51 : 4 = ??
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mình cần gấppppppppppppppppppppppppppppp
nhé
nhanh giùm mình vs!!!
ta có
\(a^4+16a^2b^2+64b^4-16a^2b^2=\left(a^2+8b^2\right)^2-\left(4ab\right)^2\)
\(=\left(a^2+8b^2-4ab\right)\left(a^2+8b^2+4ab\right)\)
Ta có: \(\frac{y}{x+y}+\frac{2y^2}{x^2+y^2}+\frac{4y^4}{x^4+y^4}+\frac{8y^8}{x^8-y^8}=4\)
\(\Leftrightarrow\frac{y}{x+y}+\frac{2y^2}{x^2+y^2}+\left[\frac{4y^4}{x^4+y^4}+\frac{8y^8}{\left(x^4-y^4\right)\left(x^4+y^4\right)}\right]=4\)
\(\Leftrightarrow\frac{y}{x+y}+\frac{2y^2}{x^2+y^2}+\frac{4y^4\left(x^4-y^4\right)+8y^8}{\left(x^4-y^4\right)\left(x^4+y^4\right)}=4\)
\(\Leftrightarrow\frac{x}{x+y}+\frac{2y^2}{x^2+y^2}+\frac{4x^4y^4+4y^8}{\left(x^4-y^4\right)\left(x^4+y^4\right)}=4\)
\(\Leftrightarrow\frac{x}{x+y}+\frac{2y^2}{x^2+y^2}+\frac{4y^4}{x^4-y^4}=4\)
\(\Leftrightarrow\frac{y}{x+y}+\left[\frac{2y^2}{x^2+y^2}+\frac{4y^4}{\left(x^2-y^2\right)\left(x^2+y^2\right)}\right]=4\)
\(\Leftrightarrow\frac{y}{x+y}+\frac{2y^2\left(x^2-y^2\right)+4y^4}{\left(x^2-y^2\right)\left(x^2+y^2\right)}=4\)
\(\Leftrightarrow\frac{y}{x+y}+\frac{2x^2y^2+2y^4}{\left(x^2-y^2\right)\left(x^2+y^2\right)}=4\)
\(\Leftrightarrow\frac{y}{x+y}+\frac{2y^2}{x^2-y^2}=4\)
\(\Leftrightarrow\frac{y\left(x-y\right)+2y^2}{\left(x-y\right)\left(x+y\right)}=4\)
\(\Leftrightarrow\frac{xy+y^2}{\left(x+y\right)\left(x-y\right)}=4\)
\(\Leftrightarrow\frac{y}{x-y}=4\)
\(\Leftrightarrow y=4x-4y\Rightarrow4x=5y\)
=> đpcm
TL:
51:4=12( dư 3)
HT
= 12 dư 3