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\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{\left(x-1\right)^2}{x^2-1}\right).\frac{x+2003}{x}\)ĐKXĐ: \(x\ne-1;0;1\)
\(A=\frac{\left(x+1\right)^2-\left(x-1\right)^2+\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}.\frac{x+2003}{x}\)
\(A=\frac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}.\frac{x+2003}{x}\)
\(A=\frac{x+1}{x-1}.\frac{x+2003}{x}\)
\(A=\frac{x^2+2004x+2003}{x^2-x}\)
a)ĐKXĐ:
\(x-1\ne0;x+1\ne0;x\ne0\)
\(\Leftrightarrow x\ne1;x\ne-1;x\ne0\)
b)\(K=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{x^2-1}\right).\frac{x+2003}{x}\)
\(=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{x+2003}{x}\)
\(=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}+\frac{x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{x+2003}{x}\)
\(=\frac{x^2+2x+1+x^2-2x+1+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)
\(=\frac{3x^2-4x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)
\(=\frac{3x^2-3x-x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)
\(=\frac{3x.\left(x-1\right)-\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)
\(=\frac{\left(x-1\right)\left(3x-1\right)}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)
\(=\frac{\left(3x-1\right)\left(x+2003\right)}{\left(x+1\right).x}\)
\(=\frac{3x^2+6008x-2003}{x^2+x}\)
câu c bí
a, dk \(x\ge0.x\ne1\)
\(\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{2\left(1-x\right)}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)=\(\left(\frac{1}{1-x}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)
=\(\left(\frac{1+x-x^2-1}{1-x^2}\right)\left(\frac{x+1}{x}\right)=\frac{x\left(1-x\right)\left(x+1\right)}{x\left(1-x\right)\left(1+x\right)}=1\)
phan b,c ban tu lam not nhe dai lam mk ko lam dau mk co vc ban rui
Ta có: \(A=\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right)^2\times\frac{x^2-1}{2}-\sqrt{1-x^2}\) \(\left(ĐK:x\ge1\right)\)
\(\Leftrightarrow A=\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x-1}.\sqrt{x+1}}\right)^2\times\frac{x^2-1}{2}-\sqrt{1-x^2}\)
\(\Leftrightarrow A=\frac{x+1+x-1+2\sqrt{x^2-1}}{x^2-1}\times\frac{x^2-1}{2}-\sqrt{1-x^2}\)
\(\Leftrightarrow A=\frac{2x+2\sqrt{x^2-1}}{2}-\sqrt{1-x^2}\)
\(\Leftrightarrow A=x+\sqrt{1-x^2}-\sqrt{1-x^2}\)
\(\Leftrightarrow A=x\)
Học tốt
ĐKXĐ : ...............
\(A=\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right)^2\times\frac{x^2-1}{2}-\sqrt{1-x^2}\)
\(A=\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{\left(x-1\right)\left(x+1\right)}}\right)^2\times\frac{x^2-1}{2}-\sqrt{1-x^2}\)
\(A=\frac{\left(\sqrt{x+1}+\sqrt{x-1}\right)^2}{x^2-1}\times\frac{x^2-1}{2}-\sqrt{1-x^2}\)
\(A=\frac{x+1+2\sqrt{x^2-1}+x-1}{2}-\sqrt{1-x^2}\)
\(A=\frac{2x+2\sqrt{x^2-1}-2\sqrt{1-x^2}}{2}\)
\(A=\frac{2x+2\sqrt{x^2-1}+2\sqrt{x^2-1}}{2}\)
\(A=\frac{2x+4\sqrt{x^2-1}}{2}\)
\(A=x+2\sqrt{x^2-1}\)
Bài 1 : Điều kiện xác định : \(x\ne\pm1\)
\(K=\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{x^2-1}{x^2}\)
\(K=\frac{2}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{x^2}=\frac{2}{x^2}\)
Nhận thấy giá trị của x càng tăng thì giá trị của M càng giảm
mặt khác , giá trị của x lại không giảm quá 0 nên ta không thể nào xác định được giá trị lớn nhất của K
