1x 4 + 2 x6 + 3x8+4x 10 + 5x12/ 5x2+10 x 3 + 15 x4+ 20 x5+ 25 x6
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2: \(=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{-\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{-\left(x+y\right)\left(x^2+y^2\right)}{x^2+xy+y^2}\)
b: \(\left(x^2+10x+8\right)^2-\left(8x+4\right)\left(x^2+8x+7\right)\)
\(=\left\lbrack\left(x^2+8x+7\right)+\left(2x+1\right)\right\rbrack^2-4\left(2x+1\right)\left(x^2+8x+7\right)\)
\(=\left(x^2+8x+7\right)^2+2\left(x^2+8x+7\right)\left(2x+1\right)+\left(2x+1\right)^2-4\left(2x+1\right)\left(x^2+8x+7\right)\)
\(=\left(x^2+8x+7\right)^2-2\left(x^2+8x+7\right)\left(2x+1\right)+\left(2x+1\right)^2\)
\(=\left(x^2+8x+7-2x-1\right)^2=\left(x^2+6x+6\right)^2\)
d: \(B=x^4+4x^3+8x^2+8x+4\)
\(=x^4+2x^3+2x^2+2x^3+4x^2+4x+2x^2+4x+4\)
\(=x^2\left(x^2+2x+2\right)+2x\left(x^2+2x+2\right)+2\left(x^2+2x+2\right)\)
\(=\left(x^2+2x+2\right)\left(x^2+2x+2\right)=\left(x^2+2x+2\right)^2\)
e: \(C=x^4-2x^3+5x^2-4x+4\)
\(=x^4-x^3+2x^2-x^3+x^2-2x+2x^2-2x+4\)
\(=x^2\left(x^2-x+2\right)-x\left(x^2-x+2\right)+2\left(x^2-x+2\right)=\left(x^2-x+2\right)\left(x^2-x+2\right)\)
\(=\left(x^2-x+2\right)^2\)
X1+X2=X3+X4=X5+X6=2
nên X1+X2+X3+X4+X5+X6=0
2+2+2=0
6=0(loại)
vậy không có giá trị nào thỏa mãn đề



\(\dfrac{1.4+2.6+3.8+4.10+5.12}{5.2+10.3+15.4+20.5+25.6}\)
\(=\dfrac{2\left(1.2+2.3+3.4+4.5+5.6\right)}{5\left(1.2+2.3+3.4+4.5+5.6\right)}\)
\(=\dfrac{2}{5}\)