1: Sửa đề: 3x-5

\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)

2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

=5x^2+14x^2+12x+8

3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)

4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)

5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)

    \(\dfrac{7}{6}\) + \(\dfrac{5}{12}\) - \(\dfrac{1}{18}\) - 1 

=  \(\dfrac{42}{36}\) + \(\dfrac{15}{36}\) - \(\dfrac{2}{36}\) - \(\dfrac{36}{36}\)

= \(\dfrac{19}{36}\)

\(\dfrac{13}{6}\) + \(\dfrac{5}{18}+3-\dfrac{7}{12}\)

= \(\dfrac{78}{36}+\dfrac{10}{36}+\dfrac{108}{36}-\dfrac{21}{36}\)

= \(\dfrac{175}{36}\)

3 + \(\dfrac{11}{4}-\dfrac{1}{12}-\dfrac{3}{16}\)

= \(\dfrac{144}{48}+\dfrac{132}{48}-\dfrac{4}{48}-\dfrac{9}{48}\)

= \(\dfrac{263}{48}\)

   \(\dfrac{2}{5}\) x \(\dfrac{7}{4}\) - \(\dfrac{2}{5}\) x \(\dfrac{3}{4}\)

=  \(\dfrac{2}{5}\) x ( \(\dfrac{7}{4}\) - \(\dfrac{3}{4}\))

= \(\dfrac{2}{5}\)  x \(\dfrac{4}{4}\)

= \(\dfrac{2}{5}\)

21:

a: \(f\left(x\right)=4x^4-x^3-4x^2+x-1\)

\(g\left(x\right)=x^4+4x^3+x-5\)

b: f(x)-g(x)

=4x^4-x^3-4x^2+x-1-x^4-4x^3-x+5

=3x^4-5x^3-4x^2+4

f(x)+g(x)

=4x^4-x^3-4x^2+x-1+x^4+4x^3+x-5

=5x^4+3x^3-4x^2+2x-6

c: g(-1)=1-4-1-5=-9

4 + 4 x 2 + 4 x 3 + 4 x 4 = ( 4 x 2 + 4 x 3 ) + (4 x 4 + 4) 

                                      =   20 + 20 

                                        = 40 

5 + 5 x 2 + 5 x 3 + 5 x 4 = (5 x 4 + 5 x 2 ) + ( 5 x 3 + 5 )

                                      = 30 + 20 

                                     = 50

x=4

=>x+1=5

A=(x+1)x^5 -(x+1)x^4+(x+1)x^3-(x+1)x^2+(x+1)x-1

  =x^6+x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2-x+1

  =x^6-x-1

  =4^6-4-1

  =4091

\(a,A=5\cdot4^5-5\cdot4^4+5\cdot4^3-5\cdot4^2+5\cdot4+1\\ A=4^4\left(20-5\right)+4^2\left(20-5\right)+\left(20-5\right)\\ A=15\left(4^4+4^2+1\right)=15\cdot273=4095\)

\(b,x=7\Leftrightarrow x+1=8\\ \Leftrightarrow B=x^{2006}-\left(x+1\right)x^{2005}+\left(x+1\right)x^{2004}-...+\left(x+1\right)x^2-\left(x+1\right)x-5\\ B=x^{2006}-x^{2006}-x^{2005}+x^{2005}+x^{2004}-...+x^3+x^2-x^2-x-5\\ B=-x-5=-12\)

b)tương tự

=x^2006-x^2006-x^2005+x^2005+x^2004-...+x^3-x^2-x^2-x-5

=-x-5

=-7-5=-12

4/5x3/7+4/5+6/7x6/7-4/5x4/14

= ( 3/7 + 6/7 + 4/14) x 4/5 

= 11/7 x 4/5 

= 44/35

44/35

4/5x3/7+4/5+6/7x6/7-4/5x4/14 ⇒ ( 3/7 + 6/7 + 4/14) x 4/5 ⇒ 11/7 x 4/5 ⇒ 44/35

a) C1 : 2/5 x 3/7 + 2/7 x 4/7

         = 2/5 x (3/7 + 4/7)

        = 2/5 x 1

        = 2/5

C2 : 2/5 x 3/7 + 2/5 x 4/7

= 6/35 + 8/35

= 2/5 

b) (2/3 - 4/7) x 5/5

= 2/21 x 5/5

= 2/21

C2 : (2/3 - 4/7) x 5/5

= 2/3 x 1 - 4/7 x 1

= 2/3 - 4/7

= 2/21

c) 3/4 x 2/5 - 3/4 x 2/7

= 3/4 x (2/5 - 2/7)

= 3/4 x 4/35

= 3/35

C2 : 3/4 x 2/5 - 3/4 x 2/7

= 3/10 - 3/14

= 3/35

a, 

Cách 1

\(\frac{2}{5}\times\frac{3}{7}+\frac{2}{5}\times\frac{4}{7}\)

\(=\frac{6}{35}+\frac{8}{35}\)

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

Cách 2 :

\(\frac{2}{5}\times\frac{3}{7}+\frac{2}{5}\times\frac{4}{7}\)

\(=\frac{2}{5}\times\left(\frac{3}{7}+\frac{4}{7}\right)\)

\(=\frac{2}{5}\times1\)

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

b,

Cách 1 :

\(\left(\frac{2}{3}-\frac{4}{7}\right)\times\frac{5}{5}\)

\(=\frac{2}{21}\times1\)

\(=\frac{2}{21}\)

Cách 2 :

\(\left(\frac{2}{3}-\frac{4}{7}\right)\times\frac{5}{5}\)

\(=\frac{2}{3}\times\frac{5}{5}-\frac{4}{7}\times\frac{5}{5}\)

\(=\frac{2}{3}-\frac{4}{7}\)

\(=\frac{2}{21}\)

c, 

Cách 1 :

\(\frac{3}{4}\times\frac{2}{5}-\frac{3}{4}\times\frac{2}{7}\)

\(=\frac{3}{10}-\frac{3}{14}\)

\(=\frac{3}{35}\)

Cách 2 :

\(\frac{3}{4}\times\frac{2}{5}-\frac{3}{4}\times\frac{2}{7}\)

\(=\frac{3}{4}\times\left(\frac{2}{5}-\frac{2}{7}\right)\)

\(=\frac{3}{4}\times\frac{4}{35}\)

\(=\frac{3}{35}\)

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