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Rút gọn M
\(M=\left(\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\right)\left(x^4+\frac{1-x^4}{1+x^2}\right)\)
\(=\frac{x^4-1-x^4+x^2-1}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\cdot\left(x^4+\frac{\left(1-x^2\right)\left(1+x^2\right)}{1+x^2}\right)\)
\(=\frac{x^2-2}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\cdot\left(x^4-x^2+1\right)\)
\(=\frac{x^2-2}{x^2+1}\)
\(M_{min}\Leftrightarrow\frac{x^2-2}{x^2+1}\) có giá trị nhỏ nhất
Biến đổi:\(M=\frac{x^2-2}{x^2+1}=\frac{x^2+1-3}{x^2+1}=1-\frac{3}{x^2+1}\)
M có giá trị nhỏ nhất khi \(\frac{3}{x^2+1}\) có giá trị lớn nhất
\(\Rightarrow x^2+1\) có giá trị nhỏ nhất
Mà \(x^2\ge0\Rightarrow x^2+1\ge1\) dấu "=" xảy ra tại x=0
Vậy.........................................
a
\(ĐKXĐ:x\in R\)
\(A=\left(\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\right)\left(x^4+\frac{1-x^4}{1+x^2}\right)\)
\(A=\left(\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\right)\left(x^4-x^2+1\right)\)
\(=\frac{\left(x^2-1\right)\left(x^4-x^2+1\right)}{x^4-x^2+1}-\frac{x^4-x^2+1}{x^2+1}\)
\(=x^2-1-\frac{x^4-x^2+1}{x^2+1}\)
\(=-1+\frac{x^4+x^2-x^4+x^2+1}{x^2+1}\)
\(=\frac{2x^2+1}{x^2+1}-1=\frac{2x^2+1-x^2-1}{x^2+1}=\frac{x^2}{x^2+1}\)
b
Xét \(x>0\Rightarrow M>0\)
Xét \(x=0\Rightarrow M=0\)
Xét \(x< 0\Rightarrow M>0\)
Vậy \(M_{min}=0\) tại \(x=0\)
a, \(A=\left(\frac{4}{2x+1}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4\left(x^2+1\right)}{\left(2x+1\right)\left(x^2+1\right)}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4x^2+4+4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\frac{\left(2x+1\right)^2}{\left(x^2+1\right)\left(2x+1\right)}\frac{x^2+1}{x^2+2}=\frac{2x+1}{x^2+2}\)
a) \(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)(với \(x\ne\pm2;x\ne-1\))
\(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{-\left(6-5x\right)}{x^2-4}\right):\frac{x+1}{x-2}\)
\(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\right):\frac{x+1}{x-2}\)
\(M=\left(\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{5x-6}{\left(x+2\right)\left(x-2\right)}\right):\frac{x+1}{x-2}\)
\(M=\frac{4\left(x-2\right)+2\left(x+2\right)-5x+6}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)
\(M=\frac{4x-8+2x+4-5x+6}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)
\(M=\frac{x+2}{\left(x+2\right)\left(x-2\right)}:\frac{x+1}{x-2}\)
\(M=\frac{1}{x-2}:\frac{x+1}{x-2}=\frac{1}{x-2}\cdot\frac{x-2}{x+1}=\frac{1}{x+1}\)
b) Với \(M=\frac{1}{4}\)ta có :
\(M=\frac{1}{x+1}\Rightarrow\frac{1}{4}=\frac{1}{x+1}\)
\(\Rightarrow1\left(x+1\right)=4\Rightarrow x+1=4\Rightarrow x=3\)
Vậy x = 3
a, \(M=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{4-x^2}\right):\frac{x+1}{x-2}\)
\(=\left(\frac{4}{x+2}+\frac{2}{x-2}-\frac{6-5x}{\left(2-x\right)\left(x+2\right)}\right):\frac{x+1}{x-2}\)
\(=\left(\frac{4\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x-2}\)
\(=\frac{4x-8+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}:\frac{x+1}{x-2}\)
\(=\frac{x+2}{\left(x-2\right)\left(x+2\right)}:\frac{x+1}{x-2}=\frac{1}{x-2}.\frac{x-2}{x+1}=\frac{1}{x+1}\)
b, Ta có : M = 1/4 hay \(\frac{1}{x+1}=\frac{1}{4}\Leftrightarrow4=x+1\Leftrightarrow x=3\)
a.
\(ĐKXĐ:x\ne\pm1;\)
Ta có:
\(P=\left(\frac{x^4+x^2-4x+1}{x^2-1}-\frac{x-1}{x+1}+\frac{x+1}{x-1}\right)\cdot\frac{x\left(x+1\right)-\left(1+x\right)}{x^3-1}\)
\(\Rightarrow P=\left(\frac{x^4+x^2-4x+1}{x^2-1}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{\left(x+1\right)\left(x-1\right)}{x^3-1}\)
\(\Rightarrow P=\left(\frac{x^4+x^2-4x+1}{x^2-1}-\frac{x^2-2x+1}{x^2-1}+\frac{x^2+2x+1}{x^2-1}\right)\cdot\frac{x^2-1}{x^3-1}\)
\(\Rightarrow P=\frac{x^4+x^2+1}{x^2-1}\cdot\frac{x^2-1}{x^3-1}\)
\(\Rightarrow P=\frac{x^4+x^2+1}{x^3-1}\)
b.
Để P là số nguyên thì \(x^4+x^2+1⋮x^3-1\)
\(\Rightarrow\left(x^4-x\right)+\left(x^2+x+1\right)⋮\left(x-1\right)\left(x^2+x+1\right)\)
\(\Rightarrow x\left(x^3-1\right)+\left(x^2+x+1\right)⋮\left(x-1\right)\left(x^2+x+1\right)\)
\(\Rightarrow x\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)⋮\left(x-1\right)\left(x^2+x+1\right)\)
\(\Rightarrow\left(x^2+x+1\right)\left(x^2-x+1\right)⋮\left(x-1\right)\left(x^2+x+1\right)\)
\(\Rightarrow x^2-x+1⋮x-1\)
\(\Rightarrow x\left(x-1\right)+1⋮x-1\)
\(\Rightarrow1⋮x-1\)
\(\Rightarrow x-1\in\left\{1;-1\right\}\)
\(\Rightarrow x=1\left(KTMĐK\right);x=0\)
Vậy x=0.
P/S:Không chắc chắn lắm đâu nha mn,nếu có j sai thì ib vs em ah.