Đáp án:
k. \({\left( {2\sqrt 5 - 3} \right)^2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0\\
{\left( {\sqrt x + 2y} \right)^3} = \sqrt {{x^3}} + 3.\sqrt {{x^2}} .2y + 3.\sqrt x .4{y^2} + 16{y^3}\\
= x\sqrt x + 6x.y + 12\sqrt x .{y^2} + 16{y^3}\\
c.{\left( {\sqrt 2 x + y} \right)^3} = {\left( {x\sqrt 2 } \right)^3} + 3.{\left( {x\sqrt 2 } \right)^2}.y + 3\left( {x\sqrt 2 } \right).{y^2} + {y^3}\\
= 2\sqrt 2 {x^3} + 6{x^2}y + 3\sqrt 2 x{y^2} + {y^3}\\
e.DK:x \ge 0\\
{\left( {2\sqrt x - y} \right)^3} = 8\sqrt {{x^3}} - 3.4.x.y + 3.2.\sqrt x .{y^2} - {y^3}\\
= 8x\sqrt x - 12xy + 6\sqrt x .{y^2} - {y^3}\\
g.x - 2\\
i.4 + 2\sqrt 3 = 3 + 2\sqrt 3 .1 + 1 = {\left( {\sqrt 3 + 1} \right)^2}\\
b.{\left( {x - 2\sqrt y } \right)^3} = {x^3} - 3.{x^2}.2\sqrt y + 3.x.4y - {y^3}\\
= {x^3} - 6{x^2}\sqrt y + 12xy - {y^3}\\
d.{\left( {x + 2\sqrt y } \right)^2} = {x^2} + 4x\sqrt y + 4y\\
f.{x^2} - 4 = \left( {x - 2} \right)\left( {x + 2} \right)\\
h.{\left( {\sqrt 3 x - \sqrt y } \right)^3} = 3\sqrt 3 {x^3} - 3.3{x^2}.\sqrt y + 3.x\sqrt 3 .{y^2} + y\sqrt y \\
= 3\sqrt 3 {x^3} - 9{x^2}\sqrt y + 3\sqrt 3 x{y^2} + y\sqrt y \\
k.29 - 12\sqrt 5 = {\left( {2\sqrt 5 } \right)^2} - 2.2\sqrt 5 .3 + 9\\
= {\left( {2\sqrt 5 - 3} \right)^2}
\end{array}\)
