Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, điều kiện xác định: x2 - 4 ≠ 0
⇔ x2 ≠ 4
⇔x ≠ 2 và x ≠ -2
b, A= \(\dfrac{x^2}{x^2-4}-\dfrac{x}{x-2}+\dfrac{2}{x+2}\)
=\(\dfrac{x^2-x\left(x+2\right)+2\left(x-2\right)}{x^2-4}\)
= \(\dfrac{x^2-x^2-2x+2x-4}{x^2-4}\)
= \(\dfrac{x^2-4}{x^2-4}\)
= 1
c, x=1 ⇒ A= \(\dfrac{1^2}{1^2-4}-\dfrac{1}{1-2}+\dfrac{2}{1+2}\)
= \(\dfrac{4}{3}\)
a) Điều kiện xác định:
A\(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\end{matrix}\right.⇔\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
b) Rút gọn:
A= \(\dfrac{x^2}{x^2-4}-\dfrac{x}{x-2}+\dfrac{2}{x+2}\).
A= \(\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x}{x-2}+\dfrac{2}{x+2}\).
A= \(\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)[do MTC là (x-2)(x+2)].
A= \(\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2+2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}\)
A= \(\dfrac{x^2-\left(x^2+2x\right)+2x-4}{\left(x-2\right)\left(x+2\right)}\)
A= \(\dfrac{x^2-x^2-2x+2x-4}{\left(x-2\right)\left(x+2\right)}\)
A= \(\dfrac{-4}{\left(x-2\right)\left(x+2\right)}\)
a, ĐKXĐ: x\(\ne\) 1;-1;2
b, A= \(\left(\frac{x}{x+1}+\frac{1}{x-1}-\frac{4x}{2-2x^2}\right):\frac{x+1}{x-2}\)
=\(\left(\frac{2x^2-2x}{2\left(x+1\right)\left(x-1\right)}+\frac{2x+2}{2\left(x+1\right)\left(x-1\right)}+\frac{4x}{2\left(x-1\right)\left(x+1\right)}\right)\times\frac{x-2}{x+1}\)
=\(\frac{2x^2-2x+2x+2+4x}{2\left(x+1\right)\left(x-1\right)}\times\frac{x-2}{x+1}\)
=\(\frac{2x^2+4x+2}{2\left(x+1\right)\left(x-1\right)}\times\frac{x-2}{x+1}\)
=\(\frac{2\left(x+1\right)^2}{2\left(x+1\right)\left(x-1\right)}\times\frac{x-2}{x+1}\)
=\(\frac{x-2}{x-1}\)
c, Khi x= -1
→A= \(\frac{-1-2}{-1-1}\)
= -3
Vậy khi x= -1 thì A= -3
Câu d thì mình đang suy nghĩ nhé, mình sẽ quay lại trả lời sau ^^
a,ĐKXĐ:x#1; x#-1; x#2
b,Ta có:
A=\(\left(\frac{x}{x+1}+\frac{1}{x-1}-\frac{4x}{2-2x^2}\right):\frac{x+1}{x-2}\)
=\(\left(\frac{x\left(x-1\right)2}{\left(x+1\right)\left(x-1\right)2}+\frac{\left(x+1\right)2}{\left(x-1\right)\left(x+1\right)2}+\frac{4x}{2\left(x-1\right)\left(x+1\right)}\right):\frac{x+1}{x-2}\)
=\(\frac{2x^2-2x+2x+2+4x}{\left(x+1\right)\left(x-1\right)2}.\frac{x-2}{x+1}\)
=\(\frac{2x^2+4x+2}{\left(x+1\right)\left(x-1\right)2}.\frac{x-2}{x+1}\)
=\(\frac{2\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)2}.\frac{x-2}{x+1}\)
=\(\frac{x-2}{x+1}\)
c,Tại x=-1 ,theo ĐKXĐ x#-1 \(\Rightarrow\)A không có kết quả
d,Để A có giá trị nguyên \(\Rightarrow\frac{x-2}{x+1}\)có giá trị nguyên
\(\Leftrightarrow x-2⋮x+1\)
\(\Leftrightarrow x+1-3⋮x+1\)
Mà \(x+1⋮x+1\Rightarrow3⋮x+1\)
\(\Rightarrow x+1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow x\in\left\{0;-2;2;-4\right\}\)
Mà theo ĐKXĐ x#2\(\Rightarrow x\in\left\{0;-2;-4\right\}\)
Vậy \(x\in\left\{0;-2;-4\right\}\)thì a là số nguyên
\(a,x\ne2;x\ne-2;x\ne0\)
\(b,A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\frac{6}{x+2}\)
\(=\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
\(=\frac{1}{2-x}\)
\(c,\)Để A > 0 thi \(\frac{1}{2-x}>0\Leftrightarrow2-x>0\Leftrightarrow x< 2\)
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.
a) ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne-2\end{cases}}\)
\(N=\frac{\left(x+2\right)^2}{x}.\left(1-\frac{x^2}{x+2}\right)-\frac{x^2+6x+4}{x}\)
\(N=\frac{\left(x+2\right)^2}{x}.\frac{x+2-x^2}{x+2}-\frac{x^2+6x+4}{x}\)
\(N=\frac{\left(x+2\right)\left(x+2-x^2\right)-x^2-6x-4}{x}\)
\(N=\frac{x^2+2x-x^3+2x+4-2x^2-x^2-6x-4}{x}\)
\(N=\frac{-x^3-2x^2-2x}{x}\)
\(N=\frac{-x\left(x^2+2x+2\right)}{x}\)
\(N=-\left(x^2+2x+2\right)\)
b) \(N=-\left(x^2+2x+2\right)\)
\(\Leftrightarrow N=-\left(x^2+2x+1+1\right)\)
\(\Leftrightarrow N=-\left(x+1\right)^2-1\le-1\)
Max N = -1 \(\Leftrightarrow x=-1\)
Vậy .......................
câu a, phân tích từng mẫu thành nhân tử (nếu cần)
rồi tìm mtc, ở đây, nhân chia cũng như cộng trừ, nên phân tích hết rồi ra mtc, đkxđ là cái mtc ấy khác 0
câu b với c tự làm
câu d thì lấy cái rút gọn rồi của câu b, rồi giải ra, để nguyên thì mẫu là ước của tử, thế thôi
a, \(ĐKXĐ\hept{\begin{cases}2-x\ne0\\2+x\ne0\end{cases}\Leftrightarrow x\ne\pm2}\)
b, Ta có: \(A=\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\)
\(=\frac{\left(2+x\right)^2}{\left(2-x\right)\left(2+x\right)}+\frac{4x^2}{\left(2-x\right)\left(2+x\right)}-\frac{\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\)
\(=\frac{4+4x+x^2+4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}\)
\(=\frac{4x^2+8x}{\left(2-x\right)\left(2+x\right)}\)
\(=\frac{4x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4x}{x-2}\)
a) ĐKXĐ: \(\hept{\begin{cases}2-x\ne0\\x^2-4\ne0\\2+x\ne0\end{cases}}\)<=>\(\hept{\begin{cases}2-x\ne0\\2+x\ne0\\\left(x-2\right)\left(x+2\right)\ne0\end{cases}}\)<=>\(x\ne\pm2\)
b)\(A=\frac{2+x}{2-x}-\frac{4x}{x^2-4}-\frac{2-x}{2+x}\)
\(\Leftrightarrow A=\frac{2+x}{2-x}+\frac{4x}{4-x^2}-\frac{2-x}{2+x}\)
\(\Leftrightarrow A=\frac{\left(2+x\right)\left(2+x\right)}{\left(2-x\right)\left(2+x\right)}+\frac{4x}{\left(2-x\right)\left(2+x\right)}-\frac{\left(2-x\right)\left(2-x\right)}{\left(2+x\right)\left(2-x\right)}\)
\(\Leftrightarrow A=\frac{x^2+4x+4+4x-x^2+4x-4}{\left(2+x\right)\left(2-x\right)}\)
\(\Leftrightarrow A=\frac{12x}{\left(2+x\right)\left(2-x\right)}\)