Question

a)A=\(\sqrt{49a^2}+3a\) với a>=0

b)B=\(\sqrt{16a^4}+6a^2\)

c)C=\(4x-\sqrt{\left(x^2-4x+4\right)}\) với x<2

d)D=\(\dfrac{a-b}{\sqrt{a}-\sqrt{b}}\) với a,b >0 ;a#b

e)E=\(\sqrt{y^2+6y+9}-\sqrt{y^2-6y+9}\) với y tuỳ ý

Answer 1

\(B=\sqrt{16a^4}+6a^2=4a^2+6a^2=10a^2\)\(A=\sqrt{49a^2}+3a=7a+3a=10a\)

\(C=4x-\sqrt{\left(x^2-4x+4\right)}=4x-\sqrt{\left(x-2\right)^2}=4x-x+2=3x+2\)

\(E=\sqrt{y^2+6y+9}-\sqrt{y^2-6y+9}=\sqrt{\left(y+3\right)^2}-\sqrt{\left(y-3\right)^2}=\left|y+3\right|-\left|y-3\right|=y+3-y+3=6\)

\(D=\dfrac{a-b}{\sqrt{a}-\sqrt{b}}=\dfrac{\left(a-b\right)\cdot\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\cdot\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{a\sqrt{a}+a\sqrt{b}-b\sqrt{a}-b\sqrt{b}}{a-b}=\dfrac{\sqrt{a}\cdot\left(a-b\right)+\sqrt{b}\cdot\left(a-b\right)}{a-b}=\dfrac{\left(a-b\right)\cdot\left(\sqrt{a}+\sqrt{b}\right)}{a-b}=\sqrt{a}+\sqrt{b}\)