Question

Tính

a) (3+√2) ²

b) (2-√3) ²

c) (√2+√3) ²

d) √a² (a≥0)

e) -2√a⁴ (a<0)

f) √x²-6x+9 (x>3)

g) √4(a-2) ² (a<2)

h) √9(x-5) ⁴

i) √b²(a²+2ab+b²) (b>0)

Answer 1

a)\((3+\sqrt{2})^2=3^2+2.3.\sqrt{2}+\sqrt{2}^2=9+6\sqrt{2}+2=11+6\sqrt{2}\)

b)\(\left(2-\sqrt{3}\right)^2=2^2-2.2.\sqrt{3}+\sqrt{3}^2=4-4\sqrt{3}+3=7-4\sqrt{3}\)

c)\(\left(2+\sqrt{3}\right)^2=2^2+2.2.\sqrt{3}+\sqrt{3}^2=4+4\sqrt{3}+3=7+4\sqrt{3}\)

d)\(\sqrt{a^2}=a\) ( vì a\(\ge0\))

e)\(-2\sqrt{a^4}=-2\sqrt{(a^2)^2}=-2.a^2\)( vì a<0 => a² >0

f)\(\sqrt{x^2-6x+9}=\sqrt{\left(x-3\right)^2}=x-3\)(vì x>3)

g)\(\sqrt{4\left(a-2\right)^2}=2.\left|a-2\right|=2\left(2-a\right)\) (vì a<2)

h)\(\sqrt{9\left(x-5\right)^4}=\sqrt{9[\left(x-5\right)^2]^2}=3\left(x-5\right)^2=3x^2-30x+75\)

i)\(\sqrt{b^2\left(a^2+2ab+b^2\right)}=\sqrt{b^2\left(a+b\right)^2}=b.\left(a+b\right)=b^2+ab\)(vì b>0)