a) ĐKXĐ: \(\left\{{}\begin{matrix}2x+3\ne0\\2x+1\ne0\\\left(2x+3\right)\left(2x+1\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{3}{2}\\x\ne-\dfrac{1}{2}\\\left(2x+3\right)\left(2x+1\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{3}{2}\\x\ne-\dfrac{1}{2}\end{matrix}\right.\)

b) \(\Rightarrow P=\dfrac{2\left(2x+1\right)+3\left(2x+3\right)-6x-5}{\left(2x+3\right)\left(2x+1\right)}\)

\(\Rightarrow P=\dfrac{4x+2+6x+9-6x-5}{\left(2x+3\right)\left(2x+1\right)}\)

\(\Rightarrow P=\dfrac{4x+6}{\left(2x+3\right)\left(2x+1\right)}\)

\(\Rightarrow P=\dfrac{2\left(2x+3\right)}{\left(2x+3\right)\left(2x+1\right)}\)

\(\Rightarrow P=\dfrac{2}{2x+1}\)

c) \(P=-1\Rightarrow\dfrac{2}{2x+1}=-1\\ \Rightarrow2=-2x-1\\ \Rightarrow2x=-3\\ \Rightarrow x=-\dfrac{3}{2}\)

Món soup từ dây xúc xích.

Có ý nghĩa là : Nói về sự mơ tưởng, mong muốn vượt đến điều tốt đẹp của cuộc sống.

Bài 6

\(a,ĐK:x\ne\pm5\\ b,P=\dfrac{x-5+2x+10-2x-10}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{1}{x+5}\\ c,P=-3\Leftrightarrow\dfrac{1}{x+5}=-3\Leftrightarrow-3\left(x+5\right)=1\Leftrightarrow x=-\dfrac{16}{3}\\ \Leftrightarrow Q=\left(3x-7\right)^2=\left[3\cdot\left(-\dfrac{16}{3}\right)-7\right]^2=529\)

Bài 7:

\(a,ĐK:x\ne\pm3\\ b,P=\dfrac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}=\dfrac{4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{4}{x-3}\\ b,P=4\Leftrightarrow4\left(x-3\right)=4\Leftrightarrow x=4\)

Điều kiện (x≠5, x≠-5)

\(P=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)

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

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

    \(=\dfrac{x-5}{\left(x+5\right)\left(x-5\right)}=\dfrac{1}{x+5}\)

a: Xét tứ giác MIPC có

K là trung điểm của MP

K là trung điểm của IC

Do đó: MIPC là hình bình hành

mà MI=PI

nên MIPC là hình thoi

Bài 1:

a) \(=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1=\left(x+1\right)\left(x+3\right)\)

f) \(=\left(x^2-4x+4\right)-9=\left(x-2\right)^2-3^2=\left(x-5\right)\left(x+1\right)\)

h) \(=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)

k) \(=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x^3-x-1\right)\)

m) \(=\left(x^4+4x^2+4\right)-9=\left(x^2+2\right)^2-9=\left(x^2-1\right)\left(x^2+5\right)=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)

Bài 2:

a) \(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)

e) \(=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)

f) giống câu a

g) \(=x^2-2xy=x\left(x-2y\right)\)

i) \(=\left(x^3-y^3\right)\left(x^3+y^3\right)=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)

k) \(=\left(x+1\right)^3-27z^3=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)

l) \(=\left(2x+1\right)^2-9y^2=\left(2x+1-3y\right)\left(2x+1+3y\right)\)