a. \(\dfrac{3}{4}\) + \(\dfrac{-4}{5}\) - \(\dfrac{1}{2}\) = \(\dfrac{-1}{20}\)  - \(\dfrac{1}{2}\) = \(\dfrac{-11}{20}\) 

b. (4 - \(\dfrac{5}{12}\) ): 2 + \(\dfrac{5}{24}\) 

= \(\dfrac{43}{12}\) : 2 + \(\dfrac{5}{24}\) 

= \(\dfrac{43}{24}\) + \(\dfrac{5}{24}\) 

=\(\dfrac{48}{24}\) = 2

2.2 

\(\dfrac{4}{7}\) .x - \(\dfrac{2}{3}\) = \(\dfrac{1}{5}\) 

  \(\dfrac{4}{7}\) x   = \(\dfrac{1}{5}\) + \(\dfrac{2}{3}\) 

  \(\dfrac{4}{7}\) . x = \(\dfrac{13}{15}\) 

        x = \(\dfrac{91}{60}\)

2.1

\(a)\dfrac{3}{4}+\dfrac{-4}{5}-\dfrac{1}{2}\\ =\dfrac{3\times5}{4\times5}+\dfrac{-4\times4}{5\times4}-\dfrac{1\times10}{2\times10}\\ =\dfrac{15}{20}+\dfrac{-16}{20}-\dfrac{10}{20}\\ =\dfrac{15-16-10}{20}\\ =\dfrac{-11}{20}\)

\(b)\left(4-\dfrac{5}{12}\right):2+\dfrac{5}{24}\\ =\left(\dfrac{4\times12}{1\times12}-\dfrac{5}{12}\right):2+\dfrac{5}{24}\\ =\left(\dfrac{48}{12}-\dfrac{5}{12}\right):2+\dfrac{5}{24}\\ =\left(\dfrac{48-5}{12}\right):2+\dfrac{5}{24}\\ =\dfrac{43}{12}:2+\dfrac{5}{24}\\ =\dfrac{43}{12}\times\dfrac{1}{2}+\dfrac{5}{24}\\ =\dfrac{43}{24}+\dfrac{5}{24}\\ =\dfrac{43+5}{24}\\ =\dfrac{48}{24}\\ =2\)

\(2.2\\ \dfrac{4}{7}\times x-\dfrac{2}{3}=\dfrac{1}{5}\\ \dfrac{4}{7}\times x=\dfrac{1}{5}+\dfrac{2}{3}\\ \dfrac{4}{7}\times x=\dfrac{1\times3}{5\times3}+\dfrac{2\times5}{3\times5}\\ \dfrac{4}{7}\times x=\dfrac{3}{15}+\dfrac{10}{15}\\ \dfrac{4}{7}\times x=\dfrac{13}{15}\\ x=\dfrac{13}{15}:\dfrac{4}{7}\\ x=\dfrac{13}{15}\times\dfrac{7}{4}\\ x=\dfrac{91}{60}\)

a: \(x^2-9-x^2\left(x^2-9\right)\)

\(=\left(x^2-9\right)-x^2\left(x^2-9\right)\)

\(=\left(x^2-9\right)\left(1-x^2\right)\)

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

b: \(x^2\left(x-y\right)+y^2\left(y-x\right)\)

\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)

\(=\left(x-y\right)\left(x^2-y^2\right)\)

\(=\left(x-y\right)\left(x-y\right)\left(x+y\right)=\left(x-y\right)^2\cdot\left(x+y\right)\)

c: \(x^3+27+\left(x+3\right)\left(x-9\right)\)

\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)

\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)\)

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

d: \(x^2+5x+6\)

\(=x^2+2x+3x+6\)

\(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)

e: \(3x^2-4x-4\)

\(=3x^2-6x+2x-4\)

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

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

g: \(x^4+64y^4\)

\(=x^4+16x^2y^2+64y^4-16x^2y^2\)

\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)

\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)

h: \(a^2+b^2+2a-2b-2ab\)

\(=a^2-2ab+b^2+2a-2b\)

\(=\left(a-b\right)^2+2\left(a-b\right)=\left(a-b\right)\left(a-b+2\right)\)

i: \(\left(x+1\right)^2-2\left(x+1\right)\left(y-3\right)+\left(y-3\right)^2\)

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

\(=\left(x-y+4\right)^2\)

k: \(x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)\)

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

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

a) \(\dfrac{5}{3}+\left(-\dfrac{7}{13}\right)+\left(-\dfrac{1}{2}\right)\cdot2\)

\(=\dfrac{5}{3}-\dfrac{7}{13}-1\)

\(=\dfrac{65}{39}-\dfrac{21}{39}-\dfrac{39}{39}\)

\(=\dfrac{5}{39}\)

b) \(\dfrac{1}{7}\cdot\dfrac{3}{9}+\left(-\dfrac{13}{9}\right)\cdot\dfrac{1}{7}\cdot\dfrac{4}{9}\)

\(=\dfrac{1}{21}-\dfrac{52}{567}=-\dfrac{25}{567}\)

a: =20/12-7/12+1/4

=13/12+3/12

=16/12

=4/3

b: =1/7(3/9-13/9*4/9)

=1/7*(3/9-52/81)

=-25/567

a: Ta có: \(-\left(-3x^2\right)^3+4x-9-27x^6\)

\(=27x^6-27x^6+4x-9\)

=4x-9

=-1

Lời giải:
a.

$A=(x+6)^2-(x+2)^2+2[(x-5)^2-(x-3)^2]$

$=(x+6-x-2)(x+6+x+2)+2[(x-5-x+3)(x-5+x-3)]$

$=4(2x+8)+2(-2)(2x-8)$

$=4(2x+8)-4(2x-8)=4[(2x+8)-(2x-8)]=4.16=64$ không phụ thuộc vào $x$

b.

$B=(x^3-2^3)-(x^3+2^3)=-16$ không phụ thuộc vào $x$

c.

$C=x^4+2x^2-[(x^2+3)^2-(2x)^2]$

$=x^4+2x^2-(x^4+6x^2-4x^2)$

$=x^4+2x^2-(x^4+2x^2)=0$ không phụ thuộc vào $x$

a) Ta có: \(A=\left(x+6\right)^2+2\left(x-5\right)^2-\left(x+2\right)^2-2\left(x-3\right)^2\)

\(=x^2+12x+36+2\left(x^2-10x+25\right)-\left(x^2+4x+4\right)-2\left(x^2-6x+9\right)\)

\(=x^2+12x+36+2x^2-20x+50-x^2-4x-4-2x^2+12x-18\)

\(=34\)

b) Ta có: \(B=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+2\right)\left(x^2-2x+4\right)\)

\(=x^3-8-x^3-8\)

=-16

c) Ta có: \(C=x^4+2x^2-\left(x^2-2x+3\right)\left(x^2+2x+3\right)\)

\(=x^4+2x^2-\left[\left(x^2+3\right)^2-4x^2\right]\)

\(=x^4+2x^2-\left(x^4+6x^2+9\right)+4x^2\)

\(=-9\)

a, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x^2+x-2\right)=0\)

\(\Leftrightarrow\left(x+1\right)=0;3x^2+x-2=0\)

=> x=-1  

với \(3x^2+x-2=0\)

ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)

Vậy  ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)

b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)

\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)

\(\Leftrightarrow3x^2=3\)

hay \(x\in\left\{1;-1\right\}\)

c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)

hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)