\(a,=\dfrac{\left(x+1\right)^2}{x\left(x+1\right)}=\dfrac{x+1}{x}\\ b,=\dfrac{-\left(x^2-5x-6\right)}{\left(x+2\right)^2}=\dfrac{-\left(x+1\right)\left(x-6\right)}{\left(x+2\right)^2}\)

Bài 1 : 

\(\left(x-2\right)^2-\left(x-3^2\right)=\left(x-2\right)^2-\left(x-9\right)\)

\(=x^2-4x+4-x+9=x^2-5x+13\)

Bài 2 : 

a, \(P=\frac{1-4x^2}{4x^2-4x+1}=\frac{\left(1-2x\right)\left(2x+1\right)}{\left(2x-1\right)^2}\)

\(=\frac{-\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)^2}=\frac{-\left(2x+1\right)}{2x-1}=\frac{-2x-1}{2x-1}\)

b, Thay x = -4 ta được : 

\(\frac{-2.\left(-4\right)-1}{2.\left(-4\right)-1}=\frac{8-1}{-8-1}=-\frac{7}{9}\)

a) \(\frac{x^2+2x+4}{4x^3-32}=\frac{x^2+2x+4}{4\left(x^3-8\right)}=\frac{x^2+2x+4}{4\left(x-2\right)\left(x^2+2x+4\right)}=\frac{1}{4\left(x-2\right)}.\)

 b) \(\frac{10x-15}{4x^2-9}=\frac{5\left(2x-3\right)}{\left(2x\right)^2-3^2}=\frac{5\left(2x-3\right)}{\left(2x-3\right)\left(2x+3\right)}=\frac{5}{2x+3}.\)

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^HAND!!!!

\(\frac{x^2+2x+4}{4x^3-32}=\frac{\left(x+2\right)^2}{4\left(x^3-8\right)}=\frac{\left(x+2\right)^2}{4\left(x-2\right)\left(x^2+2x+4\right)}=\frac{x+2}{4\left(x^2+2x+4\right)}.\)

\(\frac{10x-15}{4x^2-9}=\frac{5\left(2x-3\right)}{\left(2x\right)^2-3^2}=\frac{5\left(2x-3\right)}{\left(2x-3\right)\left(2x+3\right)}=\frac{5}{2x+3}\)

câu 1 sai ròi tớ tưởng hằng đẳng thức -.-' 

lm theo bn trc ý nhé

a: A=x+3+|x-3|

=x+3+3-x(x<=3)

=6

b:\(B=\sqrt{x^2+4x+4}-\sqrt{x^2}\)

\(=\left|x+2\right|-\left|x\right|\)

=x+2-x=2

c: \(C=\dfrac{\sqrt{x^2-2x+1}}{x-1}\)

\(=\dfrac{\left|x-1\right|}{x-1}=\dfrac{x-1}{x-1}=1\)

(x+2).(x²-2x+4)+(2x-3).(4x²+6x+9)

 =(x³+8)+(8x³-27)

=x³+8+8x³-27

=+9x³-19

Câu 2 giống câu 1

Câu 5: B

Câu 6: 

a: ĐKXĐ: \(x-2\ne0\)

=>\(x\ne2\)

b: ĐKXĐ: \(x+1\ne0\)

=>\(x\ne-1\)

8:

\(A=\dfrac{x^2+4}{3x^2-6x}+\dfrac{5x+2}{3x}-\dfrac{4x}{3x^2-6x}\)

\(=\dfrac{x^2+4-4x}{3x\left(x-2\right)}+\dfrac{5x+2}{3x}\)

\(=\dfrac{\left(x-2\right)^2}{3x\left(x-2\right)}+\dfrac{5x+2}{3x}\)

\(=\dfrac{x-2+5x+2}{3x}=\dfrac{6x}{3x}=2\)

7: 

\(\dfrac{8x^3yz}{24xy^2}\)

\(=\dfrac{8xy\cdot x^2z}{8xy\cdot3y}\)

\(=\dfrac{x^2z}{3y}\)

1 a

2c

3b

4d

5c

6c

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