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ĐK: \(x\ne1;x\ne-1\)
\(Q=\left(\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)^2}-\dfrac{1}{\left(x+1\right)}+\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}\right)\left(x-1\right)\left(x+1\right)\)
\(Q=\left(\dfrac{x-1}{x+1}-\dfrac{1}{x+1}+\dfrac{x+1}{x-1}\right)\left(x-1\right)\left(x+1\right)\)
\(Q=\left(x-1\right)^2-\left(x-1\right)+\left(x+1\right)^2\)
\(Q=x^2-2x+1-x+1+x^2+2x+1=2x^2-x+3\)
c/ \(Q=2\left(x^2-\dfrac{1}{2}x\right)+3=2\left(x^2-2.\dfrac{1}{4}x+\dfrac{1}{16}\right)-\dfrac{1}{8}+3\)
\(Q=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{23}{8}\ge\dfrac{23}{8}\)
\(\Rightarrow Q_{min}=\dfrac{23}{8}\) khi \(x=\dfrac{1}{4}\)
a: ĐKXĐ: \(x\notin\left\{1;-1;0\right\}\)
b: \(A=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{5\left(x-1\right)}{2x}=\dfrac{20\left(x-1\right)}{2x}=\dfrac{10\left(x-1\right)}{x}\)
c: Khi x=3,5 thì \(A=\dfrac{10\cdot2.5}{3.5}=\dfrac{25}{3.5}=\dfrac{50}{7}\)
d: Để A=4 thì 10x-10=4x
=>6x=10
=>x=5/3
Câu 1 :
a) ĐKXĐ : \(\hept{\begin{cases}x+1\ne0\\2x-6\ne0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)
b) Để \(P=1\Leftrightarrow\frac{4x^2+4x}{\left(x+1\right)\left(2x-6\right)}=1\)
\(\Leftrightarrow\frac{4x^2+4x-\left(x+1\right)\left(2x-6\right)}{\left(x+1\right)\left(2x-6\right)}=0\)
\(\Rightarrow4x^2+4x-2x^2+4x+6=0\)
\(\Leftrightarrow2x^2+8x+6=0\)
\(\Leftrightarrow x^2+4x+4-1=0\)
\(\Leftrightarrow\left(x+2-1\right)\left(x+2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=-3\left(TMĐKXĐ\right)\end{cases}}\)
Vậy : \(x=-3\) thì P = 1.
\(e ) Để \) \(M\)\(\in\)\(Z \) \(thì\) \(1 \)\(⋮\)\(x +3\)
\(\Leftrightarrow\)\(x + 3 \)\(\in\)\(Ư\)\((1)\)\(= \) { \(\pm\)\(1 \) }
\(Lập\) \(bảng :\)
| \(x +3\) | \(1\) | \(- 1\) |
| \(x\) | \(-2\) | \(- 4\) |
\(Vậy : Để \) \(M\)\(\in\)\(Z\) \(thì\) \(x\)\(\in\){ \(- 4 ; - 2\) }
e) Để M \(\in\)Z <=> \(\frac{1}{x+3}\in Z\)
<=> 1 \(⋮\)x + 3 <=> x + 3 \(\in\)Ư(1) = {1; -1}
Lập bảng:
| x + 3 | 1 | -1 |
| x | -2 | -4 |
Vậy ....
f) Ta có: M > 0
=> \(\frac{1}{x+3}\) > 0
Do 1 > 0 => x + 3 > 0
=> x > -3
Vậy để M > 0 khi x > -3 ; x \(\ne\)3 và x \(\ne\)-3/2
a, gt của B xđ là x\(\ne\)2,x\(\ne\)-2
b, kq \(\frac{-8}{x+2}\)
a: \(B=\left(\dfrac{x}{x\left(x-2\right)\left(x+2\right)}-\dfrac{10}{5\left(x+2\right)}+\dfrac{1}{x-2}\right):\dfrac{x^2-4+6-x^2}{x-2}\)
\(=\left(\dfrac{1}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x+2}+\dfrac{1}{x-2}\right):\dfrac{2}{x-2}\)
\(=\dfrac{1-2x+4+x+2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x-2}{2}=\dfrac{-x+7}{2\left(x+2\right)}\)
b: Ta có: |x|=1/2
=>x=1/2 hoặc x=-1/2
Thay x=1/2 vào B, ta được:
\(B=\dfrac{-\dfrac{1}{2}+7}{2\left(\dfrac{1}{2}+2\right)}=\dfrac{13}{10}\)
Thay x=-1/2 vào B, ta được:
\(B=\dfrac{\dfrac{1}{2}+7}{2\left(-\dfrac{1}{2}+2\right)}=\dfrac{5}{2}\)
a) \(ĐKXĐ:x\ne\pm3\)
b) \(A=\left(\frac{x}{x+3}+\frac{3-x}{x+3}\cdot\frac{x^2+3x+9}{x^2-9}\right):\frac{3}{x+3}\)
\(\Leftrightarrow A=\left(\frac{x}{x+3}-\frac{\left(x-3\right)\left(x^2+3x+9\right)}{\left(x+3\right)\left(x^2-9\right)}\right):\frac{3}{x+3}\)
\(\Leftrightarrow A=\left(\frac{x}{x+3}-\frac{x^2+3x+9}{\left(x+3\right)^2}\right):\frac{3}{x+3}\)
\(\Leftrightarrow A=\frac{x^2+3x-x^2-3x-9}{\left(x+3\right)^2}:\frac{3}{x+3}\)
\(\Leftrightarrow A=\frac{-9\left(x+3\right)}{3\left(x+3\right)^2}\)
\(\Leftrightarrow A=\frac{-3}{x+3}\)
c) Tại \(x=-\frac{1}{2}\)
\(\Leftrightarrow A=\frac{-3}{-\frac{1}{2}+3}\)
\(\Leftrightarrow A=\frac{-6}{5}\)
d) Để \(A>0\)
\(\Leftrightarrow\frac{-3}{x+3}>0\)
\(\Leftrightarrow x+3< 0\)(Vì -3 < 0)
\(\Leftrightarrow x< -3\)
e) +) Với \(A>\frac{-1}{2}\)
\(\Leftrightarrow\frac{-3}{x+3}>-\frac{1}{2}\)
\(\Leftrightarrow-6>-x-3\)
\(\Leftrightarrow x>3\)(tm)
+) Với \(A< -\frac{1}{2}\)
\(\Leftrightarrow\frac{-3}{x+3}< -\frac{1}{2}\)
\(\Leftrightarrow-6< -x-3\)
\(\Leftrightarrow x< 3\)(chú ý : \(x\ne-3\))
+) Với \(A=-\frac{1}{2}\)
\(\Leftrightarrow-\frac{3}{x+3}=-\frac{1}{2}\)
\(\Leftrightarrow x+3=6\)
\(\Leftrightarrow x=3\)(ktm)
Vậy \(\orbr{\begin{cases}A>-\frac{1}{2}\\A< -\frac{1}{2}\end{cases}}\)
a) ĐKXĐ: \(x\ne\mp1\)
\(Q=\dfrac{x^2}{x^4-1}\left(x^2-1\right)-\dfrac{1}{x^2+1}\)
\(Q=\dfrac{x^2}{\left(x^2-1\right)\left(x^2+1\right)}\left(x^2-1\right)-\dfrac{1}{x^2+1}\)
\(Q=\dfrac{x^2}{x^2+1}-\dfrac{1}{x^2+1}=\dfrac{x^2-1}{x^2+1}\)
b) \(Q=0\Rightarrow\dfrac{x^2-1}{x^2+1}=0\\ \Leftrightarrow x^2=1\\ \Leftrightarrow x=\mp1\left(loại\right)\)
Không tồn tại x để Q=0
d) \(Q=\dfrac{x^2-1}{x^2+1}=\dfrac{x^2+1-2}{x^2+1}=1-\dfrac{2}{x^2+1}\)
Ta có: \(x^2\ge0\Leftrightarrow x^2+1\ge1\\ \Leftrightarrow-\dfrac{2}{x^2+1}\ge-\dfrac{2}{1}=-2\\ 1-\dfrac{2}{x^2+1}\ge-1\\ Q\ge-1\)
Vậy GTNN của Q=-1 <=> x=0