a) P xác định \(\Leftrightarrow\hept{\begin{cases}x\ne0\\x+5\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}}\)

Vậy P xác định \(\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)

b) \(P=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)

\(P=\frac{x\left(x+2\right)}{2\left(x+5\right)}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)

\(P=\frac{x^2\left(x+2\right)}{2x\left(x+5\right)}+\frac{\left(x-5\right)\left(x+5\right)2}{2x\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}\)

\(P=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)

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

Có: \(P=0\)

\(\Rightarrow P=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}=0\Leftrightarrow x\left(x^2+4x-5\right)=0\Leftrightarrow x^2+4x-5=0\)

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

\(\Leftrightarrow x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

Vậy \(P=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

a)   ĐKXĐ:   \(x\ne\pm2\)

\(A=\frac{x}{x-2}-\frac{2}{x+2}\)

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

\(=\frac{x^2+2x-2x+4}{x^2-4}\)\(=\frac{x^2+4}{x^2-4}\)

b)   \(A>0\) \(\Rightarrow\)\(\frac{x^2+4}{x^2-4}>0\) 

Mà    \(x^2+4>0\)  \(\Rightarrow\)\(x^2-4>0\)

\(\Rightarrow\)\(x^2>4\)

Nếu   x   dương  thì      \(x>\sqrt{4}=2\)

Nếu   x  âm  thì   \(x< \sqrt{4}=2\)

a) P xác định \(\Leftrightarrow\hept{\begin{cases}2x+10\ne0\\x\ne0\\2x\left(x+5\right)\ne0\end{cases}\Leftrightarrow x\ne\left\{-5;0\right\}}\)

b) \(P=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)

\(P=\frac{x^2\left(x+2\right)}{2x\left(x+5\right)}+\frac{2\left(x-5\right)\left(x+5\right)}{2x\left(x+5\right)}+\frac{5\left(10-x\right)}{2x\left(x+5\right)}\)

\(P=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)

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

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

\(P=\frac{x^2\left(x+5\right)-x\left(x+5\right)}{2x\left(x+5\right)}\)

\(P=\frac{\left(x+5\right)\left(x^2-x\right)}{2x\left(x+5\right)}\)

\(P=\frac{x\left(x-1\right)}{2x}\)

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

c) Để P = 0 thì \(x-1=0\Leftrightarrow x=1\)( thỏa mãn ĐKXĐ )

Để P = 1/4 thì \(\frac{x-1}{2}=\frac{1}{4}\)

\(\Leftrightarrow4\left(x-1\right)=2\)

\(\Leftrightarrow4x-4=2\)

\(\Leftrightarrow4x=6\)

\(\Leftrightarrow x=\frac{3}{2}\)( thỏa mãn ĐKXĐ )

d) Để P > 0 thì \(\frac{x-1}{2}>0\)

Mà 2 > 0, do đó để P > 0 thì \(x-1>0\Leftrightarrow x>1\)

Để P < 0 thì \(\frac{x-1}{2}< 0\)

Mà 2 > 0, do đó để P < 0 thì \(x-1< 0\Leftrightarrow x< 1\)

sai đề

\(\left(x^4+\dfrac{1}{x^4}\right)\left(x^3+\dfrac{1}{x^3}\right)-\left(x+\dfrac{1}{x}\right)=x^7+\dfrac{x^4}{x^3}+\dfrac{x^3}{x^4}+\dfrac{1}{x^7}-x-\dfrac{1}{x}=x^7+\dfrac{1}{x^7}+x+\dfrac{1}{x}-x-\dfrac{1}{x}\)\(=x^7+\dfrac{1}{x^7}=VT\Rightarrowđpcm\)

\(b,x+\dfrac{1}{x}=7\Rightarrow\left(x+\dfrac{1}{x}\right)^2=49\)

\(\Leftrightarrow x^2+2.x.\dfrac{1}{x}+\dfrac{1}{x^2}=49\)

\(\Leftrightarrow x^2+\dfrac{1}{x^2}=49-2=47\)

\(\left(x+\dfrac{1}{x}\right)=7\Rightarrow\left(x+\dfrac{1}{x}\right)^3=343\)

\(\Leftrightarrow x^3+3x^2\dfrac{1}{x}+3x\dfrac{1}{x^2}+\dfrac{1}{x^3}=343\)

\(\Leftrightarrow x^3+\dfrac{1}{x^3}+3x\dfrac{1}{x}\left(x+\dfrac{1}{x}\right)=343\)

\(\Leftrightarrow x^3+\dfrac{1}{x^3}+3.7=343\)

\(\Leftrightarrow x^3+\dfrac{1}{x^3}=343-21=322\)

\(\left(x^2+\dfrac{1}{x^2}\right)\left(x^3+\dfrac{1}{x^3}\right)=47.322\)

\(\Leftrightarrow x^5+\dfrac{x^2}{x^3}+\dfrac{x^3}{x^2}+\dfrac{1}{x^5}=15134\)

\(\Leftrightarrow x^5+\dfrac{1}{x^5}+x+\dfrac{1}{x}=15134\)

\(\Leftrightarrow x^5+\dfrac{1}{x^5}+7=15134\)

\(\Rightarrow x^5+\dfrac{1}{x^5}=15134-7=15127\)

a. \(\left(x^4+\dfrac{1}{x^4}\right)\left(x^3+\dfrac{1}{x^3}\right)-\left(x+\dfrac{1}{x}\right)\)

\(x^7+x+\dfrac{1}{x}+\dfrac{1}{x^7}-\left(x+\dfrac{1}{x}\right)=x^7+\dfrac{1}{x^7}\)

b. Ta có:

\(\left(x+\dfrac{1}{x}\right)^2=49\)

\(\Leftrightarrow x^2+\dfrac{1}{x^2}=49-2=47\)

\(\left(x+\dfrac{1}{x}\right)^3=343\)

\(\Leftrightarrow x^3+\dfrac{1}{x^3}+3\left(x+\dfrac{1}{x}\right)=343\)

\(\Leftrightarrow x^3+\dfrac{1}{x^3}=343-3.7=322\)

\(\Rightarrow\left(x^2+\dfrac{1}{x^2}\right)\left(x^3+\dfrac{1}{x^3}\right)=47.322=15134\)

\(\Leftrightarrow x^5+\dfrac{1}{x}+x+\dfrac{1}{x^5}=15134\)

\(\Leftrightarrow x^5+\dfrac{1}{x^5}=15134-7=15127\)

Chúc bạn hok tốt

5) 3x - 1 < 8

⇔ 3x < 9

⇔ x < 3

4) -8x > 24

<=> x > 32

Do \(x>0:\)

\(x^2+\dfrac{1}{x^2}=7\Leftrightarrow x^2+2.x.\dfrac{1}{x}+\dfrac{1}{x^2}=9\Leftrightarrow\left(x+\dfrac{1}{x}\right)^3=9\Rightarrow x+\dfrac{1}{x}=3\)

\(\Rightarrow\left(x+\dfrac{1}{x}\right)^3=3^3\Leftrightarrow x^3+3x.\dfrac{1}{x}.\left(x+\dfrac{1}{x}\right)+\dfrac{1}{x^3}=27\)

\(\Leftrightarrow x^3+3.1.3+\dfrac{1}{x^3}=27\Leftrightarrow x^3+\dfrac{1}{x^3}=18\)

\(\Rightarrow\left(x^2+\dfrac{1}{x^2}\right)\left(x^3+\dfrac{1}{x^3}\right)=7.18\Leftrightarrow x^5+\dfrac{1}{x}+x+\dfrac{1}{x^5}=126\)

\(\Leftrightarrow x^5+3+\dfrac{1}{x^5}=126\Rightarrow x^5+\dfrac{1}{x^5}=123\)

Ở dòng đầu gõ nhầm xíu \(\left(x+\dfrac{1}{x}\right)^2=9\) chứ ko phải \(\left(x+\dfrac{1}{x}\right)^3=9\)

a: ĐKXĐ: x<>0; x<>-5

b: \(P=\dfrac{x^3+2x^2+2\left(x^2-25\right)+50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2+50-5x+2x^2-50}{2x\left(x+5\right)}\)

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

Để P=0 thì x-1=0

=>x=1

c: Để P=-1/4 thì x-1/2=-1/4

=>x-1=-1/2

=>x=1/2

1) \(\left(x-3\right)\left(x-5\right)+44\)

\(=x^2-3x-5x+15+44\)

\(=x^2-8x+59\)

\(=x^2-2.x.4+4^2+43\)

\(=\left(x-4\right)^2+43\ge43>0\)

\(\rightarrowĐPCM.\)

2) \(x^2+y^2-8x+4y+31\)

\(=\left(x^2-8x\right)+\left(y^2+4y\right)+31\)

\(=\left(x^2-2.x.4+4^2\right)-16+\left(y^2+2.y.2+2^2\right)-4+31\)

\(=\left(x-4\right)^2+\left(y+2\right)^2+11\ge11>0\)

\(\rightarrowĐPCM.\)

3)\(16x^2+6x+25\)

\(=16\left(x^2+\dfrac{3}{8}x+\dfrac{25}{16}\right)\)

\(=16\left(x^2+2.x.\dfrac{3}{16}+\dfrac{9}{256}-\dfrac{9}{256}+\dfrac{25}{16}\right)\)

\(=16\left[\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{256}\right]\)

\(=16\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{16}>0\)

-> ĐPCM.

4) Tương tự câu 3)

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

\(=x^2+2.x.\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{2}\)

\(=\left(x+\dfrac{1}{3}\right)^2+\dfrac{7}{18}>0\)

-> ĐPCM.

6) Tương tự câu 5)

7) 8) 9) Tương tự câu 3).

Giải rõ giúp mình với

Ta có:

\(x^2+\dfrac{1}{x^2}=7\)

\(\Rightarrow\left(x+\dfrac{1}{x}\right)^2-2=7\)

\(\Rightarrow\left(x+\dfrac{1}{x}\right)^2=9\)

\(\Rightarrow x+\dfrac{1}{x}=3\) ( Vì x > 0 )

\(\Rightarrow\left(x+\dfrac{1}{x}\right)^3=27\)

\(\Rightarrow x^3+\dfrac{1}{x^3}+3\left(x+\dfrac{1}{x}\right)=27\)

\(\Rightarrow x^3+\dfrac{1}{x^3}+3.3=27\)

\(\Rightarrow x^3+\dfrac{1}{x^3}=18\)

Ta lại có:\(\left(x+\dfrac{1}{x}\right)\left(x^4+\dfrac{1}{x^4}\right)=x^5+x^3+\dfrac{1}{x^3}+\dfrac{1}{x^5}=x^5+\dfrac{1}{x^5}+18\)

Mặt khác:

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

\(=\left(x+\dfrac{1}{x}\right)\left(7^2-2\right)\)

\(=3.47=141\)

\(\Rightarrow x^5+\dfrac{1}{x^5}+18=141\)

\(\Rightarrow x^5+\dfrac{1}{x^5}=123\)