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Ta có: \(P=\frac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\frac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}=\frac{\left(x^3+1\right)\left(x+1\right)}{x^2\left(x^2-x+1\right)+\left(x^2-x+1\right)}\)
\(=\frac{\left(x+1\right)\left(x^2-x+1\right)\left(x+1\right)}{\left(x^2-x+1\right)\left(x^2+1\right)}=\frac{\left(x+1\right)^2\left(x^2-x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)}\)
Vì \(\hept{\begin{cases}x^2+1\ge1>0\\x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\end{cases}}\)
Nên mẫu số luôn luôn khác 0
Do đó: \(P=\frac{\left(x+1\right)^2\left(x^2-x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)}=\frac{\left(x+1\right)^2}{x^2+1}\)
Vì \(\hept{\begin{cases}\left(x+1\right)^2\ge0\\x^2+1>0\end{cases}\left(\forall x\right)}\) nên \(P\ge0\left(\forall x\right)\)
\(P=\frac{x^4+x^2+x+1}{x^4-x^2+2x^2-x+1}=\frac{\left(x+1\right)^2\left(x^2-x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)}\)
Do \(\left(x^2+1\right)\left(x^2-x+1\right)\ne0\)do đó không cần điều kiện của x
Vậy \(P=\frac{\left(x+1\right)^2\left(x^2-x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)}=\frac{\left(x+1\right)^2}{x^2+1}\)
\(\hept{\begin{cases}\left(x+1\right)^2\ge0\forall x\\x^2+1>0\forall x\end{cases}\Rightarrow P\ge0\forall x}\)
\(A=\frac{1}{x+2}+\frac{1}{x-2}+\frac{x^2+1}{x^2-4}\)
\(=\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{x+2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}\)
Với \(\forall x\in\left[-2;2\right]\) thì \(\left(x-2\right)\left(x+2\right)< 0\Rightarrow\frac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}< 0\Rightarrow A< 0\)
a)\(\frac{x^2+4}{x^2}+\frac{4}{x+1}\left(\frac{1}{x}+1\right)\)
\(=\frac{x^2+4}{x^2}+\frac{4}{x+1}.\frac{x+1}{x}\)
\(=\frac{x^2+4}{x^2}+\frac{4}{x}\)
\(=\frac{x^2+4x+4}{x^2}\)
\(\left(\frac{x+2}{x}\right)^2\)
=>phép chia = 1 với mọi x # 0 và x#-1
b)Cm tương tự
M = \(\left(\frac{x}{x-3}-\frac{x+3}{3x^2-6x-9}+\frac{1}{3x+3}\right)\)\(\frac{x^2-2x-3}{x^2+x+2}\)
= \(\left(\frac{x\left(3x+3\right)}{3\left(x-3\right)\left(x+1\right)}-\frac{x+3}{3\left(x-3\right)\left(x+1\right)}+\frac{x-3}{3\left(x+1\right)\left(x-3\right)}\right)\)\(\frac{\left(x+1\right)\left(x-3\right)}{x^2+x+2}\)
= \(\frac{3\left(x^2+x-2\right)}{3\left(x-3\right)\left(x+1\right)}\)* \(\frac{\left(x+1\right)\left(x-3\right)}{x^2+x+2}\) = \(\frac{x^2+x-2}{x^2+x+2}\)
Ta thấy x2 + x - 2 < x2 + x + 2
nên M < 1
a, \(M=\left(\frac{1}{x-1}-\frac{x}{1-x^3}.\frac{x^2+x+1}{x+1}\right):\frac{1}{x^2-1}\)
\(=\left(\frac{1}{x-1}-\frac{x}{\left(1-x\right)\left(x^2+x+1\right)}.\frac{x^2+x+1}{x+1}\right):\frac{1}{x^2-1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{1}{x^2-1}\)
\(=\left(\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}\right).\left(x-1\right)\left(x+1\right)=2x+1\)
b, Thay x = 1/2 vào biểu thức trên ta được : \(2.\frac{1}{2}+1=1\)
c, Để M luôn dương hay \(2x+1\ge0\Leftrightarrow x\ge-\frac{1}{2}\)
Vậy với x \(\ge-\frac{1}{2}\)thì \(M\ge0\)
đk x khác -1
A=\(\frac{\left(x^3-x^2+x\right)+\left(3x^2-3\right)+\left(x+4\right)}{x^3+1}=\frac{\left(x^3+1\right)+2x^2+2x}{x^3+1}=1+\frac{2x}{x^2-x+1}=\frac{x^2+x+1}{x^2-x+1}\)
a) \(A=\frac{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}=\frac{\left(2x+1\right)^2+3}{\left(2x-1\right)^2+3}\) Gọn thế nào quan điểm của người chấm.
b) Tử & mẫu của A luôn lớn hơn 3 lên suy ra ta luôn dương
A = \(\frac{x}{x+1}\)\(-\)\(\frac{3-3x}{x^2-x+1}\)\(+\)\(\frac{x+4}{x^3+1}\)
= \(\frac{x\left(x^2-x+1\right)}{x^3+1}\)\(-\)\(\frac{3-3x\left(x+1\right)}{x^3+1}\)\(+\)\(\frac{x+4}{x^3+1}\)
= \(\frac{x\left(x^2-x+1\right)-\left(3x-3\right)\left(x+1\right)+\left(x+4\right)}{x^3+1}\)
đến đây cậu tự nhân phá ra rồi rút gọn nhé