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1/
a, \(4x^4+1=4x^4+4x^2+1-4x^2=\left(2x^2+1\right)^2-\left(2x\right)^2=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)\)
b, \(4x^4+y^4=4x^4+4x^2y^2+y^4-4x^2y^2=\left(2x^2+y^2\right)^2-\left(2xy\right)^2=\left(2x^2+2xy+y^2\right)\left(2x^2-2xy+y^2\right)\)
c, \(x^4+324=x^4+36x^2+324-36x^2=\left(x^2+18\right)^2-\left(6x\right)^2=\left(x^2+6x+18\right)\left(x^2-6x+18\right)\)
2/
a, \(x^2+\frac{1}{3}x+\frac{1}{36}=\left(x+\frac{1}{6}\right)^2=\left(\frac{35}{6}+\frac{1}{6}\right)^2=6^2=36\)
b, \(x^2-y^2+2y-1=x^2-\left(y-1\right)^2=\left(x+y-1\right)\left(x-y+1\right)=\left(100+1-1\right)\left(100-1+1\right)=100.100=10000\)
\(\frac{\left[35\left(27^2+2.9^{11}\right)\right]}{\left[15\left(8^{16}-12.3^{19}\right)\right]}=5,2\)
Ta có:
\(x=35\)
\(\Rightarrow\left\{{}\begin{matrix}x+1=36\\x+2=27\\2x-1=69\\x-1=34\end{matrix}\right.\) (1)
Thay (1) vào biểu thức ta được:
\(x^5-\left(x+1\right)x^4+\left(x+2\right)x^3-\left(2x-1\right)x^2-\left(x-1\right)x+15\)
\(=x^5-x^5-x^4+x^4+2x^3-2x^3+x^2-x^2+x+15\)
\(=x+15\)
\(=35+15\)
\(=50\)
Để olm giúp em em nhé!
a, \(\dfrac{x+2}{7x+42}\) = \(\dfrac{x+2}{7.\left(x+6\right)}\) = \(\dfrac{\left(x+2\right)\left(x-6\right)}{7\left(x-6\right)\left(x+6\right)}\) (đk \(x\ne\) \(\mp\) 6)
\(\dfrac{-13x}{x^2-36}\) = \(\dfrac{-13x}{\left(x-6\right)\left(x+6\right)}\) = \(\dfrac{-7.13.x}{7.\left(x-6\right).\left(x+6\right)}\) = \(\dfrac{-91x}{7.\left(x-6\right)\left(x+6\right)}\)
b, \(\dfrac{7}{4x+16}\) = \(\dfrac{7\left(x-4\right)}{4.\left(x+4\right).\left(x-4\right)}\) (đk \(x\ne\) \(\pm\) 4)
\(\dfrac{15}{x^2-16}\) = \(\dfrac{15.4}{\left(x-4\right)\left(x+4\right).4}\) = \(\dfrac{60}{4.\left(x-4\right).\left(x+4\right)}\)
a: \(2004^2-16=2000\cdot2008=4016000\)
b: \(892^2+892\cdot216+108^2=1000^2=1000000\)
c: \(36^2-52\cdot36+26^2=10^2=100\)