\(a,=\left(4x^2-25\right)-\left(6x^2+9x-4x-6\right)\\ =4x^2-25-6x^2-5x+6=-2x^2-5x-19\\ b,=4x^2-4x+1-4\left(x^2-4\right)\\ =4x^2-4x+1-4x^2+16\\ =-4x+17\)

a: \(=x^2-x^3-2+2x+x^3+27=x^2+2x+25\)

b: \(=\dfrac{2x^4-2x^3+2x^2+3x^3-3x^2+3x-2x^2+2x-2-x-1}{x^2-x+1}\)

\(=2x^2+3x-2+\dfrac{-x-1}{x^2-x+1}\)

b: \(=2x^2-3x+10x-15=2x^2+7x-15\)

\(a,=\left(6x^3+3x^2-10x^2-5x+4x+2\right):\left(2x+1\right)\\ =\left[3x^2\left(2x+1\right)-5x\left(2x+1\right)+2\left(2x+1\right)\right]:\left(2x+1\right)\\ =3x^2-5x+2\\ b,Sửa:\left(2x^3-21x^2+67x-60\right):\left(x-5\right)\\ =\left(2x^3-10x^2-11x^2+55x+12x-60\right):\left(x-5\right)\\ =\left[2x^2\left(x-5\right)-11x\left(x-5\right)+12\left(x-5\right)\right]:\left(x-5\right)\\ =2x^2-11x+12\)

a)\({\left( {\frac{2}{5} + \frac{1}{2}} \right)^2} = {\left( {\frac{4}{{10}} + \frac{5}{{10}}} \right)^2} = {\left( {\frac{9}{{10}}} \right)^2} = \frac{{81}}{{100}}\);               

 b)\({\left( {0,75 - 1\frac{1}{2}} \right)^3} = {\left( {\frac{3}{4} - \frac{3}{2}} \right)^3} = {\left( {\frac{3}{4} - \frac{6}{4}} \right)^3} = {\left( { - \frac{3}{4}} \right)^3} = \frac{{ - 27}}{{64}};\)

c)

\(\begin{array}{l}{\left( {\frac{3}{5}} \right)^{15}}:{\left( {0,36} \right)^5} = {\left( {\frac{3}{5}} \right)^{15}}:{\left( {\frac{9}{{25}}} \right)^5}\\ = {\left( {\frac{3}{5}} \right)^{15}}:{\left[ {{{\left( {\frac{3}{5}} \right)}^2}} \right]^5} = {\left( {\frac{3}{5}} \right)^{15}}:{\left( {\frac{3}{5}} \right)^{10}} = {\left( {\frac{3}{5}} \right)^5}\end{array}\)

d) \({\left( {1 - \frac{1}{3}} \right)^8}:{\left( {\frac{4}{9}} \right)^3} = {\left( {\frac{3}{3} - \frac{1}{3}} \right)^8}:{\left( ({\frac{2}{3}})^2 \right)^3}\\= {\left( {\frac{2}{3}} \right)^8}:{\left( {\frac{2}{3}} \right)^6} = {\left( {\frac{2}{3}} \right)^{8-6}}\\= {\left( {\frac{2}{3}} \right)^2} = \frac{4}{9}\)

a,\(\left(\sqrt{1\dfrac{9}{16}}-\sqrt{\dfrac{9}{16}}\right):5=\left(\sqrt{\dfrac{25}{16}}-\dfrac{3}{4}\right):5=\left(\dfrac{5}{4}-\dfrac{3}{4}\right):5\)

\(=\dfrac{1}{2}:5=\dfrac{1}{10}\)

b,\(\left(\sqrt{3}-2\right)^2\left(\sqrt{3}+2\right)^2=\left[\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\right]^2\)

\(=\left[3-4\right]^2=1\)

c,\(\left(11-4\sqrt{3}\right)\left(11+4\sqrt{3}\right)=11^2-\left(4\sqrt{3}\right)^2\)

\(=121-48=73\)

d,\(\left(\sqrt{2}-1\right)^2-\dfrac{3}{2}\sqrt{\left(-2\right)^2}+\dfrac{4\sqrt{2}}{5}+\sqrt{1\dfrac{11}{25}}.\sqrt{2}\)

\(=2-2\sqrt{2}+1-3+\dfrac{4\sqrt{2}}{5}+\sqrt{\dfrac{36}{25}.2}\)

\(=-2\sqrt{2}+\dfrac{4\sqrt{2}+6\sqrt{2}}{5}\)

\(=-2\sqrt{2}+\dfrac{10\sqrt{2}}{5}=-2\sqrt{2}+2\sqrt{2}=0\)

e,\(\left(1+\sqrt{2021}\right)\sqrt{2022-2\sqrt{2021}}\)

\(=\left(1+\sqrt{2021}\right)\sqrt{2021-2\sqrt{2021}.1+1}\)

\(=\left(1+\sqrt{2021}\right)\sqrt{\left(\sqrt{2021}-1\right)^2}\)

\(=\left(1+\sqrt{2021}\right)\left(\sqrt{2021}-1\right)\)

\(=\sqrt{2021}-1+\sqrt{2021^2}-\sqrt{2021}=2020\)

b: \(=16-2\cdot4\cdot2\sqrt{5}+20-9-4\sqrt{5}\)

=27-20căn 5

a: 2-4căn 3<0

nên biểu thức ko có giá trị

\(b,\left(4-2\sqrt{5}\right)^2-\left(\sqrt{5}+2\right)^2\\ =\left[\left(4-2\sqrt{5}\right)-\left(\sqrt{5}+2\right)\right].\left[\left(4-2\sqrt{5}\right)+\left(\sqrt{5}+2\right)\right]=\left(2-3\sqrt{5}\right)\left(6-\sqrt{5}\right)\)

`b)(4-2\sqrt5)^2-(\sqrt5+2)^2`

`=(4-2\sqrt5-\sqrt5-2)(4-2\sqrt5+\sqrt5+2)`

`=(2-3\sqrt5)(6-\sqrt5)`

$---------$

Áp dụng HĐT :

`a^2-b^2=(a-b)(a+b)`

a) \(\left( { - 3} \right).\left( { - 2} \right).\left( { - 5} \right).4\)\( = \left[ {\left( { - 3} \right).\left( { - 2} \right)} \right].\left( { - 5} \right).4\)\( = 6.\left( { - 5} \right).4 =  - 30.4 =  - 120\).

b) \(3.2.\left( { - 8} \right).\left( { - 5} \right)\)\( = 3.2.\left[ {\left( { - 8} \right).\left( { - 5} \right)} \right] = 6.40\)\( = 240\).

a)

\(\left( { - 3,5} \right).\left( {1\frac{3}{5}} \right) = \frac{{ - 7}}{2}.\frac{8}{5} = \frac{{ - 7.8}}{{2.5}} = \frac{{ - 7.4.2}}{{2.5}} = \frac{{ - 28}}{5}\)                  

b) \(\frac{{ - 5}}{9}.\left( { - 2\frac{1}{2}} \right) = \frac{{ - 5}}{9}.\frac{{ - 5}}{2} = \frac{{25}}{{18}}\)