Có 12 số 0,mà chỉ có 10 số 9.

1000000000000 - 9999999999 = 990000000001

990000000001

Ơ,thế đề bài là \(4-\frac{1}{3}=x\) à?

\(4-\frac{1}{3}=\frac{4\times3-1}{3}=\frac{11}{3}\)

Gà rén là GARENA

14x + 54 = 82

14x = 28

x = 2

175 + (30 - x) = 2001

30 - x = 1826

x = 30 - 1826

x = - 1796

1551 - 10 ( x + 1 ) = 55

10(x + 1) = 1496

x + 1 = 149,6

x = 148,6

5 . (x + 12) + 22 = 92

5 . (x + 12) = 70

x + 12 = 14

x = 2

6 . (x + 23) + 40 = 100

6 . (x + 23) = 60

x + 23 = 10

x = - 13

\(14x+54=82\)

\(\Leftrightarrow14x=82-54\)

\(\Leftrightarrow14x=28\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\)

\(175+\left(30-x\right)=2001\)

\(\Leftrightarrow30-x=2001-175\)

\(\Leftrightarrow30-x=1826\)

\(\Leftrightarrow x=-1796\)

Vậy \(x=-1796\)

\(1551-10\left(x+1\right)=55\)

\(\Leftrightarrow10.\left(x+1\right)=1551-55\)

\(\Leftrightarrow10.\left(x+1\right)=1496\)

\(\Leftrightarrow x+1=\frac{1496}{10}\)

\(\Leftrightarrow x=\frac{1486}{10}\)

Vậy \(x=\frac{1486}{10}\)

\(5.\left(x+12\right)+22=92\)

\(\Leftrightarrow5.\left(x+12\right)=70\)

\(\Leftrightarrow x+12=14\).

\(\Leftrightarrow x=2\)

Vậy \(x=2\)

\(6.\left(x+23\right)+40=100\)

\(\Leftrightarrow6.\left(x+23\right)=60\)

\(\Leftrightarrow x+23=10\)

\(\Leftrightarrow x=-13\)

Vậy \(x=-13\)

\(\left(\frac{6}{7}+\frac{1}{4}\right):\left(\frac{19}{14}-\frac{1}{4}\right)\)

\(=\frac{31}{28}:\frac{31}{28}\)

\(=1\)

125 = 125^2-1 ; - 125 = - 125^1009-1008 ; 27 = 27^{10000000000-9999999999} ; - 27 = - 27^{\(\infty+1-\infty\)}             :)

B = 64 . 16 + 81 . 84 + 17 . 16

B = (64 + 17) . 16 + 81 . 84

B = 81 . 16 + 81 . 84

B = 81 . (16 + 84)

B = 81 . 100

B = 8100

B = ( 64 . 16 + 17 . 16 ) + 81 . 84

 = 16 . ( 64 + 17 ) + 81 . 84

 = 16 . 81 + 81 . 84

 = 81 . ( 16 + 64 )

 = 81 . 100

 = 8100

\(x\times\frac{1}{2}+\frac{3}{2}\times x=\frac{4}{5}\)

\(x\times\left(\frac{1}{2}+\frac{3}{2}\right)=\frac{4}{5}\)

\(x\times2=\frac{4}{5}\)

\(x=\frac{4}{5}:2\)

\(x=\frac{2}{5}\)

\(x.\frac{1}{2}+\frac{3}{2}.x=\frac{4}{5}\)

\(x.\left(\frac{1}{2}+\frac{3}{2}\right)=\frac{4}{5}\)

\(x.2=\frac{4}{5}\)

\(x=\frac{4}{5}.2\)

\(x=\frac{2}{5}\)

1) \(x-2y=3\Rightarrow\hept{\begin{cases}x=3+2y\\y=\frac{x-3}{2}\end{cases}}\)

\(\Rightarrow A=2x\left(x+2y-3\right)-y\left(6x-3y-10\right)+x-7+\left(x-3y\right)^2\)

\(=2x^2+4xy-6x-6xy+3y^2+10y+x-7+x^2-6xy+9y^2\)

\(=3x^2+12y^2-8xy-5x+10y-7\)

\(=3.\left(3+2y\right)^2+12y^2-8\left(3+2y\right).y-5\left(3+2y\right)+10y-7\)

\(=3\left(9+12y+4y^2\right)+12y^2-8\left(3y+2y^2\right)-15-10y+10y-7\)

\(=27+36y+12y^2+12y^2-24y-16y^2-15-10y+10y-7\)

\(=8y^2+12y+5\)

\(M=\left(x^2-2x+1\right)\left(1+2x\right)-\left(x^2+2x+1\right)\left(1-3x\right)-\left(3-6x\right)\left(x^2+3x+2\right)\)

\(=x^2+2x^3-2x-4x^2+1+2x-x^2+3x^8-2x+6x^2-1+3x-3x^2-9x-6+6x^8\)\(+18x^2+12x=11x^3+17x^2+4x-6\)