a) Ta có: \(A=\sqrt{12}+2\sqrt{27}-3\sqrt{48}\)

\(=2\sqrt{3}+6\sqrt{3}-12\sqrt{3}\)

\(=-4\sqrt{3}\)

b) Ta có: \(C=\sqrt{20a}+4\sqrt{45a}-2\sqrt{125a}\)

\(=2\sqrt{5a}+12\sqrt{5a}-10\sqrt{5a}\)

\(=4\sqrt{5a}\)

a) 2√18 - 4√50 + 3√32

= 6√2 - 20√2 + 12√2

= -2√2

b) √(√8 - 4)² + √8

= 4 - √8 + √8

= 4

c) √(14 - 6√5) + √(6 + 2√5)

= √(3 - √5)² + √(√5 + 1)²

= 3 - √5 + √5 + 1

= 4

\(a,2\sqrt{18}-4\sqrt{50}+3\sqrt{32}\\ =6\sqrt{2}-20\sqrt{2}+12\sqrt{2}=-2\sqrt{2}\\ b,\sqrt{\left(\sqrt{8}-4\right)^2}+\sqrt{8}\\ =4-\sqrt{8}+\sqrt{8}\\ =4\\ c,\sqrt{14-6\sqrt{5}}+\sqrt{6+2\sqrt{5}}\\ =\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}=3+\sqrt{5}+\sqrt{5}+1\\ =4+2\sqrt{5}\)

\(A=\sqrt{\left(2\sqrt{3}\right)^2+2^2+\left(\sqrt{2}\right)^2+2.2\sqrt{3}.2+2.2.\sqrt{2}+2.2\sqrt{3}.\sqrt{2}}\)

\(=\sqrt{\left(2\sqrt{3}+2+\sqrt{2}\right)^2}=2\sqrt{3}+2+\sqrt{2}\)

`6\sqrt(2/3)-\sqrt(24)+2\sqrt(3/8)+2\sqrt(1/6)`

`=6. \sqrt6/3 - \sqrt(2^2 .6) + 2. \sqrt(24)/8 + 2. \sqrt6/6`

`=2\sqrt6-2\sqrt6+ \sqrt6/2 + \sqrt6/3`

`=\sqrt6/2+\sqrt6/3`

`=(3\sqrt6+2\sqrt6)/6`

`=(5\sqrt6)/6`

Ta có: \(6\sqrt{\dfrac{2}{3}}-\sqrt{24}+2\sqrt{\dfrac{3}{8}}+2\sqrt{\dfrac{1}{6}}\)

\(=\dfrac{6\sqrt{2}}{\sqrt{3}}-2\sqrt{6}+2\cdot\dfrac{\sqrt{3}}{2\sqrt{2}}+\dfrac{2}{\sqrt{6}}\)

\(=2\sqrt{6}-2\sqrt{6}+\dfrac{\sqrt{3}}{\sqrt{2}}+\dfrac{\sqrt{2}}{\sqrt{3}}\)

\(=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{6}}{3}=\dfrac{5\sqrt{6}}{6}\)

\(A=\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\left(\sqrt{2}+1\right)-\left(\sqrt{2}-1\right)=2\)

\(B=\sqrt{18+8\sqrt{2}}+\sqrt{18-8\sqrt{2}}=\sqrt{\left(\sqrt{2}+4\right)^2}+\sqrt{\left(4-\sqrt{2}\right)^2}=4+\sqrt{2}+4-\sqrt{2}=8\)

\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)

\(=\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}=\sqrt{6+\frac{2\sqrt{2}}{\sqrt{2}}.\sqrt{4-2\sqrt{3}}}\)

\(=\sqrt{6+2.\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)