Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, ĐKXĐ: x≠±2
A=\(\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right)\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
A=\(\left(\dfrac{x}{x^2-4}-\dfrac{2x+4}{x^2-4}+\dfrac{x-2}{x^2-4}\right)\left(\dfrac{x^2+2x}{x+2}-\dfrac{2x+4}{x+2}+\dfrac{10-x^2}{x+2}\right)\)
A=\(\left(\dfrac{-6}{x^2-4}\right)\left(\dfrac{6}{x+2}\right)\)
A=\(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\)
b, |x|=\(\dfrac{1}{2}\)
TH1z: x≥0 ⇔ x=\(\dfrac{1}{2}\) (TMĐKXĐ)
TH2: x<0 ⇔ x=\(\dfrac{-1}{2}\) (TMĐXĐ)
Thay \(\dfrac{1}{2}\), \(\dfrac{-1}{2}\) vào A ta có:
\(\dfrac{-36}{\left(\dfrac{1}{2}-2\right)\left(\dfrac{1}{2}+2\right)^2}\)=\(\dfrac{96}{25}\)
\(\dfrac{-36}{\left(\dfrac{-1}{2}-2\right)\left(\dfrac{-1}{2}+2\right)^2}\)=\(\dfrac{32}{5}\)
c, A<0 ⇔ \(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\) ⇔ (x-2)(x+2)2 < 0
⇔ {x-2>0 ⇔ {x>2
[ [
{x+2<0 {x<2
⇔ {x-2<0 ⇔ {x<2
[ [
{x+2>0 {x>2
⇔ x<2
Vậy x<2 (trừ -2)
a: \(A=\dfrac{x^2+x+x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{x^2+2x}{\left(x-1\right)}\cdot\dfrac{x+1}{2x+1}\)
\(=\dfrac{\left(x^2+2\right)\left(x+1\right)}{\left(2x+1\right)\left(x-1\right)}\)
b: Khi x=2 thì \(A=\dfrac{\left(4+2\right)\left(2+1\right)}{\left(2\cdot2+1\right)\left(2-1\right)}=\dfrac{18}{5}\)
a: \(A=\left(1+x+x^2-x\right):\dfrac{1-x^2}{x^3-x^2-x+1}\)
\(=\left(x^2+1\right)\cdot\dfrac{\left(x-1\right)\left(x^2-1\right)}{-\left(x^2-1\right)}=\left(1-x\right)\left(x^2+1\right)\)
b: Khi x=-5/3 thì \(A=\left(1+\dfrac{5}{3}\right)\left(\dfrac{25}{9}+1\right)=\dfrac{8}{3}\cdot\dfrac{34}{9}=\dfrac{272}{27}\)
c: Để A<0 thì 1-x<0
hay x>1
ĐK: \(y\ne1;y\ne3\).
Ta có \(\dfrac{y+5}{y-1}-\dfrac{y+1}{y-3}=\dfrac{-8}{\left(y-1\right)\left(y-3\right)}\)
\(\Leftrightarrow\dfrac{\left(y+5\right)\left(y-3\right)-\left(y+1\right)\left(y-1\right)}{\left(y-1\right)\left(y-3\right)}=\dfrac{-8}{\left(y-1\right)\left(y-3\right)}\)
\(\Rightarrow\left(y+5\right)\left(y-3\right)-\left(y+1\right)\left(y-1\right)=-8\Leftrightarrow\left(y^2+2y-15\right)-\left(y^2-1\right)=-8\Leftrightarrow2y-14=-8\Leftrightarrow y=3\). (loại)
Vậy không tồn tại y thỏa mãn
1)trước khi rút gọn bạn cần tìm điều kiện để có phân thức này như
+)Điều kiện: \(\left\{{}\begin{matrix}x-1\ne0\\x^2-1\ne\\x+1\ne0\end{matrix}\right.0}\)
\(\Rightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
rồi bạn rút gọn
2) với \(x=1\dfrac{1}{3}=\dfrac{4}{3}\) khi đó bạn thay x vào biểu thức A thì tìm đc giá trị
3) bạn tự làm đc :))
(\(\dfrac{x+1}{x-1}\)-- \(\dfrac{x^2+2x+9}{x^2-1}\)).\(\dfrac{x+1}{5}\)=(\(\dfrac{\left(x+1\right)^2}{x^2-1}\)--\(\dfrac{x^2+2x+9}{x^2-1}\)):\(\dfrac{x+1}{5}\)
=\(\dfrac{-8}{x^2-1}\):\(\dfrac{x+1}{5}\)=\(\dfrac{-8}{5\left(x-1\right)}\)
Cố gắng lên bạn nhé!
a, ĐKXĐ: \(x\ne1;x\ne-1\)
b, Với \(x\ne1;x\ne-1\)
\(B=\left[\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right]\cdot\dfrac{4\left(x^2-1\right)}{5}\\ =\left[\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\right]\cdot\dfrac{4\left(x^2-1\right)}{5}\\ =\dfrac{5}{x^2-1}\cdot\dfrac{4\left(x^2-1\right)}{5}\\ =4\)
=> ĐPCM
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}\cdot\left(\dfrac{x+2-2x}{1-x}\right)\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(x-2\right)}{x-1}\)
\(=\dfrac{-6}{\left(x+2\right)\left(x-1\right)}\)
b: Thay x=-4 vào A, ta được:
\(A=-\dfrac{6}{\left(-4+2\right)\left(-4-1\right)}=\dfrac{-6}{-2\cdot\left(-5\right)}=\dfrac{-6}{10}=\dfrac{-3}{5}\)
a: \(N=\left(\dfrac{1}{y-1}+\dfrac{1}{\left(y-1\right)\left(y^2+y+1\right)}\cdot\dfrac{y^2+y+1}{y+1}\right)\cdot\left(y^2-1\right)\)
\(=\dfrac{y+1+1}{\left(y-1\right)\left(y+1\right)}\cdot\left(y^2-1\right)=y+2\)
b: Thay y=1/2 vào N, ta được:
N=1/2+2=5/2
c: Để N>0 thì y+2>0
hay y>-2
Kết hợp ĐKXĐ, ta được:
\(\left\{{}\begin{matrix}y>-2\\y\notin\left\{-1;1\right\}\end{matrix}\right.\)
Lời giải:
a. ĐKXĐ: $y\neq \pm 1$
\(N=\left(\frac{1}{y-1}-\frac{1}{(1-y)(1+y+y^2)}.\frac{y^2+y+1}{y+1}\right).(y^2-1)\)
\(=(\frac{1}{y-1}-\frac{1}{(1-y)(y+1)})(y-1)(y+1)\)
\(=\frac{1}{y-1}(y-1)(y+1)-\frac{1}{-(y-1)(y+1)}.(y-1)(y+1)=y+1-(-1)=y+2\)
b. Khi $y=\frac{1}{2}$ thì:
$N=\frac{1}{2}+2=\frac{5}{2}$
c. Để $N>0\Leftrightarrow y+2>0\Leftrightarrow y>-2$
Kết hợp đkxđ suy ra $y>-2$ và $y\neq \pm 1$ thì $N$ dương.