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\(2\sqrt{\left(-5\right)^6}+3\sqrt{\left(-2\right)^8}.\)
\(=2\sqrt{\left(\left(-5\right)^3\right)^2}+3\sqrt{\left(\left(-2\right)^4\right)^2}\)
\(=2\cdot\left(-5\right)^3+3\cdot\left(-2\right)^4\)
\(=2\cdot125+3\cdot16=250+48=298\)
\(\Rightarrow2\sqrt{\left(-5\right)^6}+3\sqrt{\left(-2\right)^8}=298\)
\(2\sqrt{\left(-5\right)^6}+3\sqrt{\left(-2\right)^8}\)
\(=2\times125+3\times16\)
\(=250+48\)
\(=298\)
\(\sqrt{6+2\sqrt{5}}+\sqrt{21-8\sqrt{5}}\)
\(=\sqrt{5+2\sqrt{5}+1}+\sqrt{16-2\cdot4\sqrt{5}+5}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(4-\sqrt{5}\right)^2}\)
\(=\sqrt{5}+1+4-\sqrt{5}\)( vì \(\sqrt{5}+1>0\)và \(4-\sqrt{5}>0\))
\(=5\)
\(\sqrt{6+2\sqrt{5}}+\sqrt{21-8\sqrt{5}}\)
\(=\sqrt{5+2\sqrt{5}+1}+\sqrt{16+8\sqrt{5}+5}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(4-\sqrt{5}\right)^2}\)
\(=\left|\sqrt{5}+1\right|+\left|4-\sqrt{5}\right|\)
\(=\sqrt{5}+1+4-\sqrt{5}\)
\(=5\)
\(A=\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\left(\sqrt{2}+1\right)-\left(\sqrt{2}-1\right)=2\)
\(B=\sqrt{18+8\sqrt{2}}+\sqrt{18-8\sqrt{2}}=\sqrt{\left(\sqrt{2}+4\right)^2}+\sqrt{\left(4-\sqrt{2}\right)^2}=4+\sqrt{2}+4-\sqrt{2}=8\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}=\sqrt{6+\frac{2\sqrt{2}}{\sqrt{2}}.\sqrt{4-2\sqrt{3}}}\)
\(=\sqrt{6+2.\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
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)
\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{16-2.4\sqrt{2}+2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)
\(=\sqrt{6+2\sqrt{2}\sqrt{3-\left(\sqrt{3}+1\right)}}=\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
\(A=\sqrt{\left(2\sqrt{3}\right)^2+2^2+\left(\sqrt{2}\right)^2+2.2\sqrt{3}.2+2.2.\sqrt{2}+2.2\sqrt{3}.\sqrt{2}}\)
\(=\sqrt{\left(2\sqrt{3}+2+\sqrt{2}\right)^2}=2\sqrt{3}+2+\sqrt{2}\)
