(x+1)+(x+3)+(x+5)+....+(x+99)=0

=> (x+x+x...+x) + (1+3+5+...+99) = 0

=>  50x             + 2500                 = 0

=>   50x            = -2500

=> x                 = -50

Vậy x = -50

chịuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

    A  = 2 +  2² + 2³ +...+2⁹⁹ + 2¹⁰⁰

   2A  =       2² + 2³ +...+ 2⁹⁹ + 2¹⁰⁰ + 2¹⁰¹

2A - A =      2¹⁰¹ - 2

   A      = 2¹⁰¹ - 2

  B        =  1 + 5 + 5² + 5³ +...+ 5⁹⁹ + 5¹⁰⁰

5B        =       5 + 5² + 5³ +...+ 5⁹⁹ + 5¹⁰⁰ + 5¹⁰¹

5B - B   =  5¹⁰¹ - 1

   4B     = 5¹⁰¹ - 1

     B      = \(\dfrac{5^{101}-1}{4}\)

     B     = 

a, \(390-\left(x-7\right)=13^2:12\)

\(390-\left(x-7\right)=\) \(\dfrac{169}{12}\)

\(x-7=390-\dfrac{169}{12}\)

\(x-7=\dfrac{4511}{12}\)

\(x=\dfrac{4511}{12}+7\)

\(x=\dfrac{4595}{12}\)

Vậy ...

b, \(\left(x-35.2^2\right):7=3^3-24\)

\(\left(x-35.4\right):7=27-24\)

\(\left(x-140\right):7=3\)

\(\Leftrightarrow\left(x-140\right)=3.7\)

\(\Leftrightarrow x-140=21\)

\(\Leftrightarrow x=161\)

Vậy .....

c) \(x-6:2-\left(4^2.3-24\right):2:6=3\)

\(x-3-\left(16.3-24\right):2:6=3\)

\(x-3-\left(48-24\right):2:6=3\)

\(x-3-24:2:6=3\)

\(x-3-2=3\)

\(x=3+2+3\)

\(x=8\)

Vậy ......

d) \(4x-5=5+5^2+5^3+.....+5^{99}\)

Đặt :

\(A=5+5^2+.........+5^{99}\)

\(\Leftrightarrow5A=5^2+5^3+..........+5^{100}\)

\(\Leftrightarrow5A-A=\left(5^2+5^3+......+5^{100}\right)-\left(5+5^2+....+5^{99}\right)\)

\(\Leftrightarrow4A=5^{100}-5\)

\(\Leftrightarrow A=\dfrac{5^{100}-5}{4}\)

\(\Leftrightarrow4x+5=\dfrac{5^{100}-5}{4}\)

Đến đây thì sao nữa nhỉ ?

e) \(\left(2x-1\right)^4=625\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^4=5\\\left(2x-1\right)^4=-5\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy ....

a) Có x = 99 => x+1 = 100

A = x⁵ - (x+1)x⁴ + (x+1)x³ + (x+1)x² + (x+1)x - 9

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

= x - 9

=> A = 90

b) Chữa đề: x⁶ - 20x⁵ - 20x⁴ - 20x³ - 20x² - 20x + 3

Có: x = 21 => x-1 = 20

B = x⁶ - (x-1)x⁵ - (x-1)x⁴ - (x-1)x³ - (x-1)x² - (x-1)x + 3

= x⁶ - x⁶ + x⁵ - x⁵ + x⁴ - x⁴ + x³ - x³ + x² - x + 3

= x + 3

=> B = 24

a)64:2mũ5×30×4

= 64 : 32 x 30 x 4

= 240

b)3 mũ 2× 5 - 2 mũ 2×7+2 mũ 0 × 5

= 9 x 5 - 4 x 7 + 1 x 5

= 45 - 28 + 5

= 22

c)2 mũ 3-5 mũ 3÷5 mũ 2 + 12×2 mũ 2

= 8 - 125 : 25 + 12 x 4

= 8 - 5 + 48

= 51

d)2[(7-3 mũ 3÷3 mũ 2) chia 2 mũ 2 + 99]-100

= 2[( 7 - 27 : 9) : 4 + 99] - 100

= 2[4 : 4 + 99] - 100

= 2. 100 - 100

= 200 - 100

= 100

e)4[(3 + 3^7:3^4)chia 10 + 97]-300

= 4[( 3 + 3^3) : 10 + 97] - 300

= 4[ 30 : 10 + 97 ] - 300

= 4. 100 - 300

= 400 - 300

= 100

f)2^2 x 5 [(5 mũ 2 cộng 2 mũ 3) chia 11 - 2] - 3^2 x 2

= 4 x 5 [ (25 + 8 ) : 11 - 2] - 9 x 2

= 20 [ 33 : 11 - 2] - 18

= 20. 1 - 18

= 20 - 18

= 2

hum biết nữa

a)240

b)22

c)51

d)100

e)100

f)2

gọi biểu thức trên là A , ta có :

\(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+\dfrac{5}{3^5}-...+\dfrac{99}{3^{99}}+\dfrac{100}{3^{100}}\\ 3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\\ \Rightarrow A+3A=\left(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\right)+\left(1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\right)\\ \Rightarrow4A\cdot3=12A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}-\dfrac{1}{3^{99}}\)

từ đó ta được :

\(16A=3-\dfrac{100}{3^{99}}-\dfrac{100}{3^{100}}\\ \Rightarrow A=\dfrac{\dfrac{3-101}{3^{99}}-\dfrac{100}{3^{100}}}{16}\\ \Rightarrow A=\dfrac{3}{16}-\dfrac{\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}}{16}< \dfrac{3}{16}\)

help mik với 

1) \(2^x-15=17\)

\(\Leftrightarrow2^x=32=2^5\)

\(\Rightarrow x=5\)

2) \(\left(7x-11\right)^3=25\cdot5^2+200\)

\(\Leftrightarrow\left(7x-11\right)^3=825\)

\(\Leftrightarrow7x-11=\sqrt[3]{825}\)

\(\Leftrightarrow7x=11+\sqrt[3]{825}\)

\(\Rightarrow x=\frac{11+\sqrt[3]{825}}{7}\)

3) \(\left(x+1\right)^{100}-3\left(x+1\right)^{99}=0\)

\(\Leftrightarrow\left(x+1\right)^{99}\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^{99}=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

4) \(4x+5\left(x+3\right)=105\)

\(\Leftrightarrow9x+15=105\)

\(\Leftrightarrow9x=90\)

\(\Rightarrow x=10\)

5) \(5\cdot\left(x-2\right)+10\left(x+3\right)=170\)

\(\Leftrightarrow5\left[x-2+2\left(x+3\right)\right]=170\)

\(\Leftrightarrow3x+4=34\)

\(\Leftrightarrow3x=30\)

\(\Rightarrow x=10\)

cảm ơn bạn nhá NGUYỄN MINH ĐĂNG