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thì bạn cx cộng cả hai vế của bất đẳng thức đầu tiên vs 1/2 thì đc điều phải chứng minh
a, \(\dfrac{2-x}{2001}-1=\dfrac{1-x}{2002}-\dfrac{x}{2003}\)
\(\Leftrightarrow\dfrac{2-x}{2001}-1+2=\dfrac{1-x}{2002}-\dfrac{x}{2003}+2\)
\(\Leftrightarrow\dfrac{2-x}{2001}+1=\left(\dfrac{1-x}{2002}+1\right)+\left(\dfrac{-x}{2003}+1\right)\)
\(\Leftrightarrow\dfrac{2003-x}{2001}=\dfrac{2003-x}{2002}+\dfrac{2003-x}{2003}\)
\(\Leftrightarrow\dfrac{2003-x}{2001}-\dfrac{2003-x}{2002}-\dfrac{2003-x}{2003}=0\)
\(\Leftrightarrow\left(2003-x\right)\left(\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
Vì \(\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\)
\(\Rightarrow2003-x=0\)
\(\Rightarrow x=2003\)
Vậy : \(s=\left\{2003\right\}\)
b, \(\dfrac{x-5}{100}+\dfrac{x-4}{101}=\dfrac{x-100}{5}+\dfrac{x-101}{4}\)
\(\Leftrightarrow\dfrac{x-5}{100}+\dfrac{x-4}{101}-2=\dfrac{x-100}{5}+\dfrac{x-101}{4}-2\)
\(\Leftrightarrow\left(\dfrac{x-5}{100}-1\right)+\left(\dfrac{x-4}{101}-1\right)=\left(\dfrac{x-100}{5}-1\right)+\left(\dfrac{x-101}{4}-1\right)\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}=\dfrac{x-105}{5}+\dfrac{x-105}{4}\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}-\dfrac{x-105}{5}-\dfrac{x-105}{4}=0\)
\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}-\dfrac{1}{5}-\dfrac{1}{4}\right)=0\)
Vì \(\dfrac{1}{100}+\dfrac{1}{101}-\dfrac{1}{5}-\dfrac{1}{4}\ne0\)
\(\Rightarrow x-105=0\)
\(\Rightarrow x=105\)
Vậy : \(s=\left\{105\right\}\)
\(a,\dfrac{2-x}{2001}-1=\dfrac{1-x}{2002}-\dfrac{x}{2003}\)
\(\Leftrightarrow\)haizzz bạn cộng mỗi hạng tử ở mỗi vế cho một. Chuyển vế và giải ra x=2003
b, Tương tự bạn -1 cho mỗi vế. GIải phương trình đc x=105
5. phân tích ra : \(1+\dfrac{a}{b}+\dfrac{b}{a}+1\)
áp dụng bđ cosy
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)
=> đpcm
6. \(x^2-x+1=x^2-2.\dfrac{1}{2}.x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
hay với mọi x thuộc R đều là nghiệm của bpt
7.áp dụng bđt cosy
\(a^4+b^4+c^4+d^4\ge2\sqrt{a^2.b^2.c^2.d^2}=4abcd\left(đpcm\right)\)
a: \(=\dfrac{x+3}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x\left(x+1\right)}\)
\(=\dfrac{x^2+3x-x+1}{x\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{x\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{x\left(x-1\right)}\)
b: \(=\dfrac{24y^5}{7x^2}\cdot\dfrac{-21x}{12y^3}=2y^2\cdot\dfrac{-3}{x}=\dfrac{-6y^2}{x}\)
c: \(=\dfrac{-3\left(x-1\right)}{\left(x+1\right)^2}\cdot\dfrac{x+1}{6\left(x-1\right)\left(x+1\right)}=\dfrac{-1}{2\left(x+1\right)}\)
d: \(=\dfrac{7x+2}{3\left(2x-y\right)}\cdot\dfrac{6x\left(2x-y\right)}{2\left(7x+2\right)}=x\)
a) ĐKXĐ của A là \(x\ne1\)
\(A=\dfrac{x^2-1}{x^2-2x+1}=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}=\dfrac{x+1}{x-1}\)
ĐKXĐ của B là \(x\ne2;x\ne-2\)
\(B=\left(\dfrac{x-1}{x+2}-\dfrac{x+1}{x-2}\right):\dfrac{6}{x-2}=\left(\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\right).\dfrac{x-2}{6}=\left(\dfrac{x^2-3x+2-x^2-3x-2}{\left(x+2\right)\left(x-2\right)}\right).\dfrac{x-2}{6}=\dfrac{-6x}{\left(x+2\right)\left(x-2\right)}.\dfrac{x-2}{6}=\dfrac{-x}{x+2}\)b)
Với \(x\ne1\)
\(A>1\Leftrightarrow A-1>0\Leftrightarrow\dfrac{x+1}{x-1}>0\)
TH1 \(\left\{{}\begin{matrix}x+1>0\\x-1>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x>-1\\x>1\end{matrix}\right.\)\(\Leftrightarrow x>1\)
TH2 \(\left\{{}\begin{matrix}x+1< 0\\x-1< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x< -1\\x< 1\end{matrix}\right.\)\(\Leftrightarrow x< -1\)
c) Với \(x\ne1;x\ne2;x\ne-2\)
\(A=B\Leftrightarrow\dfrac{x+1}{x-1}=\dfrac{-x}{x+2}\)
\(\Leftrightarrow\dfrac{x+1}{x-1}+\dfrac{x}{x+2}=0\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}+\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=0\)
\(\Leftrightarrow x^2+3x+2+x^2-x=0\)
\(\Leftrightarrow2x^2-2x+2=0\)
\(\Leftrightarrow2\left(x^2-x+1\right)=0\)
\(\Leftrightarrow x^2-x+1=0\) \(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)
Với mọi x ta luôn có \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
=> ko có giá trị nào của x để A=B
\(x^2-5x+6=\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)
1, <=>x^2-x-2 = x^2-4
<=>x^2-4-x^2+x+2 = 0
<=> x-2 = 0
<=> x=2
2, <=> (x-2).(x-3)=0
<=> x-2 = 0 hoặc x-3 = 0
<=> x=2 hoặc x=3
Ta có: \(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}\ge\dfrac{2}{1+xy}\)
\(\Leftrightarrow\left(\dfrac{1}{1+x^2}-\dfrac{1}{1+y^2}\right)+\left(\dfrac{1}{1+y^2}-\dfrac{1}{xy}\right)\ge0\)
\(\Leftrightarrow\dfrac{xy-x^2}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{xy-y^2}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow\dfrac{x\left(y-x\right)}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{y\left(x-y\right)}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(y-x\right)^2\left(xy-1\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)
BĐT cuối đúng vì x.y > 0 => đpcm
Từ \(x\ge2\) cộng cả hai vế với \(\dfrac{1}{2}\) ta được
\(x+\dfrac{1}{2}\ge2+\dfrac{1}{2}=\dfrac{5}{2}\)
\(VT=x+\dfrac{1}{2}=x-2+2+\dfrac{1}{2}=\left(x-2\right)+\dfrac{5}{2}\)
\(\left\{{}\begin{matrix}x\ge2\Rightarrow x-2\ge0\\VT=\left(x-2\right)+\dfrac{5}{2}\ge\dfrac{5}{2}=VP\rightarrow dpcm\end{matrix}\right.\)