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Bài 6 :
a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)
c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)
d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)
Bài 7 :
a) \(3^x+3^{x+2}=9^{17}+27^{12}\)
\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)
\(\Rightarrow10.3^x=3^{34}+3^{36}\)
\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)
\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)
b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)
\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)
\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)
c) Bài C bạn xem lại đề
d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)
\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)
\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)
\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)
\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)
Bài 1:
(x² - 8)(x³ + 2x + 4)
= x².x³ + x².2x + x².4 - 8.x³ - 8.2x - 8.4
= x⁵ + 2x³ + 4x² - 8x³ - 16x - 32
= x⁵ - 6x³ + 4x² - 16x - 32
Bài 2
a) A(x) = -5/3 x² + 3/4 x⁴ + 2x - 7/3 x² - 2 + 4x + 1/4 x⁴
= (3/4 x⁴ + 1/4 x⁴) + (-5/3 x² - 7/3 x²) + (2x + 4x) - 2
= x⁴ - 4x² + 6x - 2
b) Bậc của A(x) là 4
Hệ số cao nhất là 1
a) \(\left(x+5\right)^3=64\)
\(\Leftrightarrow\left(x+5\right)^3=4^3\)
\(\Leftrightarrow x+5=4\)
\(\Leftrightarrow x=-1\)
Vậy x = - 1
b) \(x:\left(-\frac{3}{5}\right)^2=-\frac{3}{5}\)
\(\Leftrightarrow x=\left(-\frac{3}{5}\right)^2.\left(-\frac{3}{5}\right)\)
\(\Leftrightarrow x=\left(-\frac{3}{5}\right)^3\)
\(\Leftrightarrow x=-0,216\)
Vậy x = - 0, 216
c) \(\left(\frac{4}{7}\right)^4.x=\left(\frac{4}{7}\right)^6\)
\(\Leftrightarrow x=\left(\frac{4}{7}\right)^6:\left(\frac{4}{7}\right)^4\)
\(\Leftrightarrow x=\left(\frac{4}{7}\right)^2\)
\(\Leftrightarrow\text{x}=\frac{16}{49}\)
Vậy x = 16/49
d) \(\left(-\frac{1}{3}\right)^3x=\frac{1}{81}\)
\(\Leftrightarrow-\frac{1}{27}x=\frac{1}{81}\)
\(\Leftrightarrow x=\frac{1}{81}:\left(-\frac{1}{27}\right)\)
\(\Leftrightarrow x=-\frac{1}{3}\)
Vậy x = - 1/3
1. X : (-1/3)^2=-1/3
X : 1/9 =-1/3
X= -1/3 : 1/9
X= -3
2. X : 0,16 =9:X
X = 9 : X . 0,16
X =(9.0,16):X
X =1,44:X
X . X=1,44
X2 =1,44
=>X= 1,2
3. (-1/2)^3 . X=(-1/2)^5
-1/8 . X=-1/32
X =-1/32 : -1/8
X =0,25
4. |X+4/5| - 1/7 = 0
|X+4/5|=0+1/7
|X+4/5|=1/7
=>X+4/5=1/7 HOẶC X + 4/5 = -1/7
Nếu X+4/5=1/7
X=1/7- 4/5
X=-23/35
Nếu X+4/5 =- 1/7
X=-1/7 - 4/5
X=-33/35
a. x = {3;-3}
b. x thuộc rỗng
c. x2-4=0
x2 = 4
x={2;-2}
d. x2+1=82
x2 =83
x thuộc rỗng
e. (2x)2=6
x thuộc rỗng
f. (x-1)2=9
TH1: x-1=3=>x=4
TH2: x-1=-3=>x=-2
Vậy x={4;-2}
g.(2x+3)2=25
TH1: 2x+3=5=> x=1
Th2: 2x+3=-5=>x=-4
VẬY X={1;-4}
a, x^2= 9
=>\(\sqrt{9}=3\)
b,\(x^2=5=>x=\sqrt{5}\)
c, x^2-4=0
=>x^2=4
=>x=2
d, x^2+1=82
=>x^2=81 =>\(\sqrt{81}=9\)
3, 2x^2=6
=>x= \(\sqrt{6}\)
f, {x-1} ^2=9
=> x-1=3
=>x=2
g{ 2x+3}^2=25
=> 2x+3=5
=>2x=2
=>x=1
\(x+1+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\Rightarrow x+1+x+2+x+3+...+x+100=5750\)
\(\Rightarrow100x+1+2+3+...+100=5750\)
\(\Rightarrow100x+\left[\left(\dfrac{100-1}{1}+1\right):2\right]\left(100+1\right)=5750\)
\(\Rightarrow100x+5050=5750\)
\(\Rightarrow100x=700\Rightarrow x=7\)
\(25-\left(30+x\right)=x-\left(123-67\right)\)
\(\Rightarrow25-30+x=x-123+67\)
\(\Rightarrow-5+x=x-56\)
\(\Rightarrow x\in\varnothing\)
\(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)
\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^4=0\Rightarrow x=5\\\left(x-5\right)^2-1=0\Rightarrow\left(x-5\right)^2=1\Rightarrow x=6;4\end{matrix}\right.\)
\(\left(x^2+1\right)\left(x-3\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+1>0\Rightarrow x^2>-1\\x-3< 0\Rightarrow x< 3\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+1< 0\Rightarrow x^2< -1\\x-3>0\Rightarrow x>3\end{matrix}\right.\end{matrix}\right.\)
Vậy...
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