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Bạn vào biểu tượng \(\Sigma\) để nhập biểu thức cho chính xác nhé
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Bài 1:
a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để A=3 thì 3x-9=x+1
=>2x=10
hay x=5
Bài 2:
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)
b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1:
a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)
\(=4x^2-20x+25-4x^2-12x\)
=-32x+25
b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)
\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)
c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)
\(=\left(x-1-x-2\right)^2+5\left(2x-3\right)\)
\(=\left(-3\right)^2+5\left(2x-3\right)\)
\(=9+10x-15=10x-6\)
2:
a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)
\(=9x^2-12x+4-5x^2+20x+4x-4\)
\(=4x^2+12x\)
b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)
\(=27-x^3+x^3-9x^2+27x-27\)
\(=-9x^2+27x\)
c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)
\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)
\(=-5\left(x^2-16\right)=-5x^2+80\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)(3x+4)2-10x-(x-4)(x+4)
9x2+24x+16-10x-x2+16
8x2+14x+32
b)(x+1)(x-2)(x2+1)(x+2)(x-1)(x2+4)
(x+1)(x-1)(x+2)(x-2)(x2+1)(x2+4)
(x2-1)(x2-4)(7x2+4)
(-3x2+4)(7x2+4)
-21x2-12x2+28x2+16
16-x2
a)(3x+4)2-10x-(x-4)(x+4)
9x2+24x+16-10x-x2+16
8x2+14x+32
b)(x+1)(x-2)(x2+1)(x+2)(x-1)(x2+4)
(x+1)(x-1)(x+2)(x-2)(x2+1)(x2+4)
(x2-1)(x2-4)(7x2+4)
(-3x2+4)(7x2+4)
-21x2-12x2+28x2+16
16-x2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(P=\left(\dfrac{x^2-1}{x^4-x^2+1}+\dfrac{2}{x^6+1}-\dfrac{1}{x^2+1}\right)\cdot\left(x^2-\dfrac{x^4+x^2-1}{x^4+x^2+1}\right)\)
\(=\left(\dfrac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\dfrac{2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}-\dfrac{x^4-x^2+1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\right)\cdot\left(\dfrac{x^2\left(x^4+x^2+1\right)}{x^4+x^2+1}-\dfrac{x^4+x^2-1}{x^4+x^2+1}\right)\)
\(=\dfrac{x^4-1+2-x^4+x^2-1}{\left(x^2+1\right)\cdot\left(x^4-x^2+1\right)}\cdot\dfrac{x^6+x^4+x^2-x^4-x^2+1}{x^4+x^2+1}\)
\(=\dfrac{x^2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\cdot\dfrac{x^6+1}{x^4+x^2+1}\)
\(=\dfrac{x^2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\cdot\dfrac{\left(x^2+1\right)\left(x^4-x^2+1\right)}{x^4+x^2+1}\)
\(=\dfrac{x^2}{x^4+x^2+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2;A=\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right):\left(\frac{1-x}{x+2}\right)\)
\(ĐKXĐ:\hept{\begin{cases}x^2-4\ne0\\1-x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne\pm2\\x\ne1\end{cases}}\)
\(a,A=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right).\frac{x+2}{1-x}\)
\(A=\left(\frac{x+x-2-2x-4}{\left(x+2\right)\left(x-2\right)}\right).\frac{x+2}{1-x}\)
\(A=\frac{-6}{\left(x+2\right)\left(x-2\right)}.\frac{x+2}{1-x}=\frac{-6}{\left(x-2\right)\left(1-x\right)}\)
b, Khi x = -4
\(A=\frac{-6}{\left(-4-2\right)\left(1+4\right)}=\frac{-6}{-6.5}=\frac{1}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 4:
\(=\dfrac{a\left(a-b\right)-c\left(a-b\right)}{a\left(a+b\right)-c\left(a+b\right)}=\dfrac{a-b}{a+b}\)
a,(x-1) (x²+x²+x+1)
=(x-1)(2x2+x+1)
=2x3+2x+x-2x2-x-1
=2x3-2x2+2x-1
b, (x+1) (x4 -x3+x2-x+1)
=x5-x4+x3-x2+x+x4-x3+x2-x+1
=x5+1