
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.


11 x 10 - 10 x 9 + 9 x 8 - 8 x 7 + 7 x 6 - 6 x 5 + 5 x 4 - 4 x 3 + 3 x 2 - 2 x 1 = 60


x + ( 9 - 8 + 7 - 6 + 5 - 4 + 3 ) = 90,28
x + 6 = 90,28
x = 90,28 - 6
x = 84,28

Vế trái có tất cả (180 - 6)/3 + 1 = 59 số hạng trong ngoặc.
(x + x + ... + x) + ( 6 + 9 + .. + 180) = 6372
59.x + 3.(2 + 3 + .. + 60) = 6372
Tính riêng 2 + 3 + .. + 60 ta có:
2 + 3 + .. + 60
= (1 + 2 + 3 + .. + 60) - 1 (thêm 1 và bớt 1 thì tổng không đổi)
= (1 + 60) + (2+59) + .. + (30 + 31) - 1
= 61 + 61 + .. + 61 -1
= 61 x 30 - 1
= 1829
Thay vào phương trình:
59 . x + 3 . 1829 = 6372
59 . x + 5487 = 6372
59 . x = 6372 - 5487
59 . x = 885
x = 885 : 59
x = 15

\(9+\frac{x}{13}-x=\frac{5}{6}\)
\(\Leftrightarrow\frac{x}{13}-x=\frac{5}{6}-9\)
\(\Leftrightarrow x\left(\frac{1}{13}-1\right)=\frac{5}{6}-\frac{54}{6}\)
\(\Leftrightarrow\frac{-12}{13}x=\frac{-50}{6}\)
\(\Leftrightarrow x=\frac{-50}{6}\div\frac{-12}{13}\)
\(\Leftrightarrow x=\frac{50}{6}\times\frac{13}{12}\)
\(\Leftrightarrow x=\frac{650}{72}=\frac{325}{36}\)
\(9+\frac{x}{13}-x=\frac{5}{6}\)
\(\Rightarrow9+\frac{x}{13}-\frac{13x}{13}=\frac{5}{6}\)
\(\Rightarrow9+\frac{x-13x}{13}=\frac{5}{6}\)
\(\Rightarrow\frac{x-13x}{13}=\frac{5}{6}-9\)
\(\Rightarrow\frac{x-13x}{13}=-\frac{49}{6}\)
\(\Rightarrow x-13x=-\frac{49}{6}\cdot13=-\frac{637}{6}\)
\(\Rightarrow-12x=-\frac{637}{6}\)
\(\Rightarrow x=-\frac{637}{6}:(-12)=\frac{637}{72}\)

( x+x+x+x+x+x) + (9-8+7-6+5-4)=63,6
x x 6 +3=63,6
x x 6 = 63,6-3
x x 6 =60,6
x =60,6:6
x =10,1
nha bn
cam on bn nhieu
Ta có:
(x + 9) + (x - 8) + (x + 7) + (x - 6) + (x + 5) + (x - 4) = 63,6
x + 9 + x - 8 + x + 7 + x - 6 + x + 5 + x - 4 = 63,6
(x + x + x + x + x + x) + (9 - 8 + 7 - 6 + 5 - 4) = 63,6
6x + 3 = 63,6
6x = 63,6 - 3
6x = 60,6
x = 60,6 : 6
x = 10,1
Vậy x = 10,1

\(3+6+9+...+x=1785\)
\(\left[\left(x-3\right):3+1\right].\left(x+3\right):2=1785\)
\(\left[\left(x-3\right):3+1\right].\left(x+3\right)=3570\)
\(=>...\)
Đặt A = 3 + 6 + 9 + ...... + x = 1785
A có số số hạng là :
( x - 3 ) : 3 + 1 = \(\frac{x}{3}\)( số hạng )
Giá trị của A là :
( x + 3 ) .\(\frac{x}{3}\): 2 = 1785
=> \(\left(\frac{x^2}{3}+x\right)\): 2 = 1785
=> \(\frac{x^2}{3}+x\)= 3570
=> x2 + 3x = 10710
=> x ( x + 3 ) = 10710
=> x. ( x + 3 ) = 102.105
=> x. ( x + 3 ) = 102. ( 102 + 3 )
=> x = 102
Vậy x = 102

(x + 9) + (x - 8) + (x + 7) + (x - 6) + (x + 5) + (x - 4) + (x + 3) + (x - 2) + (x -1) = 95.9
=> (x + 9) + (x - 8) + (x + 7) + (x - 6) + (x + 5) + (x - 4) + (x + 3) + (x - 2) + (x -1) = 855
=> x + 9 + x - 8 + x + 7 + x - 6 + x + 5 + x - 4 + x + 3 + x - 2 + x -1 = 855
=> ( x + x + x + x + x + x + x + x + x ) + ( 9 - 8 + 7 - 6 + 5 - 4 + 3 - 2 - 1 ) = 855
=> 9x + 3 = 855
=> 9x = 855 - 3
=> 9x = 852
=> x = 852 : 9
=> x = \(\frac{284}{3}\)

1> a) \(\frac{5}{7}x4:\frac{5}{9}=\frac{5}{7}:\frac{5}{9}x4=\frac{5}{7}x\frac{9}{5}x4=\frac{9}{7}x4=\frac{9x4}{7}=\frac{36}{7}\)
\(b,8x\frac{2}{3}:\frac{1}{2}=8x\frac{2}{3}x\frac{2}{1}=8x2x\frac{2}{3}=16x\frac{2}{3}=\frac{32}{3}\)
\(c,6:\frac{3}{5}-\frac{7}{6}x\frac{6}{7}=6x\frac{5}{3}-1=10-1=9\)
\(\frac{21}{5}x\frac{10}{11}+\frac{57}{11}=\frac{42}{11}+\frac{57}{11}=\frac{99}{11}=9\)
2) a) \(\frac{35}{9}:x=\frac{35}{6}\)
\(x=\frac{35}{9}:\frac{35}{6}\)
\(x=\frac{35}{9}x\frac{6}{35}\)
\(x=\frac{2}{3}\)
b) \(\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}\right)x10-X=0\)
\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{5}-\frac{1}{6}\right)x10-X=0\)
\(\left(\frac{1}{1}-\frac{1}{6}\right)x10-X=10\)
\(\frac{5}{6}x10-X=0\)
\(X=\frac{5}{6}x10=\frac{25}{3}\)
Đúng nha !!!!
1/a/\(\frac{5}{7}\cdot4:\frac{5}{9}=\frac{20}{7}:\frac{5}{9}=\frac{20}{7}\cdot\frac{9}{5}=\frac{36}{7}\)
b/\(8\cdot\frac{2}{3}:\frac{1}{2}=\frac{16}{3}:\frac{1}{2}=\frac{16}{3}\cdot\frac{2}{1}=\frac{32}{3}\)
c/\(6:\frac{3}{5}-\frac{7}{6}\cdot\frac{6}{7}=6\cdot\frac{5}{3}-1=10-1=9\)
2/a/\(\frac{35}{9}:x=\frac{35}{6}\)
\(x=\frac{35}{9}:\frac{35}{6}=\frac{35}{9}\cdot\frac{6}{35}\)
\(x=\frac{2}{3}\)
b/\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\right)\cdot10-x=0\)
\(\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)\cdot10-x=0\)
\(\left(\frac{30}{60}+\frac{10}{60}+\frac{5}{60}+\frac{2}{30}\right)\cdot10-x=0\)
\(\frac{47}{60}\cdot10-x=0\)
\(\frac{47}{6}-x=0\)
\(x=\frac{47}{6}-0\)
\(x=\frac{47}{6}\)
\(\frac{x}{9}.x=6\)
\(\frac{x}{9}.\frac{x}{1}=6\)
\(\Rightarrow\frac{x^2}{9}=6\)
\(\Rightarrow x^2:9=6\)
\(\Rightarrow x^2=54\)
\(\Rightarrow\sqrt[2]{54}\)
\(\Rightarrow x=7,348469228\)hoặc x không tồn tại nếu thử \(7,348469228^2=53,9999999999\)thôi
vậy x không tồn tại
x/9 * x = 6
=> \(\frac{x^2}{9}=6\)
=> \(\frac{x^2-54}{9}=0\)
\(x^2=54\)
=> x =
Bn chỉ cần tìm \(x^2\)sao cho = 54 là được