Đáp án:

\(\begin{array}{l}
1,\\
a,\\
 - 1\\
b,\\
64\\
2,\\
49;\,50;\,51\\
3,\\
10000\\
4,\\
a,\\
6.\left( {4x - 3y + 5} \right)\\
b,\\
7.\left( {y - 8} \right).\left( {5x + 2y} \right)\\
c,\\
\left( {x - y + 7} \right).\left( {x + y + 7} \right)\\
d,\\
\left( {5x - \sqrt 2 } \right)\left( {5x + \sqrt 2 } \right)\\
e,\\
\left( {x - y} \right)\left( {x + y + 5} \right)\\
5,\\
a,\\
\dfrac{3}{4}.{x^2}y\\
b,\\
\dfrac{{x + 5}}{{2x}}\\
c,\\
\dfrac{{x + 2}}{{x + 4}}\\
d,\\
\dfrac{{x - 3}}{{x + 3}}
\end{array}\)

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
a,\\
 - 5x\left( {5x - 2} \right) + \left( {5x + 1} \right)\left( {5x - 1} \right) - 10x\\
 = \left( { - 25{x^2} + 10x} \right) + {\left( {5x} \right)^2} - {1^2} - 10x\\
 =  - 25{x^2} + 10x + 25{x^2} - 1 - 10x\\
 = \left( { - 25{x^2} + 25{x^2}} \right) + \left( {10x - 10x} \right) - 1\\
 =  - 1\\
b,\\
\left( {x - 8} \right)\left( {x - 4} \right) - x\left( {x - 12} \right) + 32\\
 = \left( {{x^2} - 4x - 8x + 32} \right) - \left( {{x^2} - 12x} \right) + 32\\
 = {x^2} - 12x + 32 - {x^2} + 12x + 32\\
 = \left( {{x^2} - {x^2}} \right) + \left( { - 12x + 12x} \right) + \left( {32 + 32} \right)\\
 = 64\\
2,\\
Gọi\,\,3\,\,số\,\,tự\,\,nhiên\,\,liên\,\,tiếp\,\,cần\,\,tìm\,\,là:\,\,x - 1;\,\,x;\,\,x + 1\,\,\,\,\left( {x \in {N^*}} \right)\\
Theo\,\,giả\,\,thiết\,\,ta\,\,có:\\
x.\left( {x + 1} \right) - \left( {x - 1} \right).x = 100\\
 \Leftrightarrow x.\left[ {\left( {x + 1} \right) - \left( {x - 1} \right)} \right] = 100\\
 \Leftrightarrow x.\left( {x + 1 - x + 1} \right) = 100\\
 \Leftrightarrow x.2 = 100\\
 \Leftrightarrow x = 100:2\\
 \Leftrightarrow x = 50\\
Vậy\,\,3\,\,số\,\,cần\,\,tìm\,\,là:\,\,\,49;\,50;\,51\\
3,\\
{89^2} + {11^2} + 22.89\\
 = {89^2} + 22.89 + {11^2}\\
 = {89^2} + 2.89.11 + {11^2}\\
 = {\left( {89 + 11} \right)^2}\\
 = {100^2}\\
 = 10000\\
4,\\
a,\\
24x - 18y + 30\\
 = 6.4x - 6.3y + 6.5\\
 = 6.\left( {4x - 3y + 5} \right)\\
b,\\
35x\left( {y - 8} \right) - 14y\left( {8 - y} \right)\\
 = 35x\left( {y - 8} \right) - 14y.\left[ { - \left( {y - 8} \right)} \right]\\
 = 35x\left( {y - 8} \right) + 14y.\left( {y - 8} \right)\\
 = \left( {y - 8} \right).\left( {35x + 14y} \right)\\
 = \left( {y - 8} \right).\left( {7.5x + 7.2y} \right)\\
 = 7.\left( {y - 8} \right).\left( {5x + 2y} \right)\\
c,\\
{x^2} + 49 + 14x - {y^2}\\
 = \left( {{x^2} + 14x + 49} \right) - {y^2}\\
 = \left( {{x^2} + 2.x.7 + {7^2}} \right) - {y^2}\\
 = {\left( {x + 7} \right)^2} - {y^2}\\
 = \left[ {\left( {x + 7} \right) - y} \right].\left[ {\left( {x + 7} \right) + y} \right]\\
 = \left( {x - y + 7} \right).\left( {x + y + 7} \right)\\
d,\\
25{x^2} - 2 = {\left( {5x} \right)^2} - {\sqrt 2 ^2} = \left( {5x - \sqrt 2 } \right)\left( {5x + \sqrt 2 } \right)\\
e,\\
{x^2} - {y^2} + 5x - 5y = \left( {{x^2} - {y^2}} \right) + \left( {5x - 5y} \right)\\
 = \left( {x - y} \right)\left( {x + y} \right) + 5.\left( {x - y} \right) = \left( {x - y} \right).\left[ {\left( {x + y} \right) + 5} \right]\\
 = \left( {x - y} \right)\left( {x + y + 5} \right)\\
5,\\
a,\\
\dfrac{{27{x^4}{y^3}}}{{36{x^2}{y^2}}} = \dfrac{{27}}{{36}}.\dfrac{{{x^4}}}{{{x^2}}}.\dfrac{{{y^3}}}{{{y^2}}} = \dfrac{3}{4}.{x^2}y\\
b,\\
\dfrac{{{x^2} + 10x + 25}}{{2{x^2} + 10x}} = \dfrac{{{x^2} + 2.x.5 + {5^2}}}{{2x.x + 2x.5}} = \dfrac{{{{\left( {x + 5} \right)}^2}}}{{2x\left( {x + 5} \right)}} = \dfrac{{x + 5}}{{2x}}\\
c,\\
\dfrac{{{x^2} + 7x + 10}}{{{x^2} + 9x + 20}} = \dfrac{{\left( {{x^2} + 2x} \right) + \left( {5x + 10} \right)}}{{\left( {{x^2} + 4x} \right) + \left( {5x + 20} \right)}}\\
 = \dfrac{{x.\left( {x + 2} \right) + 5.\left( {x + 2} \right)}}{{x\left( {x + 4} \right) + 5.\left( {x + 4} \right)}} = \dfrac{{\left( {x + 2} \right)\left( {x + 5} \right)}}{{\left( {x + 4} \right)\left( {x + 5} \right)}} = \dfrac{{x + 2}}{{x + 4}}\\
d,\\
\dfrac{{{x^2} - 9}}{{{x^2} + 9 + 6x}} = \dfrac{{{x^2} - {3^2}}}{{{x^2} + 2.x.3 + {3^2}}} = \dfrac{{\left( {x - 3} \right)\left( {x + 3} \right)}}{{{{\left( {x + 3} \right)}^2}}} = \dfrac{{x - 3}}{{x + 3}}
\end{array}\)