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\(M=\dfrac{13x^2-x^4-36}{x^3-5x^2+6x}\)
\(=\dfrac{-x^4+13x^2-36}{x\left(x^2-5x+6\right)}\)
\(=\dfrac{-x^4+9x^2+4x^2-36}{x\left(x^2-2x-3x+6\right)}\)
\(=\dfrac{-x^2\left(x^2-9\right)+4\left(x^2-9\right)}{x\cdot\left[x\left(x-2\right)-3\left(x-2\right)\right]}\)
\(=\dfrac{\left(-x^2+4\right)\left(x^2-9\right)}{x\left(x-3\right)\left(x-2\right)}\)
\(=\dfrac{\left(4-x^2\right)\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)\left(x-2\right)}\)
\(=\dfrac{\left(2-x\right)\left(2+x\right)\left(x+3\right)}{x\left(x-2\right)}\)
\(=\dfrac{-\left(x-2\right)\left(2+x\right)\left(x+3\right)}{x\left(x-2\right)}\)
\(=\dfrac{-\left(2+x\right)\left(x+3\right)}{x}\)
\(=\dfrac{-\left(2x+6+x^2+3x\right)}{x}\)
\(=\dfrac{-\left(5x+6+x^2\right)}{x}\)
\(=-\dfrac{5x+6+x^2}{x}\)
\(=\left[\dfrac{2x-3}{\left(2x-5\right)\left(2x-1\right)}-\dfrac{3}{2x-1}-\dfrac{2\left(x-4\right)}{\left(x-4\right)\left(2x-5\right)}\right].\dfrac{2x\left(2x+3\right)-\left(2x+3\right)}{-2x\left(4x-7\right)-3\left(4x-7\right)}+1\)
\(=\left[\dfrac{2x-3-6x+15-4x+2}{\left(2x-5\right)}\right].\dfrac{2\left(x+\dfrac{3}{2}\right)}{\left(-2x-3\right)\left(4x-7\right)}+1\)
\(=\dfrac{-2\left(4x-7\right)}{2x-5}.\dfrac{2\left(x+\dfrac{3}{2}\right)}{\left(-2x-3\right)\left(4x-7\right)}+1\)
\(=\dfrac{1}{2x-5}.2+1\)
\(=\dfrac{2+2x-5}{2x-5}\)
\(=\dfrac{-3+2x}{2x-5}\)
Bải 3a
\(\dfrac{-a+b+c}{2a}+\dfrac{-b+c+a}{2b}+\dfrac{-c+a+b}{2c}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{-a}{2a}+\dfrac{b+c}{2a}+\dfrac{-b}{2b}+\dfrac{c+a}{2b}+\dfrac{-c}{2c}+\dfrac{a+b}{2c}\ge\dfrac{3}{2}\)
\(\Leftrightarrow-\dfrac{3}{2}+\dfrac{b+c}{2a}+\dfrac{c+a}{2b}+\dfrac{a+b}{2c}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{b+c}{2a}+\dfrac{c+a}{2b}+\dfrac{a+b}{2c}\ge3\)
\(\Leftrightarrow\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\ge6\)
\(\Leftrightarrow\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\ge6\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{ab}{ba}}=2\\\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{bc}{cb}}=2\\\dfrac{c}{a}+\dfrac{a}{c}\ge2\sqrt{\dfrac{ca}{ac}}=2\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\ge2+2+2=6\)
\(\Leftrightarrow\dfrac{-a+b+c}{2a}+\dfrac{-b+c+a}{2b}+\dfrac{-c+a+b}{2c}\ge\dfrac{3}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
Bài 3b)
\(P=\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\)
\(P=\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\)( 1 )
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)
\(\Rightarrow\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\ge\dfrac{3\left(xy+yz+xz\right)}{2\left(xy+yz+xz\right)}=\dfrac{3}{2}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow\)\(\dfrac{x^2}{xy+xz}+\dfrac{y^2}{xy+yz}+\dfrac{z^2}{xz+yz}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\ge\dfrac{3}{2}\)
\(\Leftrightarrow P\ge\dfrac{3}{2}\)
Vậy \(P_{min}=\dfrac{3}{2}\)
Dấu " = " xảy ra khi \(a=b=c\)
Bài 2:
a) \(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}=4\)
\(\Rightarrow\left(\dfrac{x-17}{33}-1\right)+\left(\dfrac{x-21}{29}-1\right)+\left(\dfrac{x}{25}-2\right)=0\)
\(\Rightarrow\dfrac{x-50}{33}+\dfrac{x-50}{29}+\dfrac{x-50}{25}=0\)
\(\Rightarrow\left(x-50\right)\left(\dfrac{1}{33}+\dfrac{1}{29}+\dfrac{1}{25}\right)=0\)
Mà \(\dfrac{1}{33}+\dfrac{1}{29}+\dfrac{1}{25}\ne0\)
\(\Rightarrow x-50=0\)
\(\Rightarrow x=50\)
Vậy x = 50
Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)
a: \(N=\left(\dfrac{\left(1-a\right)\left(a^2+a+1\right)}{1-a}-a\right)\cdot\dfrac{a^3-a^2-a+1}{-\left(a^2-1\right)}\)
\(=\left(a^2+1\right)\cdot\dfrac{a^2\left(a-1\right)-\left(a-1\right)}{-\left(a-1\right)\left(a+1\right)}\)
\(=-\left(a^2+1\right)\cdot\dfrac{\left(a-1\right)\left(a^2-1\right)}{\left(a-1\right)\left(a+1\right)}\)
\(=-\left(a^2+1\right)\cdot\left(a-1\right)\)
b: Để N<0 thì \(-\left(a^2+1\right)\left(a-1\right)< 0\)
\(\Leftrightarrow\left(a^2+1\right)\left(a-1\right)>0\)
=>a-1>0
hay a>1
a: Để C vô nghĩa thì x+2=0
hay x=-2
Để C có nghĩa thì x+2<>0
hay x<>-2
\(C=\dfrac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{2}{x+2}\)
Để C=0 thì \(x\in\varnothing\)
Để C>0 thì x+2>0
hay x>-2
Để C<0 thì x+2<0
hay x<-2
b: \(C=\dfrac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{2}{x+2}\)