Đáp án:
\(\begin{array}{l}
4,\\
a,\,\,\,\,2{x^2} + 2{y^2}\\
b,\,\,\,4{x^2}\\
c,\,\,\,\,{x^2}\\
6,\\
a,\\
7500\,\\
b,\\
1000000\\
c,\\
1000000\\
7,\\
a,\\
\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right) + \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) = 2{a^3}\\
b,\\
{a^3} + {b^3} = \left( {a + b} \right).\left[ {{{\left( {a - b} \right)}^2} + ab} \right]\\
c,\\
\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right) = {\left( {ac + bd} \right)^2} + {\left( {ad - bc} \right)^2}\\
8,\\
a,\\
{x^2} - 6x + 10 \ge 1 > 0,\,\,\,\forall \,x\\
b,\\
4x - {x^2} - 5 < 0,\,\,\,\forall \,x
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
4,\\
a,\\
{\left( {x + y} \right)^2} + {\left( {x - y} \right)^2}\\
= \left( {{x^2} + 2xy + {y^2}} \right) + \left( {{x^2} - 2xy + {y^2}} \right)\\
= {x^2} + 2xy + {y^2} + {x^2} - 2xy + {y^2}\\
= \left( {{x^2} + {x^2}} \right) + \left( {2xy - 2xy} \right) + \left( {{y^2} + {y^2}} \right)\\
= 2{x^2} + 2{y^2}\\
b,\\
2\left( {x - y} \right)\left( {x + y} \right) + {\left( {x + y} \right)^2} + {\left( {x - y} \right)^2}\\
= {\left( {x - y} \right)^2} + 2.\left( {x - y} \right).\left( {x + y} \right) + {\left( {x + y} \right)^2}\\
= {\left[ {\left( {x - y} \right) + \left( {x + y} \right)} \right]^2}\\
= {\left( {x - y + x + y} \right)^2}\\
= {\left( {2x} \right)^2}\\
= 4{x^2}\\
c,\\
{\left( {x - y + z} \right)^2} + {\left( {z - y} \right)^2} + 2.\left( {x - y + z} \right).\left( {y - z} \right)\\
= {\left( {x - y + z} \right)^2} + 2.\left( {x - y + z} \right).\left( {y - z} \right) + {\left( {z - y} \right)^2}\\
= {\left( {x - y + z} \right)^2} + 2.\left( {x - y + z} \right).\left( {y - z} \right) + {\left( {y - z} \right)^2}\\
= {\left[ {\left( {x - y + z} \right) + \left( {y - z} \right)} \right]^2}\\
= {\left( {x - y + z + y - z} \right)^2}\\
= {x^2}\\
5,\\
Số\,\,tự\,\,nhiên\,\,a\,\,chia\,\,cho\,\,5\,\,dư\,\,4\,\,nên\,\,a = 5k + 4\,\,\,\,\left( {k \in N} \right)\\
Ta\,\,có:\,\,\\
{a^2} = {\left( {5k + 4} \right)^2} = {\left( {5k} \right)^2} + 2.5k.4 + {4^2}\\
= 25{k^2} + 40k + 16 = \left( {25{k^2} + 40k + 15} \right) + 1\\
= 5.\left( {5{k^2} + 8k + 3} \right) + 1\\
5 \vdots 5 \Rightarrow \left[ {5.\left( {5{k^2} + 8k + 3} \right)} \right]\, \vdots \,5\\
\Rightarrow {a^2}\,\,chia\,\,cho\,\,5\,\,\,dư\,\,1\\
6,\\
a,\\
{x^2} - {y^2} = \left( {x - y} \right)\left( {x + y} \right)\\
x = 87;y = 13 \Rightarrow {x^2} - {y^2} = \left( {87 - 13} \right).\left( {87 + 13} \right) = 75.100 = 7500\,\\
b,\\
{x^3} - 3{x^2} + 3x - 1\\
= {x^3} - 3{x^2}.1 + 3.x{.1^2} - {1^3}\\
= {\left( {x - 1} \right)^3}\\
x = 101 \Rightarrow {x^3} - 3{x^2} + 3x - 1 = {\left( {101 - 1} \right)^3} = {100^3} = 1000000\\
c,\\
{x^3} + 9{x^2} + 27x + 27\\
= {x^3} + 3.{x^2}.3 + 3.x.9 + 27\\
= {x^3} + 3.{x^2}.3 + 3.x{.3^2} + {3^3}\\
= {\left( {x + 3} \right)^3}\\
x = 97 \Rightarrow {x^3} + 9{x^2} + 27x + 27 = {\left( {97 + 3} \right)^3} = {100^3} = 1000000\\
7,\\
a,\\
\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right) + \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\\
= \left( {a + b} \right)\left( {{a^2} - a.b + {b^2}} \right) + \left( {a - b} \right).\left( {{a^2} + a.b + {b^2}} \right)\\
= \left( {{a^3} + {b^3}} \right) + \left( {{a^3} - {b^3}} \right)\\
= {a^3} + {b^3} + {a^3} - {b^3}\\
= 2{a^3}\\
b,\\
{a^3} + {b^3}\\
= \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)\\
= \left( {a + b} \right)\left[ {\left( {{a^2} - 2ab + {b^2}} \right) + ab} \right]\\
= \left( {a + b} \right).\left[ {{{\left( {a - b} \right)}^2} + ab} \right]\\
c,\\
\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right)\\
= {a^2}{c^2} + {a^2}{d^2} + {b^2}{c^2} + {b^2}{d^2}\\
= \left( {{a^2}{c^2} + 2abcd + {b^2}{d^2}} \right) + \left( {{a^2}{d^2} - 2abcd + {b^2}{c^2}} \right)\\
= \left[ {{{\left( {ac} \right)}^2} + 2.ac.bd + {{\left( {bd} \right)}^2}} \right] + \left[ {{{\left( {ad} \right)}^2} - 2ad.bc + {{\left( {bc} \right)}^2}} \right]\\
= {\left( {ac + bd} \right)^2} + {\left( {ad - bc} \right)^2}\\
8,\\
a,\\
{x^2} - 6x + 10 = \left( {{x^2} - 6x + 9} \right) + 1\\
= \left( {{x^2} - 2.x.3 + {3^2}} \right) + 1 = {\left( {x - 3} \right)^2} + 1\\
{\left( {x - 3} \right)^2} \ge 0,\,\,\,\forall \,x\\
\Rightarrow {\left( {x - 3} \right)^2} + 1 \ge 1,\,\,\,\forall \,x\\
\Rightarrow {x^2} - 6x + 10 \ge 1 > 0,\,\,\,\forall \,x\\
b,\\
4x - {x^2} - 5 = \left( {4x - {x^2} - 4} \right) - 1\\
= - 1 + \left( { - {x^2} + 4x - 4} \right) = - 1 - \left( {{x^2} - 4x + 4} \right)\\
= - 1 - \left( {{x^2} - 2.x.2 + {2^2}} \right)\\
= - 1 - {\left( {x - 2} \right)^2}\\
= - \left[ {1 + {{\left( {x - 2} \right)}^2}} \right]\\
{\left( {x - 2} \right)^2} \ge 0,\,\,\,\forall \,x\\
\Rightarrow 1 + {\left( {x - 2} \right)^2} \ge 1 > 0,\,\,\,\forall \,x\\
\Rightarrow - \left[ {1 + {{\left( {x - 2} \right)}^2}} \right] < 0,\,\,\,\forall \,x\\
\Rightarrow 4x - {x^2} - 5 < 0,\,\,\,\forall \,x
\end{array}\)
