\(\frac{6\sqrt[4]{2008}}{1+\sqrt[4]{2}+\sqrt[4]{4}+\sqrt[4]{8}}=\frac{6\sqrt[4]{2008}}{\left(\sqrt{2}+1\right)\left(\sqrt[4]{2}+1\right)}\)\(=\frac{6\sqrt[4]{2008}\left(\sqrt{2}-1\right)}{\sqrt[4]{2}+1}=6\sqrt[4]{2008}\left(\sqrt[4]{2}-1\right)\)4ata có:(a-1)3+a3+(a+1)3=3a3+6a=3a(a2+2)xét a chia hết cho 3=>(a-1)3+a3+(a+1)3 chia hết cho 9xét a ko chia hết cho 3=>a2 chia 3 dư 1=>a2+2 chia hết cho 3=>(a-1)3+a3+(a+1)3 chia hết cho...
Đọc tiếp
\(\frac{6\sqrt[4]{2008}}{1+\sqrt[4]{2}+\sqrt[4]{4}+\sqrt[4]{8}}=\frac{6\sqrt[4]{2008}}{\left(\sqrt{2}+1\right)\left(\sqrt[4]{2}+1\right)}\)
\(=\frac{6\sqrt[4]{2008}\left(\sqrt{2}-1\right)}{\sqrt[4]{2}+1}=6\sqrt[4]{2008}\left(\sqrt[4]{2}-1\right)\)
4a
ta có:
(a-1)3+a3+(a+1)3=3a3+6a=3a(a2+2)
xét a chia hết cho 3=>(a-1)3+a3+(a+1)3 chia hết cho 9
xét a ko chia hết cho 3=>a2 chia 3 dư 1
=>a2+2 chia hết cho 3
=>(a-1)3+a3+(a+1)3 chia hết cho 9
b,
x2+y2+z2+t2-x(y+z+t)=0
=(1/4x^2-xy+y2)+(1/4x2-xz+z2)+(1/4x2-xt+t2)+1/4x2=0
<=>(1/2x-y)2+(1/2x-z)2+(1/2x-t)2+1/4x2=0
<=>x=y=z=t=0
