=50:2
(x+5)2=25
x+5=5 hoặc x+5=-5
x=5-5 x=-5-5
x=0 x=-10
Vậy x=0 hoặc x=-10
a. 52 + (x+3) = 52
=> x + 3 = 52 - 52
=> x + 3 = 0
=> x = -3
b. 23 + (x-32) = 53 - 43
=> 8 + (x-9) = 125 - 64
=> x - 9 = 125 - 64 - 8
=> x - 9 = 53
=> x = 53 + 9
=> x = 62
\(a,4^x=64\)
\(4^x=4^3\)
\(x=3\)
\(b,3\cdot\left(5^x-16\right)=3^3\)
\(5^x-16=3^3:3=3^2\)
\(5^x-16=9\)
\(5^x=9+16=25\)
\(5^x=5^2\)
\(x=2\)
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
1. abcde4 . 4 = 4abcde
=> e = đuôi 4.4 = 16
=> e = 6 , thay e = 6 vào bt
=> d = đuôi 4.6 + 1 = 25
=> d = 5, thay d = 5 vào bt
=> c = đuôi 5.4 + 2 = 22
=> c = 2, thay d= 2 vào bt
=> b = đuôi 2.4 + 2 = 10
=> b = 0. thay b = 0 vào bt
=> a = 4.0 + 1 = 1
=> a = 1
Vậy : 102564 . 4 = 410256
=> abcde = 10256
2.
3x+2 + 3x = 90
=> 3x . 9 + 3x = 90
=> 3x. 9 = - 90 - 3x
=> 3x . 9 + 3x = 3x + 90 - 3x
=> 3x. 10 = 90
=> 3x = 9 = 32
=> x = 2
tíc mình nha!