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a) Đk \(x\ne\pm1\), sau khi rút gọn ta được: (bạn tư làm)
\(P=\frac{x}{x+1}\)
b) Khi \(\left|x-\frac{2}{3}\right|=\frac{1}{3}\) thì hoặc \(x-\frac{2}{3}=\frac{1}{3}\) hoặc \(x-\frac{2}{3}=-\frac{1}{3}\)
Hay là \(x=1\) hoặc \(x=\frac{1}{3}\)
Do để P có nghĩa thì \(x\ne\pm1\) nên \(x=\frac{1}{3}\), khi đó:
\(P=\frac{\frac{1}{3}}{\frac{1}{3}+1}=\frac{1}{4}\)
c) P > 1 khi \(\frac{x}{x+1}>1\)
\(\Leftrightarrow1-\frac{1}{x+1}>1\)
\(\Leftrightarrow\frac{1}{x+1}< 0\)
\(\Leftrightarrow x< -1\)
e) Đề không rõ ràng
ĐKXĐ: \(x\ne\pm2;x\ne0\)
\(A=\left[\frac{4x\left(x-2\right)}{x^2-4}-\frac{8x^2}{x^2-4}\right]:\left[\frac{x-1}{x\left(x-2\right)}-\frac{2\left(x-2\right)}{x\left(x-2\right)}\right]\)
\(=\frac{-4x^2-8x}{x^2-4}:\frac{-x+3}{x\left(x-2\right)}\)
\(=\frac{-4x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}.\frac{x\left(x-2\right)}{-x+3}\)
\(=\frac{4x^2}{x-3}\)
Vì \(4x^2\ge0\)với mọi x nên:
để A > 0 thì x - 3 >0 <=> x > 3
a, ĐKXĐ của B: \(\hept{\begin{cases}2x+10\ne0\\x\ne0\\2x\left(x+5\right)\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}}\)
b, \(B=\frac{\left(x^2+2x\right)x+2\left(x-5\right)\left(x+5\right)+50-5x}{2x\left(x+5\right)}=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(=\frac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}=\frac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\frac{x-1}{2}\)
\(B=0\Rightarrow\frac{x-1}{2}=0\Rightarrow x-1=0\Rightarrow x=1\)(thỏa mãn điều kiện xác định)
\(B=\frac{1}{4}\Rightarrow\frac{x-1}{2}=\frac{1}{4}\Rightarrow x-1=\frac{1}{2}\Rightarrow x=\frac{3}{2}\)(thỏa mãn)
c, \(B>0\Rightarrow\frac{x-1}{2}>0\Rightarrow x-1>0\Rightarrow x>1\)
Vậy với x > 1 thì B > 0
\(B< 0\Rightarrow\frac{x-1}{2}< 0\Rightarrow x-1< 0\Rightarrow x< 1\)
Vậy với x < 1 và \(x\ne\left\{-5;0\right\}\) thì B < 0
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 \(x+2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| x + 2 | 1 | -1 | 2 | -2 |
| x | -1 | -3 | 0 | -4 |
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
a) \(-ĐKXĐ:x\ne\pm2;1\)
Rút gọn : \(A=\left(\frac{1}{x+2}-\frac{2}{x-2}-\frac{x}{4-x^2}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)
\(=\left(\frac{1}{x+2}+\frac{-2}{x-2}+\frac{x}{x^2-4}\right).\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\left[\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{\left(-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x}{\left(x-2\right)\left(x+2\right)}\right]\)\(.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\left[\frac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\right].\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)\(=\frac{x+1}{\left(x+2\right)^2}\)
b) \(A>0\Leftrightarrow\frac{x+1}{\left(x+2\right)^2}>0\Leftrightarrow\orbr{\begin{cases}x+1< 0;\left(x+2\right)^2< 0\left(voly\right)\\x+1>0;\left(x+2\right)^2>0\end{cases}}\)
\(\Leftrightarrow x>1;x>-2\Leftrightarrow x>1\)
Vậy với mọi x thỏa mãn x>1 thì A > 0
c) Ta có : \(x^2+3x+2=0\Leftrightarrow x^2+x+2x+2=0\)
\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
Vậy x = -1;-2
a, Đẻ \(P< 1\)thì :
\(P=\left(\frac{x}{x+2}+\frac{x}{x-2}-\frac{2}{x^2-4}\right).\frac{x-2}{2x+2}< 1\)
\(=\left(\frac{x\left(x-2\right)\left(x^2-4\right)}{\left(x+2\right)\left(x-2\right)\left(x^2-4\right)}+\frac{x\left(x+2\right)\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)\left(x^2-4\right)}-\frac{2\left(x+2\right)\left(x-2\right)}{\left(x^2+4\right)\left(x+2\right)\left(x-2\right)}\right).\frac{x-2}{2x+2}\)
\(=\left(\frac{x\left(x-2\right)\left(x^2-4\right)+x\left(x+2\right)\left(x^2-4\right)-2\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)\left(x^2-4\right)}\right).\frac{x-2}{2x+2}\)
\(=\left(\frac{2x^4-10x^2+8}{x^4-8x^2+16}\right).\frac{x-2}{2x+2}=\left(2x^4-10x^2+8\right)\left(2x+2\right)=\left(x-2\right)\left(x^4-8x^2+16\right)\)
PT tương đương vs : \(\left(2x^4-10x^2+8\right)\left(2x+2\right)-\left(x-2\right)\left(x^4-8x^2+16\right)< 1\)
Khi đó pt trở thành : \(3x^5+6x^4-12x^3-36x^2+48< 1\)
Chắc vại đó ==