\(a,ĐK:x\ne\pm3\\ Sửa:M=\dfrac{x}{x+3}+\dfrac{2x}{x-3}+\dfrac{9-3x^2}{x^2-9}\\ M=\dfrac{x^2-3x+2x^2+6x+9-3x^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x-3}\\ b,x=2\Leftrightarrow M=\dfrac{3}{2-3}=-3\\ c,M\in Z\Leftrightarrow x-3\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{0;2;4;6\right\}\left(tm\right)\)

Lời giải:

a. ĐKXĐ: $x\neq 1$

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

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

b.

$x^2-x=0\Leftrightarrow x(x-1)=0$

$\Leftrightarrow x=0$ hoặc $x=1$

Vì $x\neq 1$ theo ĐKXĐ nên $x=0$

Khi đó: $P=\frac{1}{0^2+0+1}=1$
c.

Ta thấy:

$1>0$

$x^2+x+1=(x+\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}>0$ với mọi $x\neq 1$

$\Rightarrow P=\frac{1}{x^2+x+1}>0$

Hay $P$ luôn dương với mọi $x\neq 1$

Câu 1:Ta có:

a) \(\left|x-3\right|=5\Leftrightarrow\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

b) \(\left|2x+3\right|=2.\left|4-x\right|\)

+)Xét \(\left\{{}\begin{matrix}2x+3\ge0\\4-x\ge0\end{matrix}\right.\) \(\Leftrightarrow\dfrac{-3}{2}\le x\le4\)

Khi đó \(2x+3=2\left(4-x\right)\Leftrightarrow2x+3=8-2x\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\left(tm\right)\)

+) Xét \(\left\{{}\begin{matrix}2x+3\ge0\\4-x\le0\end{matrix}\right.\) \(\Leftrightarrow x\ge4\)

Khi đó: \(2x+3=2\left(x-4\right)=2x-8\Leftrightarrow0x=-11\left(vl\right)\)

+) Xét \(\left\{{}\begin{matrix}2x+3\le0\\4-x\ge0\end{matrix}\right.\) \(\Leftrightarrow x\le\dfrac{-3}{2}\)

Khi đó: \(-\left(2x+3\right)=2.\left(4-x\right)\Leftrightarrow-2x-3=8-2x\left(vl\right)\)

+)Xét \(\left\{{}\begin{matrix}2x+3\le0\\4-x\le0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{-3}{2}\\x\ge4\end{matrix}\right.\) \(\left(vl\right)\)

Vậy...

c) ĐKXĐ : \(3-x\ge0\Leftrightarrow x\le3\)

+)Xét \(x^{^2}-3x+1\ge0\)

\(\Leftrightarrow x^2-3x+1=3-x\Leftrightarrow x^2-2x-2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=\sqrt{3}\\x-1=-\sqrt{3}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1+\sqrt{3}\left(tm\right)\\x=1-\sqrt{3}\left(tm\right)\end{matrix}\right.\)

+)Xét \(x^{^2}-3x+1\le0\)

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

\(\Leftrightarrow x^2-3x+1=x-3\Leftrightarrow x^2-4x+4=0\)

\(\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\left(tm\right)\)

Vậy...

Câu 2:

 Ta có:

Phương trình \(\left(x+3\right)\left(x^2-2x+m-1\right)=0\) có một nghiệm là \(x=-3\)

\(\Rightarrow\)Phương trình \(\left(x+3\right)\left(x^2-2x+m-1\right)=0\) có ba nghiệm phân biệt khi và chỉ khi \(x^2-2x+m-1=0\) có 2 nghiệm phân biệt và khác \(-3\)

Ta có:  \(x^2-2x+m-1=0\) có 2 nghiệm phân biệt khi và chỉ khi \(\text{△}>0\Leftrightarrow8-4m>0\Leftrightarrow m< 2\)

 Gọi \(x_1\) và \(x_2\) là 2 nghiệm của phương trình \(x^2-2x+m-1=0\).Theo hệ thức Vi-ét ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{-2}{1}=2\\x_1x_2=\dfrac{m-1}{1}=m-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x_1=2-x_2\\\left(2-x_2\right).x_2=m-1\end{matrix}\right.\)

Nếu \(x_2\ne-3\) thì \(m-1\ne-15\Leftrightarrow m\ne-14\).

Do vai trò của  \(x_1\) và \(x_2\) là như nhau nên  \(x^2-2x+m-1=0\) có 2 nghiệm phân biệt và khác \(-3\) khi và chỉ khi: \(\left\{{}\begin{matrix}m< 2\\m\ne-14\end{matrix}\right.\)

a. ĐK: \(x\ne\pm2\)
\(M=\left[\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x+7}{\left(x-2\right)\left(x+2\right)}\right]:\dfrac{3-x+x-2}{x-2}\)

\(=\dfrac{x^2+2x-\left(x^2-2x+x-2\right)-2x-7}{\left(x-2\right)\left(x+2\right)}.\left(x-2\right)\)

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

b. \(\dfrac{x-5}{x+2}< 1\Leftrightarrow\dfrac{x-5}{x+2}-1< 0\)

\(\Leftrightarrow\dfrac{-7}{x+2}< 0\Leftrightarrow x+2>0\)

\(\Leftrightarrow x>-2\)
Vậy \(x>-2,x\ne2\)

ĐKXĐ: \(x>0;x\ne1\)

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

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

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

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

b.

\(P=x-\sqrt{x}+1=x-\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(P_{min}=\dfrac{3}{4}\) khi \(x=\dfrac{1}{4}\)

a) đk: \(\left\{{}\begin{matrix}\sqrt{x}+1>0\\\sqrt{x}-1>0\\x>0\end{matrix}\right.=>\sqrt{x}>\pm1\)

 rút gọn pt

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

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

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

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

b: Thay x=-1/2 vào M, ta được:

\(M=\dfrac{\dfrac{1}{2}+1}{2}=\dfrac{3}{2}:2=\dfrac{3}{4}\)

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

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

b, Thay x  =-1/2 vào ta đc 

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

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

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

b, Thay x = -1/2 vào ta đc 

\(M=\dfrac{\dfrac{1}{2}+1}{2\left(\dfrac{-1}{2}+1\right)}=\dfrac{\dfrac{3}{2}}{\dfrac{2.1}{2}}=\dfrac{3}{2}\)

1=13500

2=103500

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