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a: \(=\dfrac{x^3-x^2+x+3\left(x^2-1\right)+x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x^3-x^2+2x+4+3x^2-3}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^3+2x^2+2x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x^2+x+1}{x^2-x+1}\)
b: \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
\(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
=>A>0 với mọi x<>-1
\(Câu\text{ }1:\)
\(\text{ a) }A=\dfrac{4}{x^2+2}+\dfrac{3}{2-x^2}-\dfrac{12}{4-x^4}\\ A=\dfrac{4\left(2-x^2\right)}{\left(x^2+2\right)\left(2-x^2\right)}+\dfrac{3\left(2+x^2\right)}{\left(2-x^2\right)\left(2+x^2\right)}-\dfrac{12}{\left(2+x^2\right)\left(2-x^2\right)}\\ A=\dfrac{4\left(2-x^2\right)+3\left(2+x^2\right)-12}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{8-4x^2+6+3x^2-12}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{-x^2-2}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{-\left(x^2+2\right)}{\left(x^2+2\right)\left(2-x^2\right)}\\ A=\dfrac{-1}{2-x^2}\)
\(\text{b) }Để\text{ }A=-3\\ thì\Rightarrow\dfrac{-1}{2-x^2}=-3\\ \Leftrightarrow2-x^2=3\\ \Leftrightarrow x^2=-1\\ \Leftrightarrow x\text{ }không\text{ }có\text{ }giá\text{ }trị\left(vì\text{ }x^2\ge0\forall x\right)\\ \text{ }Vậy\text{ }để\text{ }A=-3\text{ }thì\text{ }x\text{ }không\text{ }có\text{ }giá\text{ }trị.\)
\(\text{c) }Ta\text{ }có:\text{ }A=\dfrac{-1}{2-x^2}\\ A=\dfrac{1}{x^2-2}\\ x^2\ge0\forall x\\ \Rightarrow x^2-2\ge-2\forall x\\ \Rightarrow A=\dfrac{1}{x^2-2}\le-\dfrac{1}{2}\\ Dấu\text{ }"="\text{ }xảy\text{ }khi:\\ x^2=0\\ \Leftrightarrow x=0\\\text{ }Vậy\text{ }A_{\left(Max\right)}=-\dfrac{1}{2}\text{ }khi\text{ }x=0\)
\(Câu\text{ }2:\)
\(\text{a) }B=\dfrac{1}{x}+\dfrac{1}{x+5}+\dfrac{x-5}{x\left(x+5\right)}\\ B=\dfrac{x+5}{x\left(x+5\right)}+\dfrac{x}{\left(x+5\right)x}+\dfrac{x-5}{x\left(x+5\right)}\\ B=\dfrac{x+5+x+x-5}{x\left(x+5\right)}\\ B=\dfrac{3x}{x\left(x+5\right)}\\ B=\dfrac{3}{x+5}\left(\text{*}\right)\)
\(\text{b) }Ta\text{ }có:\text{ }\left|x-1\right|=6\\ \Leftrightarrow\left[{}\begin{matrix}x-1=6\\x-1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\\ Ta\text{ }lại\text{ }có:\text{ }B=\dfrac{3}{x+5}\\ \RightarrowĐKCĐ:x+5\ne0\\ \Rightarrow x\ne-5\\ \Rightarrow x=7\text{ }thỏa\text{ }mãn\text{ }với\text{ }điều\text{ }kiện\text{ }của\text{ }biến.\\ x=-5\text{ }không\text{ }thỏa\text{ }mãn\text{ }với\text{ }điều\text{ }kiện\text{ }của\text{ }biến.\\ Thay\text{ }x=7\text{ }vào\text{ }\left(\text{*}\right),ta\text{ }được:\text{ }B=\dfrac{3}{7+5}=\dfrac{3}{12}=\dfrac{1}{4}\\ \text{ }Vậy\text{ }với\text{ }x=7\text{ }thì\text{ }B=\dfrac{1}{4}\\ với\text{ }x=-5\text{ }thì\text{ }B\text{ }không\text{ }có\text{ }giá\text{ }trị.\)
\(\text{c) }Ta\text{ }có:B=\dfrac{3}{x+5}\\ \RightarrowĐể\text{ }B\in Z\\ thì\Rightarrow3⋮x+5\\ \Rightarrow x+5\inƯ_{\left(3\right)}\\ Mà\text{ }Ư_{\left(3\right)}=\left\{\pm1;\pm3\right\}\\ Ta\text{ }lập\text{ }bảng\text{ }xét\text{ }giá\text{ }trị:\)
| \(x+5\) | \(-3\) | \(-1\) | \(1\) | \(3\) |
| \(x\) | \(-8\) | \(-6\) | \(-4\) | \(-2\) |
\(\Rightarrow x\in\left\{-8;-6;-4;-2\right\}\\ Vậy\text{ }để\text{ }B\in Z\\ thì x\in\left\{-8;-6;-4;-2\right\}\)
1/ a, \(A=\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
Vậy \(A=x\)
b/ Khi \(x=\dfrac{1}{2}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{2}}=2\)
Vậy...
2/a,
\(A=\dfrac{5x+2}{3x^2+2x}+\dfrac{-2}{3x+2}\)
\(=\dfrac{5x+2}{x\left(3x+2\right)}-\dfrac{2x}{x\left(3x+2\right)}\)
\(=\dfrac{5x+2-2x}{x\left(3x+2\right)}\)
\(=\dfrac{3x+2}{x\left(3x+2\right)}\)
\(=\dfrac{1}{x}\)
Vậy....
b/ Với \(x=\dfrac{1}{3}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{3}}=3\)
Vậy..
\(A=\left(\dfrac{1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{6x+3}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right):\left(x+2\right)\)\(A=\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x^2-x+1\right)\left(x+2\right)}\)
a) \(A=\left\{{}\begin{matrix}x\ne-1;-2\\\dfrac{1}{x^2-x+1}\end{matrix}\right.\)
b)
\(A>1;\dfrac{1}{x^2-x+1}>1\Leftrightarrow x^2-x< 0\Leftrightarrow0< x< 1\)
\(P=\dfrac{1}{x^2-x+1}.\dfrac{x^3-x^2+x}{\left(x+1\right)^2}=\dfrac{x}{\left(x+1\right)^2}\)
x>0 => P >0 đang tìm Giá trị LN => chỉ xét P>0 <=> x>0
\(\dfrac{1}{P}=\dfrac{\left(x+1\right)^2}{x}=x+2+\dfrac{1}{x}\)
áp co si hai số dương x ; 1/x
\(\dfrac{1}{P}\ge2.\sqrt{x.\dfrac{1}{x}}+2=4\Rightarrow P\le\dfrac{1}{4}\)
đẳng thức khi x =1/x => x=1 thỏa mãn đk của x
\(MaxP=\dfrac{1}{4}\)
a: \(P=\dfrac{x^3-x^2+2x-2+x^2-2x+1}{x\left(x-1\right)}\)
\(=\dfrac{x^3-1}{x\left(x-1\right)}=\dfrac{x^2+x+1}{x}\)
b: x^2+x+1=(x+1/2)^2+3/4>=3/4>0
x>0
=>P>0
Câu 3:
\(\Leftrightarrow3x^3-2x^2+6x^2-4x+9x-6>0\)
\(\Leftrightarrow\left(3x-2\right)\left(x^2+2x+3\right)>0\)
=>3x-2>0
=>x>2/3
Câu 1:
a: \(A=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{x+1+2x-2}{\left(x^2-1\right)}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\left(\dfrac{3x-1}{x^2-1}-\dfrac{3}{x}\right)\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{3x^2-x-3x^2+3}{x\left(x^2-1\right)}\cdot\dfrac{x^2-1}{x+2}\)
\(=x-2+\dfrac{6x-3}{x\left(x+2\right)}+\dfrac{-\left(x-3\right)}{x\left(x+2\right)}\)
\(=x-2+\dfrac{6x-3-x^2+3x}{x\left(x+2\right)}\)
\(=x-2+\dfrac{-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x\left(x^2-4\right)-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x^3-4x-x^2+9x-3}{x\left(x+2\right)}\)
\(=\dfrac{x^3-x^2+5x-3}{x\left(x+2\right)}\)
b: TH1: \(\left\{{}\begin{matrix}x^3-x^2+5x-3>0\\x\left(x+2\right)< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2< x< 2\\x>0.63\end{matrix}\right.\Leftrightarrow0.63< x< 2\)
TH2: \(\left\{{}\begin{matrix}x^3-x^2+5x-3< 0\\x\left(x+2\right)>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 0.63\\\left[{}\begin{matrix}x>0\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0< x< 0.63\\x< -2\end{matrix}\right.\)
a.
ĐKXĐ: \(x\ne2\)
b.
\(P=\left(\dfrac{2x}{x-2}+\dfrac{x}{2-x}\right):\dfrac{x^2+1}{x-2}\)
\(=\left(\dfrac{2x}{x-2}-\dfrac{x}{x-2}\right)\cdot\dfrac{x-2}{x^2+1}\)
\(=\dfrac{x}{x-2}\cdot\dfrac{x-2}{x^2+1}=\dfrac{x}{x^2+1}\)
c.
\(x=-1\Rightarrow P=-\dfrac{1}{\left(-1\right)^2+1}=-\dfrac{1}{2}\)
d.
\(P=\dfrac{x}{x^2+1}\cdot\dfrac{x^2+1}{x}-\dfrac{1}{P}\ge1-\dfrac{1}{P}\)
\(\Rightarrow\dfrac{P^2+1}{P}\ge1\)
\(\Rightarrow P^2+1\ge P\) \(\Rightarrow P\left(P-1\right)\ge1\)
\(\Rightarrow P\ge2\)
Dấu "=" khi x = ...................
Bài 2:
a: \(M=\dfrac{3x+1-2x-2}{\left(3x-1\right)\left(3x+1\right)}:\dfrac{3x+1-3x}{x\left(3x+1\right)}\)
\(=\dfrac{x-1}{\left(3x-1\right)\left(3x+1\right)}\cdot\dfrac{x\left(3x+1\right)}{1}=\dfrac{x\left(x-1\right)}{3x-1}\)
b: Để M=0 thì x(x-1)=0
=>x=1(nhận) hoặc x=0(loại)
c: \(P=M\cdot\left(3x-1\right)=x\left(x-1\right)=x^2-x+\dfrac{1}{4}-\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}>=-\dfrac{1}{4}\)
Dấu = xảy ra khi x=1/2
a, Rút gọn Biểu thức:
A=\(\left(\dfrac{x+2}{2x-4}-\dfrac{x-2}{2x+4}\right):\dfrac{2x}{x2+2x}\)
= \(\left(\dfrac{x+2}{2x-4}+\dfrac{-x-2}{2x+4}\right):\dfrac{2x}{x2+2x}\)
= \(\left(\dfrac{x+2+-x-2}{2x-4+2x+4}\right):\dfrac{2x}{x2+2x}\)
= 0 \(:\dfrac{2x}{x2+2x}\)
b, \(\left(\dfrac{x+2}{2x-4}-\dfrac{x-2}{2x+4}\right):\dfrac{2x}{x2+2x}\)
Thay tất cả x= -4
=> \(\left(\dfrac{-4+2}{2-4-4}-\dfrac{-4-2}{2-4+4}\right):\dfrac{2.-4}{-4.2+2.-4}\)
= -16 : \(\dfrac{1}{3}\)
= -18
\(A=\frac{x}{x+1}-\frac{3-3x}{x^2-x+1}+\frac{x+4}{x^3+1}\)
\(A=\frac{x\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{3-3x}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{x^3-x^2+x-3-3x+x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(A=\frac{1}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{1}{x^3+1}\)