8x3-8-8-8-8-8-8=-24
học tốt
8 . 3 - 8 - 8 - 8 - 8 - 8 - 8
= 24 - ( 8 . 6 )
= 24 - 48
= -24
8x3-8-8-8-8-8-8
= 24 - 8.6 (8.6 = .8-8-8-8-8-8)
=24-48
= -24
k cho mik nha
8x3-8-8-8-8-8-8=-24
học tốt
8 . 3 - 8 - 8 - 8 - 8 - 8 - 8
= 24 - ( 8 . 6 )
= 24 - 48
= -24
8x3-8-8-8-8-8-8
= 24 - 8.6 (8.6 = .8-8-8-8-8-8)
=24-48
= -24
k cho mik nha
8=8=8=8=8=8=8=8=8=8=8=8=8=8=8=8=8=8=8=?
8 + 8 + 8 + 8 + 8 + 8 = ???
\(\frac{3}{8+\frac{3}{8+\frac{3}{8+\frac{3}{8+\frac{3}{8+\frac{3}{8+\frac{3}{8+\frac{3}{8+\frac{3}{8+\frac{1}{1+x}}}}}}}}}}\)=\(\frac{381978}{382007}\)
cho \(\frac{x^2+y^2}{x^2-y^2}+\frac{x^2-y^2}{x^2+y^2}=a\) . Tính \(\frac{x^8+y^8}{x^8-y^8}+\frac{x^8-y^8}{x^8+y^8}\)theo a
\(P\left(x\right)=\sqrt[3]{\sqrt{x+8}\left(x^4+8x^3+12x\right)+6x^3+48x^2+8}\)
đặt \(A=\sqrt{x+8}\left(x^4+8x^3+12x\right)+6x^3+48x^2+8\)
\(=\sqrt{x+8}\left(x^4+8x^3\right)+6x^2\left(x+8\right)+12x\sqrt{x+8}+8\)
\(=\sqrt{\left(x+8\right)^3}x^3+3\sqrt{\left(x+8\right)^2}x^22+3\sqrt{\left(x+8\right)}x4+8\)
\(=\left(x\sqrt{x+8}+2\right)^3\)
\(\Rightarrow P\left(x\right)=x\sqrt{x+8}+2\)
cho số thực a,b thỏa mãn \(a^2\ne b^2\)đặt \(M=\frac{a^2+b^2}{a^2-b^2}+\frac{a^2-b^2}{a^2+b^2}\)tính \(N=\frac{a^8+b^8}{a^8-b^8}+\frac{a^8-b^8}{a^8+b^8}\)theo M
cho a,b ,c deu duong .cmr
\(\frac{8}{\left(a+b\right)^2+4abc}+\frac{8}{\left(b+c\right)^2+4abc}+\frac{8}{\left(a+c\right)^2+4abc}+a^2+b^2+c^2\ge\frac{8}{a+3}+\frac{8}{b+3}+\frac{8}{c+3}\)
cho số thực a,b,c>0. CMR
\(\frac{8}{\left(a+b\right)^2+4abc}+\frac{8}{\left(b+c\right)^2+4abc}+\frac{8}{\left(c+a\right)^2+4abc}+a^2+b^2+c^2\ge\frac{8}{a+3}+\frac{8}{b+3}+\frac{8}{c+3}\)