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Bài 2:
a, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}-\dfrac{3x+1}{1-x^2}\right):\dfrac{2x+1}{x^2-1}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}+\dfrac{3x+1}{x^2-1}\right).\dfrac{x^2-1}{2x+1}\)
\(P=\dfrac{\left(x-1\right)^2-x\left(x+1\right)+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{x^2-2x+1-x^2-x+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{2}{2x+1}\)
b, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P=\dfrac{3}{x-1}\Leftrightarrow\dfrac{2}{2x+1}=\dfrac{3}{x-1}\Leftrightarrow2\left(x-1\right)=3\left(2x+1\right)\)
\(\Leftrightarrow2x-2=6x+3\)\(\Leftrightarrow-4x=5\Leftrightarrow x=\dfrac{-5}{4}\)(TMĐK)
c, \(ĐKXĐ:x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P\in Z\Leftrightarrow\dfrac{2}{2x+1}\in Z\Leftrightarrow2x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
+) Với \(2x+1=1\Leftrightarrow x=0\left(TMĐK\right)\)
+) Với \(2x+1=-1\Leftrightarrow x=-1\left(KTMĐK\right)\)
+) Với \(2x+1=2\Leftrightarrow x=\dfrac{1}{2}\left(TMĐK\right)\)
+) Với \(2x+1=-2\Leftrightarrow x=\dfrac{-3}{2}\left(TMĐK\right)\)
Vậy để \(P\in Z\Leftrightarrow x\in\left\{0;\dfrac{1}{2};\dfrac{-3}{2}\right\}\)
\(A=-\dfrac{4}{x^2-4x+10}\\ =-\dfrac{4}{\left(x^2-2.x.2+4+6\right)}\\ =-\dfrac{4}{\left(x-2\right)^2+6}\)
\(\left(x-2\right)^2\ge0\\ \Rightarrow\left(x-2\right)^2+6\ge6\\ \Rightarrow\dfrac{4}{\left(x-2\right)^2+6}\le\dfrac{2}{3}\\ \Rightarrow A=-\dfrac{4}{\left(x-2\right)^2+6}\ge-\dfrac{2}{3}\)
Min A=-2/3 khi x=2
\(C=\dfrac{2}{x^2+4x+5}=\dfrac{2}{\left(x+2\right)^2+1}\)
Vì \(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2+1\ge1\)
\(\Rightarrow C\le2\)
Dấu ''='' xảy ra \(\Leftrightarrow x=-2\)
Vậy Min C = 2 kjhi x = -2
Bài 1:
a: \(P=\left(\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{4}\)
\(=\dfrac{x^2-x-2-x^2-x+2}{\left(x-1\right)\left(x+1\right)^2}\cdot\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{4}\)
\(=\dfrac{-2x}{1}\cdot\dfrac{x-1}{4}=-\dfrac{x\left(x-1\right)}{2}\)
b: Để \(\dfrac{P-4}{5}=x\) thì P-4=5x
=>P=5x+4
\(\Leftrightarrow-\dfrac{x\left(x-1\right)}{2}=5x+4\)
=>-x2+x=10x+8
=>x2-x=-10x-8
=>x2+9x+8=0
=>x=-8(nhận) hoặc x=-1(loại)
Bài 3:
Ta có:
\(81^8-1=\left(9^2\right)^8-1=\left[\left(3^2\right)^2\right]^8-1=3^{32}-1\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
Do đó:
\(A=3^4-1=80\)
1 ) \(A=\left(\dfrac{2x^3+2}{x+1}-2x\right)\left(\dfrac{x^3-1}{x-1}+x\right)\)
\(\Leftrightarrow A=\left(\dfrac{2x^3+2-2x^2-2x}{x+1}\right)\left(x^2+2x+1\right)\)
\(\Leftrightarrow A=\left(\dfrac{\left(2x^2-2\right)\left(x-1\right)}{x+1}\right)\left(x+1\right)^2\)
\(\Leftrightarrow A=\left(\dfrac{2\left(x-1\right)\left(x+1\right)\left(x-1\right)}{x+1}\right)\left(x+1\right)^2\)
\(\Leftrightarrow A=2\left(x-1\right)^2\left(x+1\right)^2\ge0\forall x\)
ta có:
A = \(\left(\dfrac{x+3}{2x+2}+\dfrac{3}{1-x^2}-\dfrac{x+1}{2x-2}\right):\dfrac{3}{2x^2-2}\)
= \(\left(\dfrac{x+3}{2\left(x+1\right)}-\dfrac{3}{x^2-1}-\dfrac{x+1}{2\left(x-1\right)}\right):\dfrac{3}{2\left(x^2-1\right)}\)
= \(\left(\dfrac{x+3}{2\left(x+1\right)}-\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{2\left(x-1\right)}\right):\dfrac{3}{2\left(x-1\right)\left(x+1\right)}\)
= \(\left(\dfrac{\left(x+3\right)\left(x-1\right)}{2\left(x+1\right)\left(x-1\right)}-\dfrac{6}{2\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}\right):\dfrac{3}{2\left(x-1\right)\left(x+1\right)}\)
= \(\left(\dfrac{x^2-x+3x-3-6-x^2-2x-1}{2\left(x+1\right)\left(x-1\right)}\right):\dfrac{3}{2\left(x-1\right)\left(x+1\right)}\)
= \(-\dfrac{10}{2\left(x+1\right)\left(x-1\right)}.\dfrac{2\left(x+1\right)\left(x-1\right)}{3}\)
= \(-\dfrac{10}{3}\)
Vậy phương trình trên ko phụ thuộc vào biến
Thanks bn