\(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)\(\left(đkcđ:x\ne\pm3;x\ne-\frac{1}{2}\right)\)

\(=\left(\frac{\left(x-1\right).\left(x-3\right)+2.\left(x+3\right)-\left(x^2+3\right)}{x^2-9}\right):\left(\frac{2x-1-\left(2x+1\right)}{2x+1}\right)\)

\(=\frac{x^2-4x+3+2x+6-x^2-3}{x^2-9}:\frac{-2}{2x+1}\)

\(=\frac{-2x-6}{x^2-9}.\frac{2x+1}{-2}\)

\(=\frac{-2\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}.\frac{2x+1}{-2}\)

\(=\frac{2x+1}{x-3}\)

b)\(\left|x+1\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x+1=\frac{1}{2}\\x+1=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\left(koTMđkxđ\right)\\x=-\frac{3}{2}\left(TMđkxđ\right)\end{cases}}}\)

thay \(x=-\frac{3}{2}\)  vào P tâ đc:   \(P=\frac{2x+1}{x-3}=\frac{2.\left(-\frac{3}{2}\right)+1}{-\frac{3}{2}-3}=\frac{4}{9}\)

c)ta có:\(P=\frac{x}{2}\Leftrightarrow\frac{2x+1}{x-3}=\frac{x}{2}\)

\(\Rightarrow2.\left(2x+1\right)=x.\left(x-3\right)\)

\(\Leftrightarrow4x+2=x^2-3x\)

\(\Leftrightarrow x^2-7x-2=0\)

\(\Leftrightarrow x^2-2.\frac{7}{2}+\frac{49}{4}-\frac{57}{4}=0\)

\(\Leftrightarrow\left(x-\frac{7}{2}\right)^2-\frac{57}{4}=0\)

\(\Leftrightarrow\left(x-\frac{7}{2}-\frac{\sqrt{57}}{2}\right).\left(x-\frac{7}{2}+\frac{\sqrt{57}}{2}\right)\)

bạn tự giải nốt nhé!!

d)\(x\in Z;P\in Z\Leftrightarrow\frac{2x+1}{x-3}\in Z\Leftrightarrow\frac{2x-6+7}{x-3}=2+\frac{7}{x-3}\in Z\)

\(2\in Z\Rightarrow\frac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

bạn tự làm nốt nhé

a, \(\left(\dfrac{x^2-4x+3+2x+6-x^2-3}{\left(x+3\right)\left(x-3\right)}\right):\left(\dfrac{2x-1-2x-1}{2x+1}\right)\)

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

b, \(\left|x+1\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}-1\\x=-\dfrac{1}{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(ktmđk\right)\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Thay x = -3/2 ta được \(\dfrac{2\left(-\dfrac{3}{2}\right)+1}{-\dfrac{3}{2}+3}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\)

a, ĐKXĐ : \(x-1\ne0\)

=> \(x\ne1\)

TH₁ : \(x-2\ge0\left(x\ge2\right)\)

=> \(\left|x-2\right|=x-2=1\)

=> \(x=3\left(TM\right)\)

- Thay x = 3 vào biểu thức P ta được :

\(P=\frac{3+2}{3-1}=\frac{5}{2}\)

TH₂ : \(x-2< 0\left(x< 2\right)\)

=> \(\left|x-2\right|=2-x=1\)

=> \(x=1\left(KTM\right)\)

Vậy giá trị của P là \(\frac{5}{2}\) .

a) \(P=\frac{x+2}{x-1}\) \(\left(ĐKXĐ:x\ne1\right)\)

Ta có: \(\left|x-2\right|=1\text{⇔}\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\text{⇔}\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) (loại x = 1 vì x ≠ 1)

Thay \(x=3\) vào P, ta có:

\(P=\frac{3+2}{3-2}=\frac{5}{1}=5\)

Vậy P = 5 tại x = 3.

b) \(Q=\frac{x-1}{x}+\frac{2x+1}{x^2+x}=\frac{x-1}{x}+\frac{2x+1}{x\left(x+1\right)}=\frac{x^2-1}{x\left(x+1\right)}+\frac{2x+1}{x\left(x+1\right)}\) (ĐKXĐ: x ≠ 0, x ≠ -1)

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

Ai giúp mk với mk đang cần gấp

Mk làm được hết

mà vẫn cứ sai hoài à

tìm mãi ko thấy lỗi sai

a, \(C=5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)-2\left(5-3x\right)^2\)

\(C=5.\left(4x^2-4x+1\right)+4\left(x^2+3x-x-3\right)-2.\left(25-75x+9x^2\right)\)

\(C=20x^2-20x+5+4x^2+8x-12-50+150x-18x^2\)

\(=\left(20x^2+4x^2-18x^2\right)+\left(-20x+8x+150x\right)+\left(5-12-50\right)\)

\(C=6x^2+138x-57\)

Chúc bạn học tốt!!! Cũng không chắc có đúng hay sai nữa do cồng kềnh quá !

ta có :

a, sửa đề : \(C=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}+\frac{1}{2-x}\)ĐK : \(x\ne-3;2\)

\(=\frac{\left(x+2\right)\left(x-2\right)-5-x-3}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-12-x}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)

b, Ta có : \(x^2-x=2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x+1\right)\left(x-2\right)=0\Leftrightarrow x=-1;x=2\)

Kết hợp với giả thiết vậy x = -1 

Thay x = -1 vào biểu thức C ta được : \(\frac{-1-4}{-1-2}=-\frac{5}{-3}=\frac{5}{3}\)

c, Ta có : \(C=\frac{1}{2}\Rightarrow\frac{x-4}{x-2}=\frac{1}{2}\Rightarrow2x-8=x-2\Leftrightarrow x=6\)( tm )

d, \(C>1\Rightarrow\frac{x-4}{x-2}>1\Rightarrow\frac{x-4}{x-2}-1>0\Leftrightarrow\frac{x-4-x+2}{x-2}>0\Leftrightarrow\frac{-2}{x-2}>0\)

\(\Rightarrow x-2< 0\Leftrightarrow x< 2\)vì -2 < 0 

e, tự làm nhéee 

f, \(C< 0\Rightarrow\frac{x+4}{x+2}< 0\)

mà x + 4 > x + 2 

\(\hept{\begin{cases}x+4>0\\x+2< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-4\\x< -2\end{cases}\Leftrightarrow-4< x< -2}}\)

Vì \(x\inℤ\Rightarrow x=-3\)( ktmđk )

Vậy ko có x nguyên để C < 0 

g, Ta có :  \(\frac{x+4}{x+2}=\frac{x+2+2}{x+2}=1+\frac{2}{x+2}\)

Để C nguyên khi \(x+2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

h, Ta có : \(D=C\left(x^2-4\right)=\frac{x+4}{x+2}.\frac{\left(x-2\right)\left(x+2\right)}{1}=x^2+2x-8\)

\(=\left(x+1\right)^2-9\ge-9\)

Dấu ''='' xảy ra khi x = -1 

Vậy GTNN D là -9 khi x = -1 

\(\left(x+5\right)\left(x^2-5x+25\right)\)

\(=\left(x+5\right)\left(x^2-5.x+5^2\right)\)

\(=x^3+5^3\)

\(=x^3+125\)

3) \(27-y^3\)

\(=3^3-y^3\)

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

bn ơi cho mk hỏi tại sao lại ko nhận 3 vậy !!!