2x-x⁰=3⁵:3³=3²=9

=>2x-1=9

=>2x=9+1=10

=>x=10:2=5.

\(2x-x^0=3^5:3^3\\ 2x-1=3^{5-3}\\ 2x-1=3^2\\ 2x-1=9\\ 2x=9+1\\ 2x=10\\ x=\dfrac{10}{2}\\ x=5\)

Bài này mới ra mà bạn? vậy chắc bạn chờ một tí rồi thử lại xem nhé.

Đó là bài của bộ sách nào vậy em, em nói cụ thể ra thì cô mới test được

\(\dfrac{28^7.5^8.15^9}{14^8.25^9.6^{12}}=\dfrac{2^7.14^7.5^8.5^9.3^9}{14^7.14.\left(5^2\right)^9.3^{12}.2^{12}}\\ =\dfrac{14^7.5^{17}.3^9.2^7}{14^7.14.5^{17}.5.3^9.3^3.2^7.2^5}\\ =\dfrac{1}{14.5.3^3.2^5}=\dfrac{1}{14.5.27.32}\\ =\dfrac{1}{60480}\)

\(\dfrac{28^7.5^8.15^9}{14^8.25^9.6^{12}}\\ =\dfrac{2^7.14^7.5^8.5^9.3^9}{14^8.5^9.5^9.3^{12}.2^{12}}\\ =\dfrac{1}{14.5.3^3.2^5}\\ =\dfrac{1}{60480}\)

16⁵+2¹⁵=(2⁴)⁵+2¹⁵=2²⁰+2¹⁵=2¹⁵(1+2⁵)=2¹⁵x33⋮33

=>ĐPCM

Ok e

a) \(3^{54}\)

\(2^{200}=4^{100}>3^{54}\)

\(\Rightarrow3^{54}< 2^{200}\)

b) \(15^{12}=3^{12}.5^{12}\)

\(1^3.125^3=\left(5^3\right)^3=5^9< 3^{12}.5^{12}\)

\(\Rightarrow15^{12}>1^3.125^3\)

c) \(78^{12}-78^{11}=78^{11}.\left(7-1\right)=78^{11}.6\)

\(78^{11}-78^{10}=78^{10}.\left(7-6\right)=78^{10}.6< 78^{11}.6\)

\(\Rightarrow78^{12}-78^{11}>78^{11}-78^{10}\)

d) \(72^{45}-72^{44}=72^{44}.\left(72-1\right)=72^{44}.72>27^{44}\)

\(\Rightarrow72^{45}-72^{44}>27^{44}\)

e) \(3^{39}=\left(3^3\right)^{13}=27^{13}>11^{11}\)

\(\Rightarrow3^{39}>11^{11}\)

Giải chi tiết dùm mik nhé.

1. ours->our

2. its their->they're their

3. my's->mine

4. he's->his

5. your->yours

6. mine->me

7. they're->their

8. dog water->dog's water

So sánh

\(A=\dfrac{1999^{1999}+1}{1999^{1998}+1}\) ; \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}\)

Ta có: \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}>1\) ( vì tử > mẫu )

Do đó: \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}>\dfrac{1999^{2000}+1+1998}{1999^{1999}+1+1998}=\dfrac{1999^{2000}+1999}{1999^{1999}+1999}=\dfrac{1999.\left(1999^{1999}+1\right)}{1999.\left(1999^{1998}+1\right)}=\dfrac{1999^{1999}+1}{1999^{1998}+1}=A\)

Vậy B > A

Chúc bạn học tốt

\(B=1\cdot2\cdot3+2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100\)

\(\Rightarrow4B=4\cdot\left(1\cdot2\cdot3+2\cdot3\cdot4+...+98\cdot99\cdot100\right)\)

\(\Rightarrow4B=1\cdot2\cdot3\cdot\left(4-0\right)+2\cdot3\cdot4\cdot\left(5-1\right)+3\cdot4\cdot5\cdot\left(6-2\right)+...+98\cdot99\cdot100\cdot\left(101-97\right)\)

\(\Rightarrow4B=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4-....+98\cdot99\cdot100\cdot101-97\cdot98\cdot99\cdot100\)

\(\Rightarrow4B=98\cdot99\cdot100\cdot101\)

\(\Rightarrow B=\dfrac{98\cdot99\cdot100\cdot101}{4}\)

\(\Rightarrow B=25\cdot98\cdot99\cdot101\)

B=1x2x3+2x3x4+...+98x99x100

=>4B=1x2x3x(4-0)+2x3x4x(5-1)+...+98x99x100x(101-97)

4B=1x2x3x4+2x3x4x5-1x2x3x4+...+98x99x100x101-97x98x99x100

4B=98x99x100x101

=>B=\(\dfrac{98\cdot99\cdot100\cdot101}{4}\)=24497550.

Có ai giải được câu này không?

ko khó lắm

(3\(x\) - 2)(\(x+4\)) - (1- \(x\))(2-\(x\)) =(\(x+1\))(\(x-2\))

3\(x^2\) + 12\(x\) - 2\(x\) - 8  - (\(x+1\))(\(x-2\)) -  [-(\(x-2\))](1- \(x\)) = 0 

                    3\(x^2\) + 10\(x\) - 8  -  (\(x-2\))( \(x\) + 1 - 1 + \(x\)) = 0

                    3\(x^2\) + 10\(x\) - 8 - (\(x-2\)). 2\(x\)  = 0

                    3\(x^2\) + 10\(x\) - 8 - 2\(x^2\) + 4\(x\)     = 0

                      \(x^2\) + 14\(x\) -  8 = 0

                    \(x^2\) + 7\(x\) + 7\(x\) + 49 - 57 = 0 

                     \(x\)( \(x\) + 7) + 7(\(x\) + 7) = 57

                       (\(x+7\))(\(x\) + 7) =57

                       (\(x+7\))² = 57

                        \(\left[{}\begin{matrix}x+7=\sqrt{57}\\x+7=-\sqrt{57}\end{matrix}\right.\)

                         \(\left[{}\begin{matrix}x=-7+\sqrt{57}\\x=-7-\sqrt{57}\end{matrix}\right.\)

Vậy \(x\) \(\in\) { -7 - \(\sqrt{57}\); - 7 + \(\sqrt{57}\)}