\(3^6\cdot45^4-15^{13}\cdot5^{-9}\)

\(=3^6\cdot\left(3^2\cdot5\right)^4-\left(3\cdot5\right)^{13}\cdot5^{-9}\)

\(=3^6\cdot3^8\cdot5^4-3^{13}\cdot5^{13}\cdot5^{-9}=3^{14}\cdot5^4-3^{13}\cdot5^4=3^{13}\cdot5^4\left(3-1\right)=3^{13}\cdot5^4\cdot2\)

\(27^4\cdot25^3+45^6\)

\(=\left(3^3\right)^4\cdot\left(5^2\right)^3+\left(3^2\cdot5\right)^6\)

\(=3^{12}\cdot5^6+3^{12}\cdot5^6=2\cdot3^{12}\cdot5^6\)

Ta có: \(\frac{3^6\cdot45^4-15^{13}\cdot5^{-9}}{27^4\cdot25^3+45^6}\)

\(=\frac{3^{13}\cdot5^4\cdot2}{2\cdot3^{12}\cdot5^6}\)

\(=\frac{3}{5^2}=\frac{3}{25}\)

KHO THE

\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)

\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)

\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)

b: \(=\dfrac{3a-9-2a-6-6}{\left(a+3\right)\left(a-3\right)}=\dfrac{a-15}{a^2-9}\)

3.(⅓x - ¼)² = ⅓ 

=> (\(\dfrac{1}{3x}\)- \(\dfrac{1}{4}\) )² = \(\dfrac{1}{9}\)

=>\(\left[{}\begin{matrix}\dfrac{1}{3x}-\dfrac{1}{4}=\dfrac{-1}{3}\\\dfrac{1}{3x}-\dfrac{1}{4}=\dfrac{1}{3}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\dfrac{1}{3x}=\dfrac{-1}{12}\\\dfrac{1}{3x}=\dfrac{7}{12}\end{matrix}\right.\)        => \(\left[{}\begin{matrix}x=-4\\x=\dfrac{12}{21}=\dfrac{4}{7}\end{matrix}\right.\)

Vậy, tập nghiệm x thỏa mãn là S=\(\left\{-4;\dfrac{4}{7}\right\}\)

a) x+3/22=-27/121.11/9

⇒ x+3/22=-3/11

⇒ x=-9/22

b) (3/2/20-2/1/20)x+2/7=3/5

⇒ (31/10-41/20)x=11/35

⇒ 21/20x=11/35

⇒x=44/147

c) |3/2x-5|-1/2=-1/4

⇒ |3/2x-5|=1/4

⇒ hoặc 3/2x-5=1/4⇒3/2x=21/4⇒x=7/2

hoặc 3/2x-5=-1/4⇒3/2x=19/4⇒x=19/6

Lời giải:
$22+23-25+27-29+31-33$

$=22+(23-25)+(27-29)+(31-33)$

$=22+(-2)+(-2)+(-2)=22+(-2).3=22-6=16$

x^3 + x=0

x (x^2 +1) =0

Th1:

x=1

Th2:

x^2 +1 =0

x^2 = -1

=> x thuộc rỗng

Vậy x=0

x = 0 => 0^3 + 0 = 0 

Viết 2 bức xong trả lời câu hỏi giúp mình luôn nha <3  

Giúp mình với mình đang cần gấp ạ :<