1. Tìm x

a) 1+2+3+...+x = 210

=> \(\frac{x\left(x+1\right)}{2}=210\)

=> x = 20

b) \(32.3^x=9.3^{10}+5.27^3\)

=>\(32.3^x=9.3^{10}+5.3^9\)(\(27^3=\left(3^3\right)^3=3^9\))

=>\(32.3^x=9.3.3^9+5.3^9\)

=>\(32.3^x=3^9\left(9.3+5\right)\)

=>\(32.3^x=3^9.32\)

=>x = 9

2.

Ta có 2A = 3A - A

=> 2A = \(3\left(1+3+3^2+3^3+....+3^{10}\right)\)\(-\)\(1-3-3^2-3^3-....-3^{10}\)

=> 2A = \(3+3^2+3^3+.....+3^{11}-\)\(1-3-3^2-3^3-...-3^{10}\)

=> 2A = \(3^{11}-1\)

=> 2A+1 = \(3^{11}-1+1\)=\(3^{11}\)

=> n = 11

Ta có : a)1 + 2 + 3 + ... + x = 210

=> \(\frac{x\left(x+1\right)}{2}=210\)

=> x(x + 1) = 420

=> x(x + 1) = 20.21

=> x = 20

mn vào đề 6 nha

mình vào rồi,thấy 1 đề là có 4 hoặc 5 bài

a ) ( - 14 ) + 12 + ( - 12 ) . ( 2³ - 7 )

= - 2 + ( - 12 ) . ( 8 - 7 )

= - 2 + ( - 12 ) . 1

= - 2 + ( - 12 )

= - 14

12000 - ( 1500 . 2 + 1800 . 3 + 1800 . 2 : 3 ) 

= 12000 - ( 3000 + 5400 + 3600 : 3 )

= 12000 - ( 3000 + 5400 + 1200 ) 

= 12000 - 9600

= 2400

12 000 - (1500 . 2 + 1800 . 3 + 1800 . 2 : 3) 

=12 000 - (3000 + 5400 + 3600 : 3)

= 12 000 - (3000 + 5400 + 1200)

= 12 000 - 9600 = 2400

e: Ta có: \(\left(x+1\right)\left(x+2\right)=444222\)

\(\Leftrightarrow x^2+3x-444220=0\)

\(\text{Δ}=3^2-4\cdot1\cdot\left(-444220\right)=1776889\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là 

\(\left\{{}\begin{matrix}x_1=\dfrac{-3-1333}{2}=-668\\x_2=\dfrac{-3+1333}{2}=665\end{matrix}\right.\)

lớp 6 be like ;-;

Đề bài sai òi, bn chép nhầm hoặc cô viết nhầm, phải sửa 1430 thành 1450 mới đúng ko thì có dư

(x + 1) + (2x + 3) + (3x + 5) + ... + (20x + 39) = 1450

(x + 2x + 3x + ... + 20x) + (1 + 3 + 5 + ... + 39) = 1450

x.(1 + 2 + 3 + ... + 20) + (1 + 39).20 : 2 = 1450

x.(1 + 20).20:2 + 40.10 = 1450

x.21.10 + 400 = 1450

x.210 = 1450 - 400

x.210 = 1050

x = 1050 : 210

x = 5

Vậy x = 5

Ủng hộ mk nha ^_-

(x+1)+(2x+3)+(3x+5)+...+(20x+39)=1430

=> x+1+2x+3+3x+5+...+20x+39=1430

=> (x+2x+3x+...+20x)+(1+3+5+...+39)=1430

=> x(1+2+3+...+20)+[(39-1):2+1].(39+1):2=1430

=>x.(20.21:2)+[38:2+1].40:2=1430

   x.210+[19+1].20=1430

    x.210+20.20=1430

    x.210+400=1430

=>x.210=1430-400

    x.210=1030

=> x=1030:210

     x=103/21

Vậy x=103/21

=\(\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)

=\(\frac{2^{19}.3^9+5.2^{18}.3.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)

\(=\frac{2^{19}.3^9+2^{18}.3^9.5}{2^{19}.3^9+2^{20}.3^{10}}\)

=\(\frac{2^{18}.3^9\left(2+5\right)}{2^{19}.3^9\left(1+2.3\right)}\)

\(=\frac{2^{18}.3^9.7}{2^{19}.3^9.7}=\frac{1}{2}\)