Cho biểu thức \(P=\left(-x+2\right)^2-\left(x-3\right)\left(3+x\right)\)
a, Tìm tất cả các giá trị của x để P là 1 số nguyên dương nhỏ nhất
b, CMR với mọi giá trị của \(x\ne2\)thì giá trị của biểu thức \(P+x^2\)bao giờ cũng vượt qua 9
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
3 tháng 4 2020
\(C=\frac{2\left(x-1\right)^2+1}{\left(x-1\right)^2+2}\)
a, Ta thấy \(\left(x-1\right)^2\ge0\forall x\Rightarrow\hept{\begin{cases}2\left(x-1\right)^2+1\ge1>0\\\left(x-1\right)^2+2\ge2>0\end{cases}}\)
\(\Rightarrow C>0\forall x\)(đpcm)
b, \(C=\frac{2\left(x-1\right)^2+1}{\left(x-1\right)^2+2}=\frac{2\left(x-1\right)^2+4-3}{\left(x-1\right)^2+2}=2-\frac{3}{\left(x-1\right)^2+2}\)
\(C\in Z\Leftrightarrow2-\frac{3}{\left(x-1\right)^2+2}\in Z\)
\(\Leftrightarrow\frac{3}{\left(x-1\right)^2+2}\in Z\)Lại do \(\left(x-1\right)^2+2\ge2\)
\(\Leftrightarrow\left(x-1\right)^2+2\inƯ\left(3\right)=\left\{3\right\}\)
\(\Leftrightarrow\left(x-1\right)^2\in\left\{1\right\}\)
\(\Leftrightarrow x\in\left\{0\right\}\)
....
c, \(C=2-\frac{3}{\left(x-1\right)^2+2}\)
Ta có : \(\left(x-1\right)^2+2\ge2\Rightarrow\frac{3}{\left(x-1\right)^2+2}\le\frac{3}{2}\)
\(\Rightarrow C=2-\frac{3}{\left(x-1\right)^2+2}\ge2-\frac{3}{2}=\frac{1}{2}\)
Dấu "=" xảy ra khi \(x-1=0\Leftrightarrow x=1\)
:33
14 tháng 5 2020
\(A=\left(\frac{2X-1}{x^2-4}+\frac{x+2}{x^2-x-2}\right):\frac{x-2}{x^2+3x+2}ĐK:x\ne\left\{2,-2,-1\right\}\)
a) \(A=\left[\frac{\left(2x-1\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x+2}{\left(x+1\right)\left(x-2\right)}\right]:\frac{x-2}{\left(x+2\right)\left(x+1\right)}\)
\(A=\left[\frac{\left(2x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}\frac{\left(x+2\right)\left(x+2\right)}{\left(x+1\right)\left(x-2\right)\left(x+2\right)}\right].\frac{\left(x+2\right)\left(x+1\right)}{x-2}\)
\(A=\frac{2x^2+x-1+x^2+4x.4}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}.\frac{\left(x+2\right)\left(x+1\right)}{\left(x-2\right)}\)
\(A=\frac{3x^2+5x+3}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}.\frac{\left(x+2\right)\left(x+1\right)}{\left(x-2\right)}\)
\(A=\frac{3x^2+5x+3}{\left(x-2\right)^2}\)
Ta có :\(3x^2+5x+3\)
\(=3\left(x^2+\frac{5}{3}x+1\right)\)
\(=3\left[x^2+2.\frac{5}{6}x+\frac{25}{36}+\frac{9}{36}\right]\)
\(=3\left[\left(x+\frac{5}{6}\right)^2+\frac{9}{36}\right]>0\)
Mà \(\left(x-2\right)^2>0\)
\(\Rightarrow A>0\left(dpcm\right)\)
\(b,A=11\Leftrightarrow\frac{3x^2+5x+3}{\left(x-2\right)^2}=11\)
\(\Rightarrow3x^2+5x+3=11.\left(x-2\right)^2\)
\(\Rightarrow3x^2+5x+3=11.\left(x^2-4x+4\right)\)
\(\Rightarrow8x^2-49x+41=0\)
\(\Rightarrow8x^2-8x-41x+41=0\)
\(\Rightarrow8x\left(x-1\right)-41\left(x-1\right)=0\)
\(\Rightarrow\left(8x-41\right)\left(x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}8x-41=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{41}{8}\\x=1\end{cases}}}\)(Thỏa mãn)
TD
31 tháng 3 2017
2.
a/\(A=5-I2x-1I\)
Ta thấy: \(I2x-1I\ge0,\forall x\)
nên\(5-I2x-1I\le5\)
\(A=5\)
\(\Leftrightarrow5-I2x-1I=5\)
\(\Leftrightarrow I2x-1I=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy GTLN của \(A=5\Leftrightarrow x=\frac{1}{2}\)
b/\(B=\frac{1}{Ix-2I+3}\)
Ta thấy : \(Ix-2I\ge0,\forall x\)
nên \(Ix-2I+3\ge3,\forall x\)
\(\Rightarrow B=\frac{1}{Ix-2I+3}\le\frac{1}{3}\)
\(B=\frac{1}{3}\)
\(\Leftrightarrow B=\frac{1}{Ix-2I+3}=\frac{1}{3}\)
\(\Leftrightarrow Ix-2I+3=3\)
\(\Leftrightarrow Ix-2I=0\)
\(\Leftrightarrow x=2\)
Vậy GTLN của\(A=\frac{1}{3}\Leftrightarrow x=2\)
24 tháng 4 2023
a: ĐKXĐ: x<>-1
b: \(P=\left(1-\dfrac{x+1}{x^2-x+1}\right)\cdot\dfrac{x^2-x+1}{x+1}\)
\(=\dfrac{x^2-x+1-x-1}{x^2-x+1}\cdot\dfrac{x^2-x+1}{x+1}=\dfrac{x^2-2x}{x+1}\)
c: P=2
=>x^2-2x=2x+2
=>x^2-4x-2=0
=>\(x=2\pm\sqrt{6}\)
28 tháng 9 2021
\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\left(đk:x\ge0,x\ne1\right)\)
\(=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2.2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)
Để A nguyên thì: \(x+\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Mà \(x+\sqrt{x}+1=\left(x+\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
\(\Rightarrow x+\sqrt{x}+1\in\left\{1;2\right\}\)
+ Với \(x+\sqrt{x}+1=1\)
\(\Leftrightarrow\sqrt[]{x}\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow x=0\left(tm\right)\left(do.\sqrt{x}+1\ge1>0\right)\)
+ Với \(x+\sqrt{x}+1=2\)
\(\Leftrightarrow\left(x+\sqrt{x}+\dfrac{1}{4}\right)=\dfrac{5}{4}\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+\dfrac{1}{2}=\dfrac{\sqrt{5}}{2}\\\sqrt{x}+\dfrac{1}{2}=-\dfrac{\sqrt{5}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{5}-1}{2}\\\sqrt{x}=-\dfrac{\sqrt{5}+1}{2}\left(VLý\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{3-\sqrt{5}}{2}\left(tm\right)\)
Vậy \(S=\left\{1;\dfrac{3-\sqrt{5}}{2}\right\}\)
14 tháng 12 2018
a,ĐK: \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)
c, Với x = 4 thỏa mãn ĐKXĐ thì
\(A=\frac{-3}{4-3}=-3\)
d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)
\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)
Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)