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a: \(A=\dfrac{2x-5+x^2-4+x^2-9}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x^2+2x-18}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x+6}{x-3}\)
b: Để A/2=x+3/x-3 là số nguyên thì \(x-3+6⋮x-3\)
=>\(x-3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
hay \(x\in\left\{4;51;6;0;9;-3\right\}\)
c: Để A=1/x-1 thì \(\dfrac{2x+6}{x-3}=\dfrac{1}{x-1}\)
=>2x^2-2x+6x-6=x-3
=>2x^2+5x-6-x+3=0
=>2x^2+4x-3=0
hay \(x=\dfrac{-2\pm\sqrt{10}}{2}\)
\(A=\left(\dfrac{1}{x-1}+\dfrac{x}{x^3-1}.\dfrac{x^2+x+1}{x+1}\right):\dfrac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x^2+x+1\right)}.\dfrac{x^2+x+1}{x+1}\right):\dfrac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}+\dfrac{x}{\left(x+1\right)\left(x-1\right)}\right).\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{2x+1}{\left(x+1\right)\left(x-1\right)}.\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{x+1}{x-1}\)
Vậy \(A=\dfrac{x+1}{x-1}\)
Giả sử tìm được \(x\in Z\) để \(A\in Z\)
\(x\in Z\Leftrightarrow\left\{{}\begin{matrix}x+1\in Z\\x-1\in Z\end{matrix}\right.\)
\(A=\dfrac{x+1}{x-1}=\dfrac{x-1+2}{x-1}=1+\dfrac{2}{x-1}\)
\(\Leftrightarrow2⋮x-1\Leftrightarrow x-1\inƯ\left(2\right)\)
Ta có các trường hợp :
+) \(x-1=1\Leftrightarrow x=2\)
+) \(x-1=2\Leftrightarrow x=3\)
+) \(x-1=-1\Leftrightarrow x=0\)
+) \(x-1=-2\Leftrightarrow x=-1\)
Vậy..
a) \(A = \frac{2x^2 - 16x+43}{x^2-8x+22}\) = \(\frac{2(x^2-8x+22)-1}{x^2-8x+22}\) = \(2 - \frac{1}{x^2-8x+22}\)
Ta có : \(x^2-8x+22 \) = \(x^2-8x+16+6 = ( x-4)^2 +6 \)
Vì \((x-4)^2 \ge 0 \) với \( \forall x\in R\) Nên \(( x-4)^2 +6 \ge 6 \)
\(\Rightarrow \) \(x^2-8x+22 \) \( \ge 6\)\(\Rightarrow \) \(\frac{1}{x^2-8x+22} \) \(\le \frac{1}{6}\) \(\Rightarrow \) - \(\frac{1}{x^2-8x+22} \) \(\ge - \frac{1}{6}\)
\(\Rightarrow \) A = \(2 - \frac{1}{x^2-8x+22}\) \( \ge 2-\frac{1}{6}\) = \(\frac{11}{6}\) Dấu "=" xảy ra khi và chỉ khi x=4
Vậy GTNN của A = \(\frac{11}{6}\) khi và chỉ khi x=4
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
a) \(A=\left(\dfrac{1}{3}+\dfrac{3}{x^2-3x}\right):\left(\dfrac{x^2}{27-3x^2}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3\left(x^2-3x\right)}:\left(\dfrac{x^2}{3\left(9-x^2\right)}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\left(\dfrac{x^2}{3.\left(3-x\right).\left(3+x\right)}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\dfrac{x^2+3.\left(3-x\right)}{3.\left(3-x\right).\left(3+x\right)}\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\dfrac{x^2+9-3x}{3.\left(3-x\right).\left(3+x\right)}\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}.\dfrac{3.\left(3x-x\right).\left(3+x\right)}{x^2+9-3x}\)
\(\Rightarrow A=\dfrac{1}{x.\left(x-3\right)}.\left(-\left(x-3\right)\right).\left(3+x\right)\)
\(\Rightarrow A=\dfrac{1}{x}.\left(-1\right).\left(3+x\right)\)
\(\Rightarrow A=-\dfrac{1}{x}.\left(3+x\right)\)
\(=\left[\dfrac{2x-3}{\left(2x-5\right)\left(2x-1\right)}-\dfrac{3}{2x-1}-\dfrac{2\left(x-4\right)}{\left(x-4\right)\left(2x-5\right)}\right].\dfrac{2x\left(2x+3\right)-\left(2x+3\right)}{-2x\left(4x-7\right)-3\left(4x-7\right)}+1\)
\(=\left[\dfrac{2x-3-6x+15-4x+2}{\left(2x-5\right)}\right].\dfrac{2\left(x+\dfrac{3}{2}\right)}{\left(-2x-3\right)\left(4x-7\right)}+1\)
\(=\dfrac{-2\left(4x-7\right)}{2x-5}.\dfrac{2\left(x+\dfrac{3}{2}\right)}{\left(-2x-3\right)\left(4x-7\right)}+1\)
\(=\dfrac{1}{2x-5}.2+1\)
\(=\dfrac{2+2x-5}{2x-5}\)
\(=\dfrac{-3+2x}{2x-5}\)
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: Sửa đề: \(A=\dfrac{x^3+2x^2+6x+8}{x+1}\)
Để A là số nguyên thì \(x^3+x^2+x^2+x+5x+5+3⋮x+1\)
\(\Leftrightarrow x+1\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{0;-2;2;-4\right\}\)
b: Để \(\dfrac{2x^2+x-2}{x-3}\) là số nguyên thì \(2x^2-6x+7x-21+19⋮x-3\)
\(\Leftrightarrow x-3\in\left\{1;-1;19;-19\right\}\)
hay \(x\in\left\{4;2;22;-16\right\}\)