Bài 1: Phân thức đại số.

Vì x + y = 1 => \(\left\{{}\begin{matrix}x=1-y\\y=1-x\end{matrix}\right.\)

Khi đó, ta có \(\dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}=\dfrac{1-y}{\left(y-1\right)\left(y^2+y+1\right)}-\dfrac{1-x}{\left(x-1\right)\left(x^2+x+1\right)}=-\dfrac{y-1}{\left(y-1\right)\left(y^2+y+1\right)}+\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=-\dfrac{1}{y^2+y+1}+\dfrac{1}{x^2+x+1}\)

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

Lại có \(\left(x^2+x+1\right)\left(y^2+y+1\right)=x^2y^2+xy^2+y^2+x^2y+xy+y+x^2+x+1=x^2y^2+xy\left(x+y+1\right)+x^2+y^2+x+y+1=x^2y^2+2xy+x^2+y^2+2=x^2y^2+\left(x+y\right)^2+2=x^2y^2+3\)

Từ đó, ta có \(P=\dfrac{2\left(y-x\right)}{x^2y^2+3}+\dfrac{2\left(x-y\right)}{x^2y^2+3}=0\)

Vậy P = 0

Đây là lần đầu tiên mình biết gõ công thức, thật là một trải nghiệm khó quên

a: \(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)

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

b: Để A>0 thì x-1>0

hay x>1

c: Để A<0 thì x-1<0

hay x<1

d: Ta có: |3x+2|=1

=>3x+2=1 hoặc 3x+2=-1

=>x=-1/3(nhận) hoặc x=-1(loại)

Thay x=-1/3 vào A, ta được:

\(A=\dfrac{1}{9}:\left(-\dfrac{1}{3}-1\right)=\dfrac{1}{9}:\dfrac{-4}{3}=\dfrac{1}{9}\cdot\dfrac{3}{-4}=-\dfrac{1}{12}\)

C

B là đẳng thức sai

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

A

\(B=-5\left(x+\dfrac{2}{5}\right)^2+\dfrac{9}{5}\le\dfrac{9}{5}\)

Vậy A là đáp án đúng

\(x\left(x-3\right)\left(x-4\right)\left(x-7\right)\\ =\left(x^2-7x\right)\left(x^2-7x+12\right)\\ =\left(x^2-7x+6-6\right)\left(x^2-7x+6+6\right)\\ =\left(x^2-7x+6\right)^2-36\)

Nhận xét:

\(\left(x^2-7x+6\right)^2\ge0\\ \Rightarrow\left(x^2-7x+6\right)^2-36\ge0\\ \Rightarrow C\)

3x(6x+4)
= 18x²+12x

\(=18x^2+12x\)

Chọn B

\(\dfrac{2019x}{xy+2019x+2019}+\dfrac{y}{yz+y+2019}+\dfrac{z}{xz+z+1}\)

\(=\dfrac{2019xz}{xyz+2019xz+2019z}+\dfrac{xz.y}{xz.yz+xz.y+2019xz}+\dfrac{z}{xz+z+1}\)

\(=\dfrac{2019xz}{2019+2019xz+2019z}+\dfrac{2019}{2019z+2019+2019xz}+\dfrac{z}{xz+z+1}\)

\(=\dfrac{xz}{1+xz+z}+\dfrac{1}{z+1+xz}+\dfrac{z}{xz+z+1}\)

\(=\dfrac{xz+z+1}{xz+z+1}=1\)

a: ĐKXĐ: \(x\notin\left\{0;3;1\right\}\)

b: \(P=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\cdot\dfrac{x}{2x-2}\)

\(=\dfrac{-6x+18}{x-3}\cdot\dfrac{1}{2\left(x-1\right)}=\dfrac{-6\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}=\dfrac{-3}{x-1}\)