Cho biểu thức \(N=\left(\frac{1}{y-1}-\frac{y}{1-y^3}.\frac{y^2+y+1}{y+1}\right):\frac{1}{y^2-1}\)
a, Rút gọn biểu thức N
b, Tìm giá trị của N khi y =1/2
c. Tìm giá trị của y để N luôn dương
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a) ĐKXĐ : \(y\ne\pm1\)
\(N=\left(\frac{1}{y-1}-\frac{y}{1-y^3}.\frac{y^2+y+1}{y+1}\right)\div\frac{1}{y^2-1}\)
\(=\left(\frac{1}{y-1}+\frac{y}{\left(y-1\right)\left(y^2+y+1\right)}.\frac{y^2+y+1}{y+1}\right)\div\frac{1}{y^2-1}\)
\(=\left(\frac{1}{y-1}+\frac{y}{\left(y-1\right)\left(y+1\right)}\right)\div\frac{1}{y^2-1}\)
\(=\frac{y+1+y}{\left(y-1\right)\left(y+1\right)}\div\frac{1}{\left(y-1\right)\left(y+1\right)}\)
\(=\frac{2y+1}{\left(y-1\right)\left(y+1\right)}.\left(y-1\right)\left(y+1\right)\)
\(=2y+1\)
Vậy \(N=2y+1\)khi \(y\ne\pm1\)
b) Với \(y=\frac{1}{2}\); phương trình N trở thành :
\(N=2.\frac{1}{2}+1=2\)
Vậy N=2 khi \(y=\frac{1}{2}\)
c) Để N luôn dương
\(\Leftrightarrow2y+1>0\)
\(\Leftrightarrow2y>-1\)
\(\Leftrightarrow y>\frac{-1}{2}\)
Kết hợp ĐKXĐ ta có : \(y>\frac{-1}{2};y\ne\pm1\)
Vậy N luôn dương khi \(y>\frac{-1}{2};y\ne\pm1\)
a., đk y khác cộng trừ 1
N=\(\left(\frac{1}{y-1}+\frac{y}{\left(y^3-1\right)}.\frac{y^2+y+1}{y+1}\right):\frac{1}{\left(y-1\right)\left(y+1\right)}\)
N=\(\left(\frac{1}{y-1}+\frac{y}{\left(y-1\right)\left(y+1\right)}\right).\left(y-1\right)\left(y+1\right)\)
N=\(\frac{y+1+y}{\left(y-1\right)\left(y+1\right)}.\left(y-1\right)\left(y+1\right)\)
N= \(2y+1\)
Vậy N=2y+1 với y khác cộng trừ 1
b, Thay y= \(\frac{1}{2}\) ( t/m đk y khác cộng trừ 1 )vào biểu thức N ta được:
N= \(2.\frac{1}{2}+1=1+1=2\)
Vậy N=2 với y = 1/2
c, Để N luôn dương thì: 2y+1>0
<=> 2y>-1
<=>y>\(\frac{-1}{2}\)( t/ m đk y khác cộng trừ 1)
Vậy với y>-1/2 thì N luôn dương
a, \(N=\left(\frac{1}{y-1}-\frac{y}{1-y^3}.\frac{y^2+y+1}{y+1}\right):\frac{1}{y^2-1}\)
\(N=\left(\frac{1}{y-1}+\frac{y}{y^3-1}.\frac{y^2+y+1}{y+1}\right):\frac{1}{y^2-1}\)
\(N=\left(\frac{1}{y-1}+\frac{y}{\left(y-1\right)\left(y^2+y+1\right)}.\frac{y^2+y+1}{y+1}\right):\frac{1}{y^2-1}\)
\(N=\left(\frac{1}{y-1}+\frac{y}{\left(y-1\right)\left(y+1\right)}\right):\frac{1}{y^2-1}\)
\(N=\left(\frac{y+1}{\left(y-1\right)\left(y+1\right)}+\frac{y}{\left(y-1\right)\left(y+1\right)}\right):\frac{1}{\left(y-1\right)\left(y+1\right)}\)
\(N=\frac{y+1+y}{\left(y-1\right)\left(y+1\right)}:\frac{1}{\left(y-1\right)\left(y+1\right)}\)
\(N=\frac{2y+1}{\left(y-1\right)\left(y+1\right)}.\left(y-1\right)\left(y+1\right)\)
\(N=2y+1\)
b, Tại \(y=\frac{1}{2}\) ta có :
\(N=2.\frac{1}{2}+1\)
\(\Rightarrow N=1+1=2\)
c, Để N luôn có giá trị dương thì \(y\in N\).
ĐKXĐ: \(\hept{\begin{cases}y>0\\y\ne1\end{cases}}\)
a/ Ta có: \(A=\left[\frac{\sqrt{y}^3-1}{\sqrt{y}\left(\sqrt{y}-1\right)}-\frac{\sqrt{y}^3+1}{\sqrt{y}\left(\sqrt{y}+1\right)}\right]:\frac{2\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}\)
\(=\left[\frac{\left(\sqrt{y}-1\right)\left(y+\sqrt{y}+1\right)}{\sqrt{y}\left(\sqrt{y}-1\right)}-\frac{\left(\sqrt{y}+1\right)\left(y-\sqrt{y}+1\right)}{\sqrt{y}\left(\sqrt{y}+1\right)}\right].\frac{\sqrt{y}+1}{2\left(\sqrt{y}-1\right)}\)
\(=\left(\frac{y+\sqrt{y}+1-y+\sqrt{y}-1}{\sqrt{y}}\right).\frac{\sqrt{y}+1}{2\left(\sqrt{y}-1\right)}\)
\(=\frac{2\sqrt{y}}{\sqrt{y}}.\frac{\sqrt{y}+1}{2\left(\sqrt{y}-1\right)}=\frac{\sqrt{y}+1}{\sqrt{y}-1}\)
b/ \(A=\frac{\sqrt{y}+1}{\sqrt{y}-1}=1+\frac{2}{\sqrt{y}-1}\)
Để \(A\in Z\Rightarrow\left(\sqrt{y}-1\right)\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Với \(\sqrt{y}-1=1\Rightarrow\sqrt{y}=2\Rightarrow y=4\)
Với \(\sqrt{y}-1=-1\Rightarrow\sqrt{y}=0\Rightarrow y=0\)(loại)
Với \(\sqrt{y}-1=2\Rightarrow\sqrt{y}=3\Rightarrow y=9\)
Với \(\sqrt{y}-1=-2\Rightarrow\sqrt{y}=-1\) (loại)
Vậy y = 4 , y = 9
1,
\(A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\left(đk:a\ne0;1;2;a\ge0\right)\)
\(=\frac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{a^2-a}.\frac{a-2}{a+2}\)
\(=\frac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{a\left(a-1\right)}.\frac{a-2}{a+2}\)
\(=\frac{2a\left(a-1\right)\left(a-2\right)}{a\left(a-1\right)\left(a+2\right)}=\frac{2\left(a-2\right)}{a+2}\)
Để \(A=1\)\(=>\frac{2a-4}{a+2}=1< =>2a-4-a-2=0< =>a=6\)
2,
a, Điều kiện xác định của phương trình là \(x\ne4;x\ge0\)
b, Ta có : \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)
\(=\frac{2\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}-\frac{\sqrt{x}-2}{x-4}\)
\(=\frac{2\sqrt{x}+2+2}{x-4}=\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{2}{\sqrt{x}-2}\)
c, Với \(x=3+2\sqrt{3}\)thì \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
a, \(N=\left(\frac{1}{y-1}-\frac{y}{1-y^3}.\frac{y^2+y+1}{y+1}\right):\frac{1}{y^2-1}\)
\(=\left(\frac{1}{y-1}-\frac{y}{\left(1-y\right)\left(1+y+y^2\right)}.\frac{y^2+y+1}{y+1}\right):\frac{1}{\left(y-1\right)\left(y+1\right)}\)
\(=\left(\frac{1}{y-1}+\frac{y\left(y^2+y+1\right)}{\left(y+1\right)^2\left(y^2+y+1\right)}\right):\frac{1}{\left(y-1\right)\left(y+1\right)}\)
\(=\left(\frac{1}{y-1}+\frac{y}{\left(y+1\right)^2}\right):\frac{1}{\left(y-1\right)\left(x+1\right)}\)
\(=\left(\frac{\left(y+1\right)^2+y\left(y-1\right)}{\left(y-1\right)\left(y+1\right)^2}\right).\frac{\left(y-1\right)\left(y+1\right)}{1}=\frac{y^2+2y+1+y^2-y}{y+1}=\frac{2y^2+y+1}{y+1}\)
b, Thay y = 1/2 ta có :
\(\frac{2.\left(\frac{1}{2}\right)^2+\frac{1}{2}+1}{\frac{1}{2}+1}=\frac{\frac{1}{2}+\frac{1}{2}+\frac{2}{2}}{\frac{1}{2}+\frac{2}{2}}=\frac{\frac{5}{2}}{\frac{3}{2}}=\frac{5}{12}\)