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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
\(a,2\sqrt{18}-4\sqrt{50}+3\sqrt{32}\\ =6\sqrt{2}-20\sqrt{2}+12\sqrt{2}=-2\sqrt{2}\\ b,\sqrt{\left(\sqrt{8}-4\right)^2}+\sqrt{8}\\ =4-\sqrt{8}+\sqrt{8}\\ =4\\ c,\sqrt{14-6\sqrt{5}}+\sqrt{6+2\sqrt{5}}\\ =\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}=3+\sqrt{5}+\sqrt{5}+1\\ =4+2\sqrt{5}\)
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\(A=\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\left(\sqrt{2}+1\right)-\left(\sqrt{2}-1\right)=2\)
\(B=\sqrt{18+8\sqrt{2}}+\sqrt{18-8\sqrt{2}}=\sqrt{\left(\sqrt{2}+4\right)^2}+\sqrt{\left(4-\sqrt{2}\right)^2}=4+\sqrt{2}+4-\sqrt{2}=8\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}=\sqrt{6+\frac{2\sqrt{2}}{\sqrt{2}}.\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2.\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
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)
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\(E=2\sqrt{3}+3\sqrt{3^3}-\sqrt{100.3}\\ =2\sqrt{3}+9\sqrt{3}-10\sqrt{3}\\ =\left(2+9-10\right)\sqrt{3}=\sqrt{3}\)
\(F=\sqrt{3^2.2}+4\sqrt{18}=\sqrt{18}+4\sqrt{18}=\left(1+4\right)\sqrt{18}=5\sqrt{18}\)
\(G=2\sqrt{3}-4\sqrt{3^3}+5\sqrt{4^2.3}=2\sqrt{3}-12\sqrt{3}+20\sqrt{3}=\left(2-12+20\right)\sqrt{3}=10\sqrt{3}\)
\(H=\left(3\sqrt{25.2}-5\sqrt{9.2}+3\sqrt{2^3}\right)\sqrt{2}\\ =\left(15\sqrt{2}-15\sqrt{2}+6\sqrt{2}\right)\sqrt{2}\\ =6\sqrt{2}.\sqrt{2}=6\)
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a: Ta có: \(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{3}-\sqrt{5}-1\)
\(=\sqrt{3}-1\)
b: Ta có: \(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\)
\(=3-2\sqrt{2}+3\sqrt{2}+1\)
\(=4+\sqrt{2}\)
c: Ta có: \(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\)
\(=2\sqrt{2}-2+2\sqrt{2}+1\)
\(=4\sqrt{2}-1\)
a)
\(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{5+2\sqrt{5}\cdot\sqrt{3}+3}-\sqrt{5+2\sqrt{5}\cdot\sqrt{1}+1}\\ =\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{1}\right)^2}\\ =\sqrt{5}+\sqrt{3}-\sqrt{5}-\sqrt{1}\\ =\sqrt{3}-\sqrt{1}\)
b)
\(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\\ =\sqrt{9-2\sqrt{9}\cdot\sqrt{8}+8}+\sqrt{18+2\sqrt{18}\cdot\sqrt{1}+1}\\ =\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}+1\right)^2}\\ =3-2\sqrt{2}+3\sqrt{2}+1\\ =4+\sqrt{2}\)
c)
\(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\\ =\sqrt{8-2\sqrt{8}\cdot\sqrt{4}+4}+\sqrt{8+2\sqrt{8}\cdot\sqrt{1}+1}\\ =\sqrt{\left(2\sqrt{2}-2\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}\\ =2\sqrt{2}-2+2\sqrt{2}+1\\ =4\sqrt{2}-1\)
\(A=\sqrt{\left(2\sqrt{3}\right)^2+2^2+\left(\sqrt{2}\right)^2+2.2\sqrt{3}.2+2.2.\sqrt{2}+2.2\sqrt{3}.\sqrt{2}}\)
\(=\sqrt{\left(2\sqrt{3}+2+\sqrt{2}\right)^2}=2\sqrt{3}+2+\sqrt{2}\)