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P(x) + Q(x)= ( x^5 - 2x^2 + 7x^4 - 9x^3 - 1/4x) + ( 5x^4 - x^5 + 4x^2 - 2x^3 - 1/4)
= x^5 - 2x^2 + 7x^4 - 9x^3 - 1/4x + 5x^4 - x^5 + 4x^2 - 2x^3 - 1/4
= ( x^5 - x^5 ) - ( 2x^2 + 4x^2) + ( 7x^4 + 5x^4) - ( 9x^3 - 2x^3) - 1/4x - 1/4
= 6x^2 + 12x^4 - 6x^3 - 1/4x - 1/4
P(x) - Q(x)= ( x^5 - 2x^2 + 7x^4 - 9x^3 -1/4x) - ( 5x^4 - x^5 + 4x^2 - 2x^3 -1/4)
= x^5 - 2x^2 + 7x^4 - 9x^3 - 1/4x - 5x^4 + x^5 - 4x^2 + 2x^3 + 1/4
= ( x^5 + x^5) - ( 2x^2 - 4x^2) + ( 7x^4 - 5x^4) - ( 9x^3 + 2x^3) - 1/4x + 1/4
= 2x^5 - (-2)x^2 + 2x^4 - 11x^3 - 1/4x + 1/4
P(x)=x^5+ 7x^4- 9x^3+ 2x^2-1/4x-0
Q(x)=(-x^5+5x^4- 2x^3+ 4x^2+0x-1/4
= 12x^4-11x^3+ 6x^2-1/4x-1/4
P(x) = x^5 - 2x^2 + 7x^4 - 9x^3 - 1/4x
=x5+7x4-9x3-2x2-1/4x
Q(x) = 5x^4 - x^5 + 4x^2 - 2x^3 - 1/4
=-x5+5x4-2x3+4x2-1/4
P(x)+Q(x)=x5+7x4-9x3-2x2-1/4x -x5+5x4-2x3+4x2-1/4
=x5-x5+7x4+5x4-9x3-2x3-2x2+4x2-1/4x-1/4
=12x4-11x3+2x2-1/4x-1/4
P(x)-Q(x)=x5+7x4-9x3-2x2-1/4x +x5-5x4+2x3-4x2+1/4
=x5+x5+7x4-5x4-9x3+2x3-2x2-4x2-1/4x-1/4
=2x5+2x4-7x3-6x2-1/4x-1/4
Xin lỗi biết làm câu 1 thôi,thông cảm
Ta có A=:
\(=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+...+\frac{100^2-1}{100^2}\)
\(=\frac{2^2}{2^2}+\frac{3^2}{3^2}+...+\frac{100^2}{100^2}-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)
\(=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)
Mà \(\left(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{100^2}\right)< |\frac{100}{101}\)(tự tính)
\(\Rightarrow C>98\left(đpcm\right)\)
\(|x+1|+|x+2|+|x+3|=4x-4\left(1\right)\)
Vì \(\hept{\begin{cases}|x+1|\ge0;\forall x\\|x+2|\ge0;\forall x\\|x+3|\ge0;\forall x\end{cases}}\)
\(\Rightarrow|x+1|+|x+2|+|x+3|\ge0;\forall x\)
Mà \(|x+1|+|x+2|+|x+3|=4x-4\)
\(\Rightarrow4x-4\ge0\)
\(\Rightarrow4x\ge4\)
\(\Rightarrow x\ge1>0\)
\(\Rightarrow\hept{\begin{cases}|x+1|=x+1\\|x+2|=x+2\\|x+3|=x+3\end{cases}}\)(2)
Thay (2) vào (1) ta được :
\(x+1+x+2+x+3=4x-4\)
\(\Leftrightarrow3x+6=4x-4\)
\(\Leftrightarrow3x-4x=-4-6\)
\(\Leftrightarrow-x=-10\)
\(\Leftrightarrow x=10\)
Vậy x=10