a: ĐKXĐ: \(x\notin\left\{1;0;-2\right\}\)

\(\dfrac{1-x^2}{x^2+2x}:\dfrac{2-2x}{3x}\)

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

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

b: ĐKXĐ: \(x\ne1\)

\(\dfrac{x^3+1}{x-1}:\left(x^2-x+1\right)\)

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

\(=\dfrac{x+1}{x-1}\)

c: ĐKXĐ: \(x\notin\left\{-1;-2;0;2\right\}\)

\(\dfrac{x^2-x-2}{x^2+3x+2}:\dfrac{x^2-4x+4}{x^2+2x}\)

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

\(=\dfrac{x}{x-2}\)

d: \(\dfrac{x-2y}{x^2-xy+y^2}:\dfrac{x^2-4xy+4y^2}{x^3+y^3}\)

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

\(=\dfrac{x+y}{x-2y}\)

a) \(\dfrac{1-x^2}{x^2+2x}:\dfrac{2-2x}{3x}\left(x\ne1;0;-2\right)\)

\(=\dfrac{\left(1+x\right)\left(1-x\right)}{x\left(x+2\right)}:\dfrac{2\left(1-x\right)}{3x}\)

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

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

\(=\dfrac{3x+3}{2x+4}\)

b) \(\dfrac{x^3+1}{x-1}:\left(x^2-x+1\right)\left(x\ne1\right)\)

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

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

\(=\dfrac{x+1}{x-1}\)

c) \(\dfrac{x^2-x-2}{x^2+3x+2}:\dfrac{x^2-4x+4}{x^2+2x}\left(x\ne-1;-2;0;2\right)\)

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

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

\(=\dfrac{x}{x-2}\)

d) \(\dfrac{x-2y}{x^2-xy+y^2}:\dfrac{x^2-4xy+4y^2}{x^3+y^3}\left(x\ne2y;-y\right)\)

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

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

\(=\dfrac{x+y}{x-2y}\)

a: \(A=\dfrac{x-2}{x+2}+\dfrac{6x-4}{x^2-4}\)

\(=\dfrac{x-2}{x+2}+\dfrac{6x-4}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{\left(x-2\right)^2+6x-4}{\left(x-2\right)\left(x+2\right)}\)

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

b: \(P=A+B=\dfrac{x}{x-2}+\dfrac{x+1}{x-2}=\dfrac{2x+1}{x-2}\)

Để P nguyên thì \(2x+1⋮x-2\)

=>\(2x-4+5⋮x-2\)

=>\(5⋮x-2\)

=>\(x-2\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{3;1;7;-3\right\}\)

a.

\(A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{6x-4}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^2-4x+4+6x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x}{x-2}\)

b.

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

Để P nguyên \(\Rightarrow\dfrac{5}{x-2}\in Z\)

\(\Rightarrow x-2=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\)

\(\Rightarrow x=\left\{-3;1;3;7\right\}\)

a: ĐKXĐ: \(x\notin\left\{0;4;3;-3\right\}\)

\(\dfrac{x^2+3x}{x-4}:P=\dfrac{x^2-9}{x^2-4x}\)

=>\(P=\dfrac{x^2+3x}{x-4}:\dfrac{x^2-9}{x^2-4x}\)

\(=\dfrac{x\left(x+3\right)}{x-4}\cdot\dfrac{x\left(x-4\right)}{\left(x-3\right)\cdot\left(x+3\right)}\)

=>\(P=\dfrac{x^2}{x-3}\)

b: ĐKXĐ: \(x\notin\left\{2;-2;-\dfrac{3}{2}\right\}\)

\(Q:\dfrac{x-2}{2x+3}=\dfrac{4x^2+12x+9}{x^2-4}\)

=>\(Q=\dfrac{4x^2+12x+9}{x^2-4}\cdot\dfrac{x-2}{2x+3}\)

=>\(Q=\dfrac{\left(2x+3\right)^2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x-2}{2x+3}\)

=>\(Q=\dfrac{2x+3}{x+2}\)

a) \(\dfrac{x^2+3x}{x-4}:P=\dfrac{x^2-9}{x^2-4x}\left(x\ne4;0;\pm3\right)\)

\(\Leftrightarrow P=\dfrac{x\left(x+3\right)}{x-4}:\dfrac{\left(x+3\right)\left(x-3\right)}{x\left(x-4\right)}\)

\(\Leftrightarrow P=\dfrac{x\left(x+3\right)}{x-4}\cdot\dfrac{x\left(x-4\right)}{\left(x+3\right)\left(x-3\right)}\)

\(\Leftrightarrow P=\dfrac{x^2}{x-3}\)

b) \(Q:\dfrac{x-2}{2x+3}=\dfrac{4x^2+12x+9}{x^2-4}\left(x\ne-\dfrac{3}{2};\pm2\right)\)

\(\Leftrightarrow Q=\dfrac{4x^2+12x+9}{x^2-4}\cdot\dfrac{x-2}{2x+3}\)

\(\Leftrightarrow Q=\dfrac{\left(2x+3\right)^2}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x-2}{2x+3}\)

\(\Leftrightarrow Q=\dfrac{2x+3}{x+2}\)

a) \(\dfrac{x+4}{x+5}:\dfrac{x+5}{x+6}:\dfrac{x+6}{x+4}\)

\(=\dfrac{x+4}{x+5}:\left(\dfrac{x+5}{x+6}\cdot\dfrac{x+6}{x+4}\right)\)

\(=\dfrac{x+4}{x+5}:\dfrac{x+5}{x+4}\)

\(=\dfrac{x+4}{x+5}\cdot\dfrac{x+4}{x+5}\)

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

b) \(\dfrac{x-7}{x+8}:\left(\dfrac{x-7}{x-9}:\dfrac{x+8}{x-9}\right)\)

\(=\dfrac{x-7}{x+8}:\left(\dfrac{x-7}{x-9}\cdot\dfrac{x-9}{x+8}\right)\)

\(=\dfrac{x-7}{x+8}:\dfrac{x-7}{x+8}\)

\(=\dfrac{x-7}{x-8}\cdot\dfrac{x+8}{x-7}\)

\(=1\)

a: ĐKXĐ:\(x\notin\left\{-4;-5;-6\right\}\)

\(\dfrac{x+4}{x+5}:\dfrac{x+5}{x+6}:\dfrac{x+6}{x+4}\)

\(=\dfrac{x+4}{x+5}\cdot\dfrac{x+6}{x+5}\cdot\dfrac{x+4}{x+6}\)

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

b: ĐKXĐ:\(x\notin\left\{7;-8;9\right\}\)

\(\dfrac{x-7}{x+8}:\left(\dfrac{x-7}{x-9}:\dfrac{x+8}{x-9}\right)\)

\(=\dfrac{x-7}{x+8}:\left(\dfrac{x-7}{x-9}\cdot\dfrac{x-9}{x+8}\right)\)

\(=\dfrac{x-7}{x+8}:\dfrac{x-7}{x+8}\)

=1

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

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

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

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

b: ĐKXĐ: \(\left\{{}\begin{matrix}x\notin\left\{0;\dfrac{3}{2}\right\}\\y\ne0\end{matrix}\right.\)

\(\dfrac{xy}{2x-3}:\dfrac{x^2y^2}{6-4x}\)

\(=\dfrac{xy}{2x-3}\cdot\dfrac{-4x+6}{x^2y^2}\)

\(=\dfrac{xy}{2x-3}\cdot\dfrac{-2\left(x-3\right)}{x^2y^2}\)

\(=\dfrac{-2}{xy}\)

c: ĐKXĐ: \(x\notin\left\{0;1;-2;2\right\}\)

\(\dfrac{x^2+2x}{x^2-2x+1}:\dfrac{x^2-4}{x^2-x}\)

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

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

d: ĐKXĐ: \(x\ne2;y\ne-\dfrac{2x}{3}\)

\(\dfrac{2x+3y}{2-x}:\dfrac{4x^2+12xy+9y^2}{x^3-8}\)

\(=\dfrac{2x+3y}{-\left(x-2\right)}\cdot\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{4x^2+12xy+9y^2}\)

\(=\dfrac{-\left(2x+3y\right)}{1}\cdot\dfrac{x^2+2x+4}{\left(2x+3y\right)^2}\)

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

a. Đây là 1 đề bài ko rõ ràng, ko biết là \(x^2-9:\dfrac{2x+6}{x-3}\) hay \(\left(x^2-9\right):\dfrac{2x+6}{x-3}\)

b.

\(\dfrac{xy}{2x-3}:\dfrac{x^2y^2}{6-4x}=\dfrac{xy}{2x-3}.\dfrac{-2\left(2x-3\right)}{x^2y^2}=\dfrac{-2}{xy}\)

c.

\(\dfrac{x^2+2x}{x^2-2x+1}:\dfrac{x^2-4}{x^2-x}=\dfrac{x\left(x+2\right)}{\left(x-1\right)^2}.\dfrac{x\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)

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

d.

\(\dfrac{2x+3y}{2-x}:\dfrac{4x^2+12xy+9y^2}{x^3-8}=\dfrac{2x+3y}{-\left(x-2\right)}.\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(2x+3y\right)^2}\)

\(=-\dfrac{x^2+2x+4}{2x+3y}\)

Em chắc là đề này đúng chứ? Kết quả ra rất xấu, nên có lẽ 1 vài hệ số trong đề ko chính xác.

Em kiểm tra lại rồi, đề ghi đúng như vậy thầy ạ.

\(\Leftrightarrow x^3+3x-5=y\left(x^2+2\right)\)

\(\Leftrightarrow y=\dfrac{x^3+3x-5}{x^2+2}=x+\dfrac{x-5}{x^2+2}\)

\(\Rightarrow\dfrac{x-5}{x^2+2}\in Z\)

\(\dfrac{x-5}{x^2+2}=\dfrac{x^2+2-x^2+x-3}{x^2+2}=1-\dfrac{x^2-x+3}{x^2+2}\)

Do \(\left\{{}\begin{matrix}x^2-x+3=\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\\x^2+2>0\end{matrix}\right.\)

\(\Rightarrow1-\dfrac{x^2-x+3}{x^2+2}< 1\)

\(\dfrac{x-5}{x^2+2}=\dfrac{-3\left(x^2+2\right)+3x^2+x+1}{x^2+2}=-3+\dfrac{3\left(x+\dfrac{1}{6}\right)^2+\dfrac{11}{12}}{x^2+2}>-3\)

\(\Rightarrow1< \dfrac{x-5}{x^2+2}< -3\)

Đến đây xét các trường hợp: \(\dfrac{x-5}{x^2+2}=\left\{-2;-1;0\right\}\) là được

Em cảm ơn nhiều ạ :3

\(\dfrac{1}{x^2+2x-3}=\dfrac{1}{\left(x+1\right)^2}+\dfrac{1}{48}\left(x\ne-3;\pm1\right)\)

\(\Leftrightarrow\dfrac{1}{\left(x-1\right)\left(x+3\right)}=\dfrac{1}{\left(x+1\right)^2}+\dfrac{1}{48}\)

\(\Leftrightarrow\dfrac{48\left(x+1\right)^2}{48\left(x+1\right)^2\left(x-1\right)\left(x+3\right)}=\dfrac{48\left(x-1\right)\left(x+3\right)}{48\left(x+1\right)^2\left(x-1\right)\left(x+3\right)}+\dfrac{\left(x+1\right)^2\left(x-1\right)\left(x+3\right)}{48\left(x+1\right)^2\left(x-1\right)\left(x+3\right)}\)

\(\Leftrightarrow48\left(x^2+2x+1\right)=48\left(x^2+2x-3\right)+\left(x+1\right)^2\left(x-1\right)\left(x+3\right)\)

\(\Leftrightarrow48x^2+96x+48=48x^2+96x-144+\left(x^4+4x^3+2x^2-4x-3\right)\)

\(\Leftrightarrow48+144=x^4+4x^3+2x^2-4x-3\)

\(\Leftrightarrow x^4+4x^3+2x^2-4x-3-192=0\)

\(\Leftrightarrow x^4+4x^3+2x^2-4x-195=0\)

\(\Leftrightarrow x^4+7x^3+23x^2+65x-3x^3-21x-69x-195=0\)

\(\Leftrightarrow x\left(x^3+7x^3+23x+65\right)-3\left(x^3+7x^3+23x+65\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^3+7x^3+23x+65\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^3+5x^2+2x^2+10x+13x+65\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[x^2\left(x+5\right)+2x\left(x+5\right)+13\left(x+5\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+5\right)\left(x^2+2x+13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+5=0\\x^2+2x+13=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\left(tm\right)\) (vì `x^2+2x+13=0` vô nghiệm) 

ĐKXĐ:...

Đặt \(\left(x+1\right)^2=t\)

\(\Rightarrow\dfrac{1}{t-4}=\dfrac{1}{t}+\dfrac{1}{48}\)

\(\Rightarrow48t=48\left(t-4\right)+t\left(t-4\right)\)

\(\Rightarrow...\)

Ta có: \(D\left(x\right)=2x^2+3y^2+4z^2-2\left(x+y+z\right)+2\)

\(=2x^2+3y^2+4z^2-2x-2y-2z+2\)

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

\(=2\left(x^2-x\right)+3\left(y^2-\dfrac{2}{3}y\right)+4\left(z^2-\dfrac{1}{2}z\right)+2\)

\(=2\left[x^2-2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^2\right]+3\left[y^2-2\cdot y\cdot\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{3}\right)^2\right]+4\left[z^2-2\cdot z\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2-\left(\dfrac{1}{4}\right)^2\right]+2\)\(=2\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{2}+3\left(y-\dfrac{1}{3}\right)^2-\dfrac{1}{3}+4\left(z-\dfrac{1}{4}\right)^2-\dfrac{1}{4}+2\)

\(=2\left(x-\dfrac{1}{2}\right)^2+3\left(y-\dfrac{1}{3}\right)^2+4\left(z-\dfrac{1}{4}\right)^2+\dfrac{11}{12}\)

Mà: \(\left\{{}\begin{matrix}2\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\\3\left(y-\dfrac{1}{3}\right)^2\ge0\forall y\\4\left(y-\dfrac{1}{4}\right)^2\ge0\forall z\end{matrix}\right.\)

\(\Rightarrow D\left(x\right)=2\left(x-\dfrac{1}{2}\right)^2+3\left(y-\dfrac{1}{3}\right)^2+4\left(z-\dfrac{1}{4}\right)^2+\dfrac{11}{12}\ge\dfrac{11}{12}\forall x,y,z\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y-\dfrac{1}{3}=0\\z-\dfrac{1}{4}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{3}\\z=\dfrac{1}{4}\end{matrix}\right.\) 

Vậy: ... 

$D(x)=2\left(x^2-x\right)+\left(3y^2-2y\right)+\left(4z^2-2z\right)+2$

$=2\left(x^2-x+\genfrac{}{}{0ex}{}{1}{4}\right)+3\left(y^2-\genfrac{}{}{0ex}{}{2}{3}y+\genfrac{}{}{0ex}{}{1}{9}\right)+\left[(2z{)}^2-2z+\genfrac{}{}{0ex}{}{1}{4}\right]+2-\genfrac{}{}{0ex}{}{1}{2}-\genfrac{}{}{0ex}{}{1}{3}-\genfrac{}{}{0ex}{}{1}{4}$

$=2{\left(x-\genfrac{}{}{0ex}{}{1}{2}\right)}^2+3{\left(y-\genfrac{}{}{0ex}{}{1}{3}\right)}^2+{\left(2z-\genfrac{}{}{0ex}{}{1}{2}\right)}^2+\genfrac{}{}{0ex}{}{11}{2}\ge \genfrac{}{}{0ex}{}{11}{2}$

Vậy giá trị nhỏ nhất của $D$ là: $\genfrac{}{}{0ex}{}{11}{2}$ tại $(x,y,z)=\left(\genfrac{}{}{0ex}{}{1}{2};\genfrac{}{}{0ex}{}{1}{3};\genfrac{}{}{0ex}{}{1}{4}\right)$.

a: Gọi I là trung điểm của MC

=>\(MI=IC=\dfrac{MC}{2}\)

mà \(AM=\dfrac{MC}{2}\)

nên AM=MI=IC

Vì AM=MI nên M là trung điểm của AI

Xét ΔBMC có

D,I lần lượt là trung điểm của CB,CM

=>DI là đường trung bình của ΔBMC

=>DI//BM và \(DI=\dfrac{BM}{2}\)

DI//BM nên OM//DI

Xét ΔADI có

M là trung điểm của AI

MO//DI

Do đó: O là trung điểm của AD

b: Xét ΔADI có

O,M lần lượt là trung điểm của AD,AI

=>OM là đường trung bình của ΔADI

=>\(OM=\dfrac{1}{2}DI=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot BM=\dfrac{1}{4}BM\)

a: Gọi I là trung điểm của MC

=>��=��=��2MI=IC=MC:2

​

m��=��2AM=MC:2

​

=> AM=MI=IC

Vì AM=MI => M là trung điểm của AI

Xét ΔBMC có:

D,I lần lượt là trung điểm của CB,CM

=>DI là đường trung bình của ΔBMC

=>DI//BM , ��=��2DI=BM:2

​

DI//BM => OM//DI

Xét ΔADI có:

M là trung điểm của AI

MO//DI

=> O là trung điểm của AD

b) Xét ΔADI có

O,M lần lượt là trung điểm của AD,AI

=>OM là đường trung bình của ΔADI

=>��=12��=12⋅12⋅��=14��OM=

​DI:2=BM:4(đpcm)