ĐKXĐ: \(x\ne-1\)

\(\left(x+2\right)^2+\left(\dfrac{x+2}{x+1}\right)^2+2\dfrac{\left(x+2\right)^2}{x+1}-\dfrac{2\left(x+2\right)^2}{x+1}=8\)

\(\Leftrightarrow\left(x+2+\dfrac{x+2}{x+1}\right)^2-\dfrac{2\left(x+2\right)^2}{x+1}=8\)

\(\Leftrightarrow\left(\dfrac{\left(x+2\right)^2}{x+1}\right)^2-\dfrac{2\left(x+2\right)^2}{x+1}=8\)

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

\(\Rightarrow t^2-2t-8=0\Rightarrow\left[{}\begin{matrix}t=4\\t=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{\left(x+2\right)^2}{x+1}=4\\\dfrac{\left(x+2\right)^2}{x+1}=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+4x+4=4x+4\\x^2+4x+4=-2x-2\end{matrix}\right.\)

\(\Rightarrow...\)

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

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

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

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

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

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

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

1 giờ 30 phút = \(1,5giờ\)

Gọi độ dài quãng đường AB là a\(\left(km\right)\), ta có:

\(\frac{a}{40}-\frac{a}{50}=1,5\left(giờ\right)\)

\(\Rightarrow\frac{5a}{200}+\frac{4\left(-a\right)}{200}=1,5\left(giờ\right)\)

\(\Rightarrow\frac{5a+4\left(-a\right)}{200}=1,5\left(giờ\right)\)

\(\Rightarrow\frac{5a-4a}{200}=1,5\left(giờ\right)\)

\(\Rightarrow\frac{a}{200}=1,5\left(giờ\right)\)

\(\Rightarrow a=1,5\text{x}200\)

\(\Rightarrow a=300\)

Vậy quãng đường AB dài \(300km\).

Gọi quãng đường AB là x ( x > 0 ) 

Theo bài ra ta có pt 

\(\dfrac{x}{40}-\dfrac{x}{50}=1+\dfrac{1}{2}=\dfrac{3}{2}\Rightarrow x=300\left(tm\right)\)

viết lại đề :\(\frac{1}{x}+\frac{1}{y}=\frac{1}{3}\)

vì vai trò của x,y là như nhau,ko mất tính tổng quát,ta giả sử: \(1\le x\le y\left(do..x;y\inℕ^∗\right)\)

thay x,y bởi x ta có:\(\frac{1}{x}+\frac{1}{x}\ge\frac{1}{x}+\frac{1}{y}=\frac{1}{3}\)(do \(x\le y\) nên theo TC của phân số thì \(\frac{1}{x}+\frac{1}{x}\ge\frac{1}{x}+\frac{1}{y}\))

\(\Leftrightarrow\frac{2}{x}\ge\frac{1}{3}\Leftrightarrow x\le6\Rightarrow1\le x\le6\)

ta lại có:\(\frac{1}{x}+\frac{1}{y}=\frac{1}{3}\Leftrightarrow\frac{1}{x}>\frac{1}{3}\Leftrightarrow3< x\le6\)

ta có 3TH:

\(x=4\Leftrightarrow\frac{1}{4}+\frac{1}{y}=\frac{1}{3}\Leftrightarrow\frac{1}{y}=\frac{1}{12}\Leftrightarrow y=12\left(TM\right)\)

\(x=5\)bạn thay vào thì y=2/15(ko TM)

\(x=6\),ta sẽ tìm đc y=6(TM)

vậy(x;y)cần tìm là  (4;12),(6;6) và (12;4)

viết lại đề:\(\left(x+y+1\right)^2=3.\left(x^2+y^2+1\right)\)

AD bất dẳng thức bunhia cho 3 số la có:

\(3.\left(x^2+y^2+1\right)=\left(1+1+1\right).\left(x^2+y^2+1\right)\ge\left(1.x+1.y+1.1\right)^2=\left(x+y+1\right)^2\)

đấu bằng xảy ra <=>  \(x=y=1\) (thử lại thì ta thấy TM)

vậy \(x=y=1\)

(x - 2) (x - 1) (x + 1) (x + 2) = 4

x (-2 - 2 + 1 + 2) = 4

x . 0 = 4

0 = 4

\(\rightarrow\)vô nghiệm

ủa nhầm

x ( -2 - 1 + 1 + 2)

\(\dfrac{x+3}{x}-1< 0\Leftrightarrow\dfrac{x+3-x}{x}< 0\Leftrightarrow\dfrac{3}{x}< 0\Rightarrow x< 0\)

132121

132121 nha bn

123564

123564 nha bn