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1) Có 3 = (22 - 1)
=> BT = (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 +1)
= (24 - 1)(24 + 1)(28 + 1)(216 +1)
= (28 - 1)(28 + 1)(216 +1)
= (216 - 1)(216 +1)
= 232 - 1
a)\(\frac{x^2+xy}{x^2-y^2}=\frac{x\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}=\frac{x}{x-y}\)
b) \(\frac{4}{x+2}+\frac{3}{x-2}+\frac{-5x-2}{x^2-4}\)
\(=\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{-5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4x-8+3x+6-5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x-4}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2}{x+2}\)
Bài 1 : hđt bạn tự làm nhé
Bài 2 :
\(\left(x-1\right)\left(x^2+x+1\right)-\left(x-4\right)^2x\)
\(=x^3-1-x\left(x^2-8x+16\right)=x^3-1-x^3+8x^2-16x\)
\(=8x^2-16x-1\)
\(\left(x+7\right)\left(x^2-7x+49\right)-\left(5-x\right)\left(5+x\right)\left(x-1\right)\)
\(=x^3+343-\left(25-x^2\right)\left(x-1\right)=x^3+343-\left(25x-25-x^3+x^2\right)\)
\(=x^3+343+x^3-x^2-25x+25=2x^3-x^2-25x+368\)
\(\left(x-3\right)^3-\left(x+3\right)^3\)
\(=\left(x-3-x-3\right)\left(\left(x-3\right)^2+\left(x-3\right)\left(x+3\right)+\left(x+3\right)^2\right)\)
\(=-6\left(\left(x-3\right)^2+\left(x^2-9\right)+\cdot\left(x+3\right)^2\right)\)
ta có B= (x-y)^3 +(y+x)^3 +(y-x)^3 -3xy(x+y)
<=> B= (x-y)^3 -(x-y)^3 +(y+x)( y^2+2xy+x^2 -3xy)
<=>B= (y+x)(y^2-xy +x^2)
<=>B=y^3+x^3
(x-3)3-(x+3)3=(a3-9x2+27x-27)-(x3+9x2+27x+27)
=x3-9x2+27x-27-x3-9x2-27x-27
=-18x2-54