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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{\left(x^2+a\right)\left(1+a\right)+a^2x^2+1}{\left(x^2-a\right)\left(1-a\right)+a^2x^2+1}\)
\(=\dfrac{x^2+ax^2+a+a^2+a^2x^2+1}{x^2-ax^2-a+a^2+a^2x^2+1}\)
\(=\dfrac{\left(x^2+ax^2+a^2x^2\right)+\left(a+a^2+1\right)}{\left(x^2-ax^2+a^2x^2\right)+\left(a^2-a+1\right)}\)
\(=\dfrac{x^2\left(1+a+a^2\right)+\left(a+a^2+1\right)}{x^2\left(1-a+a^2\right)+\left(a^2-a+1\right)}\)
\(=\dfrac{\left(x^2+1\right)\left(a^2+a+1\right)}{\left(x^2+1\right)\left(a^2-a+1\right)}\)
\(=\dfrac{a^2+a+1}{a^2-a+1}\)
=> Biểu thức không phụ thuộc vào giá trị biến x
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\dfrac{\left(x^2+a\right)\left(1+a\right)+a^2x^2+1}{\left(x^2-a\right)\left(1-a\right)+a^2x^2+1}\)
\(=\dfrac{x^2+a+a^2x+a^2+a^2x^2+1}{x^2-a-ax^2+a^2+a^2x^2+1}\)
\(=\dfrac{a\left(x^2+1\right)+a^2\left(x^2+1\right)+x^2+1}{-a\left(x^2+1\right)+a^2\left(x^2+1\right)+x^2+1}\)
\(=\dfrac{\left(x^2+1\right)\left(a^2+a+1\right)}{\left(x^2+1\right)\left(a^2-a+1\right)}\)
\(=\dfrac{a^2+a+1}{a^2-a+1}\)
Vậy phân thức M ko phụ thuộc vào giá trị của x
![](https://rs.olm.vn/images/avt/0.png?1311)
M =\(\frac{\left(x^2+a\right)\left(1+a\right)+a^2x^2+1}{\left(x^2-a\right)\left(1-a\right)+a^2x^2+1}\) = \(\frac{x^2+a+ax^2+a^2+a^2x^2+1}{x^2+a^2-a-ax^2+a^2x^2+1}\)=\(\frac{x^2\left(a^2+a+1\right)+\left(a^2+a+1\right)}{x^2\left(a^2-a+1\right)+\left(a^2-a+1\right)}\)
=\(\frac{\left(x^2+1\right)\left(a^2+a+1\right)}{\left(x^2+1\right)\left(a^2-a-1\right)}\). Mà x2>= 0 => x2+1 >0
M= \(\frac{a^2+a+1}{a^2-a+1}\)
Vậy M không phụ thuốc vào giá trị của x
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\frac{\left(x^2+a\right)\left(1+a\right)+a^2x^2+1}{\left(x^2-a\right)\left(1-a\right)+a^2x^2+1}\)
\(=\frac{x^2+x^2a+a+a^2+a^2x^2+1}{x^2-x^2a-a+a^2+a^2x^2+1}\)
\(=\frac{\left(x^2+1\right)+\left(x^2a+a\right)+\left(a^2+a^2x\right)}{\left(x^2+1\right)-\left(x^2a+a\right)+\left(a^2+a^2x^2\right)}\)
\(=\frac{\left(x^2+1\right)+a\left(x^2+1\right)+a^2\left(x^2+1\right)}{\left(x^2+1\right)-a\left(x^2+1\right)+a^2\left(x^2+1\right)}\)
\(=\frac{\left(x^2+1\right)\left(a^2+a+1\right)}{\left(x^2+1\right)\left(a^2-a+1\right)}=\frac{a^2+a+1}{a^2-a+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=\)\(\frac{x^2+a+ax^2+a^2+x^2a^2+1}{x^2-a-ax^2+a^2+a^2x^2+1}\)\(=\frac{x^2\left(a^2+a+1\right)+a^2+a+1}{x^2\left(a^2-a+1\right)+a^2-a+1}\)
\(=\frac{\left(x^2+1\right)\left(a^2+a+1\right)}{\left(x^2+1\right)\left(a^2-a+1\right)}=\frac{a^2+a+1}{a^2-a+1}\)
Từ đó suy ra đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=6x^2-9x+14x-21-4x^2+20x-25-2x\left(x+6\right)+5-31x\)
\(=2x^2-6x-41-2x^2-12x\)
=-18x-41
b: \(=2x^2-6x-2x^2+6x+14=14\)
c: \(=x^3+1-x^3+1=2\)
\(\frac{\left(x^2+a\right)\left(1+a\right)+a^2x^2+1}{\left(x^2-a\right)\left(1-a\right)+a^2x^2+1}\)
\(=\frac{x^2+ax^2+a+a^2+a^2x^2+1}{x^2-ax^2-a+a^2+a^2x^2+1}\)
\(=\frac{x^2+1+a\left(x^2+1\right)+a^2\left(x^2+1\right)}{x^2+1-a\left(x^2+1\right)+a^2\left(x^2+1\right)}\)
\(=\frac{\left(x^2+1\right)\left(a^2+a+1\right)}{\left(x^2+1\right)\left(a^2-a+1\right)}=\frac{a^2+a+1}{a^2-a+1}\)