\(C=\dfrac{\sqrt{12}+3}{\sqrt{3}}=2+\sqrt{3}\)

Ta có: \(A=\left(x-3\right)^2+\left(11-x\right)^2\)

\(=x^2-6x+9+x^2-22x+121\)

\(=2x^2-28x+130\)

\(=2\left(x^2-14x+49+16\right)\)

\(=2\left(x-7\right)^2+32\ge32\forall x\)

Dấu '=' xảy ra khi x=7

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

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

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

\(=0\)

d) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2\)

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

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

\(=0+6\left(x+1\right)^2\)

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

c. (x - 1)³ - (x - 1)(x² + x + 1)

= (x - 1)³ - (x³ - 1)

= x³ - 2x² + 2x - 1 - x³ + 1

= x³ - x³ - 2x² + 2x - 1 + 1

= -2x² + 2x

d. (x - 3)³ - (x - 3)(x² + 3x + 9) + 6(x + 1)²

= x³ - 6x² + 18x - 27 - (x³ - 27) + 6(x² + 2x + 1)

= x³ - 6x² + 18x - 27 - x³ + 27 + 6x² + 12x + 6

= x³ - x³ - 6x² + 6x² + 18x + 12x - 27 + 27 + 6

= 30x + 6

\(a.\dfrac{7}{8}+\dfrac{1}{2}\text{=}\dfrac{11}{8}\)

\(b.\dfrac{5}{6}+\left(-2\right)\text{=}\dfrac{-7}{6}\)

\(c.\dfrac{2}{5}+\dfrac{-3}{8}\text{=}\dfrac{1}{40}\)

\(d.\dfrac{5}{7}-\dfrac{3}{8}\text{=}\dfrac{19}{56}\)

\(e.\dfrac{3}{4}-\dfrac{1}{2}+\dfrac{7}{6}\text{=}\dfrac{17}{12}\)

`a. 7/8 + 1/2 = 7/8 + 4/8=(7+4)/8=11/8`

`b. 5/6 + (-2) =5/6 + (-12/6)=-7/6`

`c. 2/5 + -3/8   = 16/40 + (-15/40)=1/40`

`d. 5/7  - 3/8  = 40/56 - 21/56=19/56`

`e. 3/4 - 1/2 +7/6= 3/4 - 2/4 + 7/6= 1/4 + 7/6= 6/24 + 28/24= 34/24=17/12`

1:

a: =>(|x|+4)(|x|-1)=0

=>|x|-1=0

=>x=1; x=-1

b: =>x^2-4>=0

=>x>=2 hoặc x<=-2

d: =>|2x+5|=2x-5

=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0

=>x=0(loại)

\(\left(x+1\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=2\)

\(\Leftrightarrow x^2+4x+3-x^2-3x+10=2\)

\(\Leftrightarrow x=-11\)

\(a,x+\dfrac{1}{4}=\dfrac{5}{8}\)

\(\Leftrightarrow x=\dfrac{5}{8}-\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{5}{8}-\dfrac{2}{8}\)

\(\Leftrightarrow x=\dfrac{3}{8}\)

\(b,x-\dfrac{3}{5}=\dfrac{1}{10}\)

\(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{3}{5}\)

\(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{6}{10}\)

\(\Leftrightarrow x=\dfrac{7}{10}\)

\(c,x\times\dfrac{2}{7}=\dfrac{6}{11}\)

\(\Leftrightarrow x=\dfrac{6}{11}:\dfrac{2}{7}\)

\(\Leftrightarrow x=\dfrac{6}{11}\times\dfrac{7}{2}\)

\(\Leftrightarrow x=\dfrac{42}{22}\)

\(\Leftrightarrow x=\dfrac{21}{11}\)

\(d,x:\dfrac{3}{2}=\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{1}{4}\times\dfrac{3}{2}\)

\(\Leftrightarrow x=\dfrac{3}{8}\)

\(đ,\dfrac{7}{4}-x=\dfrac{5}{7}\)

\(\Leftrightarrow x=\dfrac{7}{4}-\dfrac{5}{7}\)

\(\Leftrightarrow x=\dfrac{49}{28}-\dfrac{20}{28}\)

\(\Leftrightarrow x=\dfrac{29}{28}\)

\(e,\dfrac{5}{4}:x=\dfrac{1}{8}\)

\(x=\dfrac{5}{4}:\dfrac{1}{8}\)

\(\Leftrightarrow x=\dfrac{5}{4}\times8\)

\(\Leftrightarrow x=\dfrac{40}{4}\)

\(\Leftrightarrow x=10\)

N = 101 - 100 - 99 - 98 - ... - 3 - 2 - 1 ( có 101 số hạng )

N = ( 101 - 100 ) - ( 99 - 98 ) - ... - ( 3 - 2 ) - 1 ( có 50 cặp và số 1 )

N = 1 - 1 - ... - 1 - 1 ( có 50 số -1 và 1 số +1 )

N = 1 - ( -1 x 50 )

N = 1 - ( - 50 )

N = 1 + 50

N = 51