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![](https://rs.olm.vn/images/avt/0.png?1311)
\(1.=5^x\left(1+5^2\right)=5^x.26=3250\)
\(< =>5^x=125=>x=3\)
2. Để P có giá trị lớn nhất thì -/x-3/ có giá trị bé nhất...
=> x-3 có bé nhất hay x=3;
=>ĐPCM
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Đk \(x\ne\pm1\), sau khi rút gọn ta được: (bạn tư làm)
\(P=\frac{x}{x+1}\)
b) Khi \(\left|x-\frac{2}{3}\right|=\frac{1}{3}\) thì hoặc \(x-\frac{2}{3}=\frac{1}{3}\) hoặc \(x-\frac{2}{3}=-\frac{1}{3}\)
Hay là \(x=1\) hoặc \(x=\frac{1}{3}\)
Do để P có nghĩa thì \(x\ne\pm1\) nên \(x=\frac{1}{3}\), khi đó:
\(P=\frac{\frac{1}{3}}{\frac{1}{3}+1}=\frac{1}{4}\)
c) P > 1 khi \(\frac{x}{x+1}>1\)
\(\Leftrightarrow1-\frac{1}{x+1}>1\)
\(\Leftrightarrow\frac{1}{x+1}< 0\)
\(\Leftrightarrow x< -1\)
e) Đề không rõ ràng
![](https://rs.olm.vn/images/avt/0.png?1311)
Phần a,b mình vừa trả lời r bạn xem lại nha
c) Với\(x\ne0;x\ne1;x\ne-1\)
Để \(\)A nhận giá trị nguyên thì \(\dfrac{3}{x+1}\) nguyên
\(\Rightarrow x+1\in\)ước nguyên của 3
\(\Rightarrow x+1\in\left\{1;-1;3;-3\right\}\)
Ta có bảng:
x+1 | -3 | -1 | 1 | 3 |
x | -4 | -2 | 0 | 2 |
(tm) | (tm) | (ktm) | (tm) |
Vậy...
![](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.` Với `x≠-2; +2`
Để `|A|=A` thì `A>0`
`=>` \(\dfrac{x+2}{x-2}>0\)
trường hợp `1:` \(\left\{{}\begin{matrix}x+2>0\\x-2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-2\\x>2\end{matrix}\right.\Leftrightarrow x>2\)
trường hợp `2:` \(\left\{{}\begin{matrix}x+2< 0\\x-2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -2\\x< 2\end{matrix}\right.\Leftrightarrow x< -2\)
Vậy \(x>2\) hoặc `x< -2`
`c.` xét phương trình `A=m`
\(\Leftrightarrow\dfrac{x+2}{x-2}=m\\ \Leftrightarrow x+2=m\left(x-2\right)\\ \Leftrightarrow x+2=mx-2m\\ \Leftrightarrow x-mx=-2m-2\\ \Leftrightarrow\left(1-m\right)x=-2m-2\\\)
để phương trình có nghiệm thì `1-m≠0 => m≠1`
b) \(x>2\).
\(\left(x+1\right).A=\left(x+1\right).\dfrac{x+2}{x-2}=\dfrac{x^2+3x+2}{x-2}=\dfrac{x^2-2x+5x-10+12}{x-2}=\dfrac{x\left(x-2\right)+5\left(x-2\right)+12}{x-2}=x+5+\dfrac{12}{x-2}=x-2+\dfrac{12}{x-2}+7\ge2\sqrt{\left(x-2\right).\dfrac{12}{\left(x-2\right)}}+7=2\sqrt{12}+7\)\(\left(x+1\right).A=2\sqrt{12}+7\Leftrightarrow x=2+\sqrt{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ : \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\\9-x^2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
a, \(A=\dfrac{x-5}{x-3}-\dfrac{2x}{x+3}-\dfrac{2x^2-x+15}{9-x^2}\)
\(=\dfrac{\left(x-5\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x^2-x+15}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2-2x-15-2x^2+6x+2x^2-x+15}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2-3x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x}{x+3}\)
b, \(\left|x-1\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\left(kot/m\right)\\x=-1\left(t/m\right)\end{matrix}\right.\)
Thay x =- 1 vào biểu thức A ,có :
\(\dfrac{-1}{-1+3}=\dfrac{-1}{2}\)
Vậy tại x = -1 gtri của bt A là -1/2
Vậy tại x = 3 biểu thức A ko có giá trị
c,\(\dfrac{x}{x+3}=\dfrac{x+3-3}{x+3}=1-\dfrac{3}{x+3}\)
Để A có giá trị nguyên
\(\Leftrightarrow\dfrac{3}{x+3}\) là số nguyên
\(\Leftrightarrow3⋮x+3\)
\(\Leftrightarrow x+3\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(x+3\) | 1 | -1 | 3 | -3 |
x | -2 (t/m) | -4(t/m) | 0 (t/m) | -6(t/m) |
Vậy \(x\in\left\{0;-2;-4;-6\right\}\) thì A có giá trị nguyên
![](https://rs.olm.vn/images/avt/0.png?1311)
Rút gọn P
a) \(P=\left(\dfrac{x+1}{x-1}+\dfrac{4x^2}{x^2-1}+\dfrac{x-1}{x+1}\right):\dfrac{x^2+x}{x^3-x}\)
\(P=\left(\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{4x^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{x^2+x}{x\left(x^2-1\right)}\)
\(P=\left(\dfrac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}+\dfrac{4x^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{x^2-x}{x\left(x-1\right)\left(x+1\right)}\)
\(P=\left(\dfrac{x^2+2x+1+4x^2+x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{x^2-x}{x\left(x-1\right)\left(x+1\right)}\)
\(P=\left(\dfrac{6x^2+2}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{x^2-x}{x\left(x-1\right)\left(x+1\right)}\)
\(P=\dfrac{2\left(3x^2+1\right)}{\left(x-1\right)\left(x+1\right)}.\dfrac{x\left(x-1\right)\left(x+1\right)}{x\left(x-1\right)}\)
\(P=\dfrac{2x\left(3x^2+1\right)}{x}=2\left(3x^2+1\right)\)
1) \(Q=-x\) khi:
\(\dfrac{x-3}{x+1}=-x\)
\(\Leftrightarrow x-3=-x\left(x+1\right)\)
\(\Leftrightarrow x-3=-x^2-x\)
\(\Leftrightarrow x-3+x^2+x\)
\(\Leftrightarrow x^2+2x-3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
2) \(Q< 1\) khi:
\(\dfrac{x-3}{x+1}< 1\)
\(\Leftrightarrow x-3< x+1\)
\(\Leftrightarrow x-x< 1+3\)
\(\Leftrightarrow0< 4\) (luôn đúng)
Vậy \(Q< 0\) với mọi x
3) \(Q=m\) khi:
\(\dfrac{x-3}{x+1}=m\)
\(\Leftrightarrow x-3=m\left(x+1\right)\)
\(\Leftrightarrow x-3=mx+m\)
\(\Leftrightarrow x-mx=m+3\)
\(\Leftrightarrow x\left(1-m\right)=m+3\)
\(\Leftrightarrow1-m\ne0\)
\(\Leftrightarrow m\ne1\)