Ta có :

\(M=5+5^2+5^3+...+5^{60}\)

\(\Leftrightarrow\)\(5M=5^2+5^3+5^4+...+5^{61}\)

\(\Leftrightarrow\)\(5M-M=\left(5^2+5^3+5^4+...+5^{61}\right)-\left(5+5^2+5^3+...+5^{60}\right)\)

\(\Leftrightarrow\)\(4M=5^{61}-5\)

\(\Leftrightarrow\)\(M=\frac{5^{61}-5}{4}\)

Vậy \(M=\frac{5^{61}-5}{4}\)

Sai ùi

nâng cao phát triển có đấy

a) 5M=5(\(5+5^2++.......+5^{60}\)

5M=\(5^2+5^3+...+5^{61}\)

5M-M=\(\left(5^2+5^3+...+5^{61}\right)-\left(5+5^2+5^3+...+5^{60}\right)\)

4M=\(5^{61}-5\)

M=\(\left(5^{61}-5\right):4\)

b) \(\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{59}+5^{60}\right)\)

\(5\left(1+5\right)+5^3\left(1+5\right)+...+5^{59}\left(1+5\right)\)

\(5\cdot6+5^3\cdot6+...+5^{59}\cdot6\)

\(6\left(5+5^3+5^5+...+5^{59}\right)\)

\(\Rightarrow M⋮6\)

a) M = 5 + 5² + 5³ + .... + 5⁶⁰

=> 5M = 5 . 5 + 5² . 5 + 5³ . 5 + ... + 5⁶⁰ . 5

=> 5M = 5² + 5³ + 5⁴ + .... + 5⁶¹

=> 5M - M = 5⁶¹ - 5

=> 4M = 5⁶¹ - 5

=> M   = \(\frac{\text{5^{61} - 5}}{4}\)\(\frac{5^{61}-5}{4}\)

b) M = 5 + 5² + 5³ + .... + 5⁶⁰

=> M = ( 5 + 5² ) + ( 5³ + 5⁴ ) + .... + ( 5⁵⁹ + 5⁶⁰ )

=> M = 5 . ( 5⁰ + 5¹ ) + 5³ . ( 5⁰ + 5¹ ) + ... + 5⁵⁹ . ( 5⁰ + 5¹ )

=> M = 5 . 6 + 5³ . 6 + ... + 5⁵⁹ . 6

=> M = 6 . ( 5 + 5³ + ... + 5⁵⁹ ) \(⋮\)6 => đpcm

a/Ta có: M(x)+N(x) = (2x⁵ - 4x³ + 2x² + 10x - 1) + (-2x⁵ + 2x⁴ + 4x³ + x² + x - 10)

                              = 2x⁵ - 2x⁵ - 4x³ + 4x³ + 2x⁴ + 2x² + x² + 10x + x -1 - 10

                              = 2x⁴ + 3x² + 11x - 11

b/ Ta có: A(x) = N(x)-M(x) = (-2x⁵ + 2x⁴ + 4x³ + x² + x - 10) - (2x⁵ - 4x³ + 2x² + 10x - 1)

                                         = -2x⁵ - 2x⁵ + 2x⁴ + 4x³ + 4x³ + x² - 2x² + x - 10x -10 + 1

                                         = -2x⁵ + 2x⁴ + 8x³ - x² - 9x -9

ta có 

M    = 5 + 5² + 5³ + ... + 5¹⁰⁰

Mx5 = 5 x ( 5 + 5² + 5³ + ... + 5¹⁰⁰ )

Mx5 = 5² + 5³ + 5⁴ + ... + 5¹⁰¹

Mx5-M= ( 5² + 5³ + 5⁴ + ... + 5¹⁰¹ ) - ( 5 + 5² + 5³ + ... + 5¹⁰⁰ )

Mx4 = 5¹⁰¹ - 5

=> Mx4+5=5¹⁰¹

mà 4M+5=5ⁿ

=> n = 101 

minh biet lam chi toi hoi day dong

kb nha xong minh giai cho

Bài 1 :

\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)

\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)

\(\Rightarrow M< N\)

Bài 3 :

a) \(t^2+5t-8\) khi \(t=2\)

\(=5^2+2.5-8\)

\(=25+10-8\)

\(=27\)

b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)

\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)

\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)

c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)

\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)

\(\left(1\right)=1^3=1\)

Bài 2 :

a) \(...=2^3\left(19-14\right)+1=8.5+1=41\)

b) \(...=100-\left[60:\left(5^2-15\right)\right]=100-\left[60:10\right]=100-6=94\)

c) \(...=160:\left[17+\left(9.5-\left(14+2^3\right)\right)\right]=160:\left[17+\left(45-22\right)\right]=160:\left[17+23\right]=160:40=4\)

d) \(...=798+100\left[16-2\left(25-22\right)\right]=798+100\left[16-2.3\right]=798+100.10=798+1000=1798\)

1: A(2;0); B(-3;4); C(1;-5)

Tọa độ vecto AB là:

\(\left\{{}\begin{matrix}x=-3-2=-5\\y=4-0=4\end{matrix}\right.\)

=>\(\overrightarrow{AB}=\left(-5;4\right)\)

Tọa độ vecto AC là:

\(\left\{{}\begin{matrix}x=1-2=-1\\y=-5-0=-5\end{matrix}\right.\)

Vậy: \(\overrightarrow{AC}=\left(-1;-5\right)\)

\(\overrightarrow{AB}=\left(-5;4\right)\)

Vì \(\left(-1\right)\cdot\left(-5\right)=5< >-20=-5\cdot4\)

nên A,B,C không thẳng hàng

=>A,B,C là ba đỉnh của một tam giác

2: Tọa độ trọng tâm G của ΔABC là:

\(\left\{{}\begin{matrix}x=\dfrac{2-3+1}{3}=\dfrac{0}{3}=0\\y=\dfrac{0+4-5}{3}=-\dfrac{1}{3}\end{matrix}\right.\)

3:

\(\overrightarrow{AB}=\left(-5;4\right);\overrightarrow{DC}=\left(1-x;-5-y\right)\)

ABCD là hình bình hành

nên \(\overrightarrow{AB}=\overrightarrow{DC}\)

=>\(\left\{{}\begin{matrix}1-x=-5\\-5-y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=1+5=6\\y=-5-4=-9\end{matrix}\right.\)

Vậy: D(6;-9)

4: \(\overrightarrow{MA}=\left(2-x;-y\right);\overrightarrow{MB}=\left(-3-x;4-y\right);\overrightarrow{MC}=\left(1-x;-5-y\right)\)

\(2\overrightarrow{MA}+\overrightarrow{MB}+3\overrightarrow{MC}=\overrightarrow{0}\)

=>\(\left\{{}\begin{matrix}2\left(2-x\right)+\left(-3-x\right)+3\left(1-x\right)=0\\2\left(-y\right)+\left(4-y\right)+3\left(-5-y\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4-2x-3-x+3-3x=0\\-2y+4-y-15-3y=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-6x+4=0\\-6y-11=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-6x=-4\\-6y=11\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{11}{6}\end{matrix}\right.\)

vậy: \(M\left(\dfrac{2}{3};-\dfrac{11}{6}\right)\)

5:

A(2;0); B(-3;4); C(1;-5); N(x;y)

A là trọng tâm của ΔBNC

=>\(\left\{{}\begin{matrix}x_A=\dfrac{x_B+x_N+x_C}{3}\\y_A=\dfrac{y_B+y_N+y_C}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2=\dfrac{-3+1+x}{3}\\0=\dfrac{4-5+y}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2=6\\y-1=0\end{matrix}\right.\)

=>x=8 và y=1

Vậy: N(8;1)

6: A là trung điểm của BE

=>\(\left\{{}\begin{matrix}x_B+x_E=2\cdot x_A\\y_B+y_E=2\cdot y_A\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-3+x_E=2\cdot2=4\\4+y_E=2\cdot0=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x_E=7\\y_E=-4\end{matrix}\right.\)

Vậy: E(7;-4)