Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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.
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:
a: =>28x-8=9x+3
=>19x=11
=>x=11/19
b: =>(3x-1)(x-1)=(2x+1)(x+1)
=>3x^2-4x+1=2x^2+3x+1
=>x^2-7x=0
=>x=0 hoặc x=7
\(\left(\dfrac{x}{x+1}+\dfrac{x-1}{x}\right):\left(\dfrac{x}{x+1}-\dfrac{x-1}{x}\right)\) \(\left(đk:x\ne0;-1\right)\)
\(=\dfrac{x^2+\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}:\left(\dfrac{x^2-\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}\right)\)
\(=\dfrac{x^2+x^2-1}{x\left(x+1\right)}.\dfrac{x\left(x+1\right)}{x^2-x^2+1}\)
\(=\dfrac{\left(2x^2-1\right)x\left(x+1\right)}{x\left(x+1\right)}=2x^2-1\)
\(a,=\left(a+5+\dfrac{1}{2}-a\right)^2=\left(\dfrac{11}{2}\right)^2=\dfrac{121}{4}\\ b,=\dfrac{\left(x+y\right)^2-16}{3x\left(x-4+y\right)}=\dfrac{\left(x+y-4\right)\left(x+y+4\right)}{3x\left(x+y-4\right)}=\dfrac{x+y+4}{3x}\)
a, \(\left(a+5\right)^2+2\left(a+5\right)\left(\dfrac{1}{2}-a\right)+\left(\dfrac{1}{2}-a\right)^2=\left(a+5+\dfrac{1}{2}-a\right)^2=\left(\dfrac{11}{2}\right)^2=\dfrac{121}{4}\)
b,\(\dfrac{x^2-16+2xy+y^2}{3x^2-12x+3xy}=\dfrac{\left(x^2+2xy+y^2\right)-4^2}{3x\left(x-4+y\right)}=\dfrac{\left(x+y-4\right)\left(x+y+4\right)}{3x\left(x+y-4\right)}=\dfrac{x+y+4}{3x}\)
Bạn thử nhân liên hợp cả tử và mẫu theo (A+B)×(A-B)=A^2 -B^2 xem s
\(a,\dfrac{y-1}{y+1}=\dfrac{\left(y-1\right)^2}{\left(y-1\right)\left(y+1\right)};\dfrac{y+1}{y-1}=\dfrac{\left(y+1\right)^2}{\left(y-1\right)\left(y+1\right)};\dfrac{1}{y^2-1}=\dfrac{1}{\left(y-1\right)\left(y+1\right)}\\ b,\dfrac{2}{y^2-4y}=\dfrac{2\left(y+4\right)}{y\left(y-4\right)\left(y+4\right)};\dfrac{y}{y^2-16}=\dfrac{y^2}{y\left(y-4\right)\left(y+4\right)}\)