tìm x để B=\(\frac{x+3}{x-1}\)<0
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\(B=\frac{x-1-4\sqrt{x}+\sqrt{x}+1}{x-1}.\frac{x-1}{x-2\sqrt{x}}\)
\(=\frac{x-3\sqrt{x}}{x-2\sqrt{x}}\)
\(=\frac{\sqrt{x}-3}{\sqrt{x}-2}\)
a.Ta co:
\(\frac{\sqrt{x}-3}{\sqrt{x}-2}< 1\left(x\ge0,x\ne4\right)\)
\(\Leftrightarrow\sqrt{x}-3< \sqrt{x}-2\)
\(\Leftrightarrow3>2\)
Vay \(B< 1\left(\forall x\ge0,x\ne4\right)\)
Lát mình giải 2 câu kia,di ăn com cái
b.Ta co:
\(\frac{\sqrt{x}-3}{\sqrt{x}-2}< \frac{3}{2}\)
\(\Leftrightarrow2\sqrt{x}-6< 3\sqrt{x}-6\)
\(\Leftrightarrow x>0\)
Vay \(B< \frac{3}{2}\left(\forall x>0,x\ne4\right)\)
c.Ta co:
\(\frac{\sqrt{x}-3}{\sqrt{x}-2}>\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-3>x-3\sqrt{x}+2\)
\(\Leftrightarrow x-4\sqrt{x}+5< 0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+1< 0\) (vo ly)
Vay khong co gia tri nao cua x thoa man \(B>\sqrt{x}-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐKXĐ:\hept{\begin{cases}x\ne\pm3\\1-\frac{1}{x+3}\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne\pm3\\x\ne-2\end{cases}}}\)
a ) \(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)
\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{x-4}{x-3}-\frac{x-1}{x+3}\right):\left(1-\frac{1}{x+3}\right)\)
\(=\frac{21+\left(x-4\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}:\frac{x+3-1}{x+3}\)
\(=\frac{21+x^2-x-12-\left(x^2-4x+3\right)}{\left(x-3\right)\left(x+3\right)}:\frac{x+2}{x+3}\)
\(=\frac{3x+6}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}\)
\(=\frac{3.\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}\)
\(=\frac{3}{x-3}\)
b ) \(B=-\frac{3}{5}\Leftrightarrow\frac{3}{x-3}=-\frac{3}{5}\)
\(\Leftrightarrow x-3=-5\Leftrightarrow x=-2\) ( do \(x\ne\pm3;x\ne-2\) )
c ) \(B< 0\Leftrightarrow\frac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow\) \(\hept{\begin{cases}x< 3\\x\ne-2\\x\ne-3\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(B=\left(\frac{9-3x}{x^2+4x-5}-\frac{x+5}{1-x}-\frac{x+1}{x+5}\right):\frac{7x-14}{x^2-1}\)
\(=\left(\frac{9-3x}{\left(x-1\right)\left(x+5\right)}+\frac{\left(x+5\right)^2}{\left(x-1\right)\left(x+5\right)}-\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+5\right)}\right):\frac{7\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{9-3x+x^2+10x+25-x^2+1}{\left(x-1\right)\left(x+5\right)}.\frac{\left(x-1\right)\left(x+1\right)}{7\left(x-2\right)}\)
\(=\frac{35+7x}{x+5}\frac{x+1}{7\left(x-2\right)}=\frac{7\left(x+5\right)\left(x+1\right)}{7\left(x+5\right)\left(x-2\right)}=\frac{x+1}{x-2}\)
b, Ta có : \(\left(x+5\right)^2-9x-45=0\)
\(\Leftrightarrow x^2+10x+25-9x-45=0\Leftrightarrow x^2+x-20=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
TH1 : Thay x = 4 vào biểu thức ta được : \(\frac{4+1}{4-2}=\frac{5}{2}\)
TH2 : THay x = 5 vào biểu thức ta được : \(\frac{5+1}{5-2}=\frac{6}{3}=2\)
c, Để B nhận giá trị nguyên khi \(\frac{x+1}{x-2}\inℤ\Rightarrow x-2+3⋮x-2\)
\(\Leftrightarrow3⋮x-2\Rightarrow x-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
x - 2 | 1 | -1 | 3 | -3 |
x | 3 | 1 | 5 | -1 |
d, Ta có : \(B=-\frac{3}{4}\Rightarrow\frac{x+1}{x-2}=-\frac{3}{4}\)ĐK : \(x\ne2\)
\(\Rightarrow4x+4=-3x+6\Leftrightarrow7x=2\Leftrightarrow x=\frac{2}{7}\)( tmđk )
e, Ta có B < 0 hay \(\frac{x+1}{x-2}< 0\)
TH1 : \(\hept{\begin{cases}x+1< 0\\x-2>0\end{cases}\Rightarrow\hept{\begin{cases}x< -1\\x>2\end{cases}}}\)( ktm )
TH2 : \(\hept{\begin{cases}x+1>0\\x-2< 0\end{cases}}\Rightarrow\hept{\begin{cases}x>-1\\x< 2\end{cases}\Rightarrow-1< x< 2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\left(ĐK:x\ne-3;x\ne2\right)\)
\(=\frac{x+2}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{1}{x-2}\)
\(=\frac{\left(x+2\right)\left(x-2\right)-5-\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}=\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{x-4}{x-2}\)
Để \(A=-\frac{3}{4}\)
\(\Leftrightarrow\frac{x-4}{x-2}=-\frac{3}{4}\)
\(\Leftrightarrow4\left(x-4\right)=-3\left(x-2\right)\)
\(\Leftrightarrow4x-16=-3x+6\)
\(\Leftrightarrow7x=22\Leftrightarrow x=\frac{22}{7}\left(tm\right)\)
Vậy \(x=\frac{22}{7}\) thì \(A=-\frac{3}{4}\)
b) \(A=\frac{x-4}{x-2}=\frac{\left(x-2\right)-2}{x-2}=1-\frac{2}{x-2}\)
Để \(A\in Z\Rightarrow\frac{2}{x-2}\in Z\Rightarrow x-2\inƯ\left(2\right)\)
Mà: \(Ư\left(2\right)=\left\{1;-1;2;-2\right\}\)
=> \(x-2\in\left\{1;-1;2;-2\right\}\)
+) \(x-2=1\Rightarrow x=3\left(tm\right)\)
+) \(x-2=-1\Rightarrow x=1\left(tm\right)\)
+) \(x-2=2\Rightarrow x=4\left(tm\right)\)
+) \(x-2=-2\Rightarrow x=0\left(tm\right)\)
Vậy \(x\in\left\{0;1;3;4\right\}\) thì \(A\in Z\)
A=x+2/x+3-5/(x-2)(x+3)-1/x-2
A=(x+2)(x-2)-5-x-3/(x-2)(x+3)
A=x^2-4-5-x-3/(x-2)(x+3)
A=x^2-x-12/(x-2)(x+3)
A=(x+3)(x-4)/(x-2)(x+3)
A=x-4/x-2
Để A=-3/4 thì x-4/x-2=-3/4
Từ đó suy ra (x-4)4=-3(x-2)
4x-16=-3x+6
7x=22
x=22/7
b,Do A nguyên nên x-4/x-2 nguyên(x#2)
suy ra x-4-x+2 chia hết cho x-2
nên 2 chia hết cho x-2
mà ước 2=-2;-1;1;2
nên x=0;1;3;4
=> x - 1 > 0 và x - 3 < 0 hoặc x - 1 < 0 và x - 3 > 0
=> x > 1 và x < 3 hoặc x < 1 và x > 3 ( loại )
=> 1 < x < 3
Vậy 1 < x < 3