Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(B=\frac{x}{x+1}+\frac{2x-3}{x-1}-\frac{2x^2-x-3}{x^2-1}\)
\(B=\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{\left(2x-3\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2x^2-x-3}{\left(x-1\right)\left(x+1\right)}\)
\(B=\frac{\left(x^2-x\right)+\left(2x^2+2x-3x-3\right)-\left(2x^2-x-3\right)}{\left(x+1\right)\left(x-1\right)}\)
\(B=\frac{x^2-x+2x^2-x-3-2x^2+x+3}{\left(x+1\right)\left(x-1\right)}\)
\(B=\frac{x^2-x}{\left(x+1\right)\left(x-1\right)}\)
\(B=\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(B=\frac{x}{x+1}\)
MÌnh nghĩ đề câu b là với x>-4 mới đúng chứ
\(B=\frac{x}{x+1}+\frac{2x-3}{x-1}-\frac{2x^2-x-3}{\left(x^2-1\right)}.\)
\(=\frac{x\left(x-1\right)+\left(2x-3\right)\left(x+1\right)-2x^2+x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2-x+2x^2-x-3-2x^2+x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2-x}{\left(x-1\right)\left(x+1\right)}=\frac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x}{x+1}\)
\(\Rightarrow A.B=\frac{x}{\left(x+1\right)}.\frac{x\left(x+1\right)}{\left(x-2\right)}=\frac{x^2}{\left(x-2\right)}=\frac{x^2-4+4}{\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x+2\right)+4}{\left(x-2\right)}=x+2+\frac{4}{x-2}=x-2+\frac{4}{x-2}+4\)
Áp dụng BĐT Cô - Si cho 2 số dương \(x-2;\frac{4}{x-2}\)ta có :
\(x-2+\frac{4}{x-2}\ge2\sqrt{\frac{\left(x-2\right).4}{x-2}}=2\sqrt{4}=4\)
\(\Rightarrow x-2+\frac{4}{x-2}\ge4\Rightarrow x-2+\frac{4}{x-2}+4\ge8\)
Hay \(S_{min}=4\Leftrightarrow x-2=\frac{4}{x-2}\)
\(\Rightarrow\frac{\left(x-2\right)^2}{\left(x-2\right)}=\frac{4}{x-2}\Rightarrow x^2+4x+4=4\)
\(\Rightarrow x^2+4x=0\Rightarrow x\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=-4\left(ktm\right)\end{cases}}\)\(\Rightarrow...\)
\(B=\frac{2\left(x^2-4x+5\right)}{\left(x+1\right)^2\left(x-3\right)}\)
ĐKXĐ: \(x\ne-1;3\)
\(A=\frac{\left(x+1\right)^2}{\left(x-2\right)^2+1}\ge0\) do \(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\\\left(x-2\right)^2+1>0\end{matrix}\right.\) \(\forall x\)
\(\Rightarrow A_{min}=0\) khi \(x=-1\)
Để AB>0 \(\Leftrightarrow\left\{{}\begin{matrix}A\ne0\\B>0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ne-1\\\frac{2\left[\left(x-2\right)^2+1\right]}{\left(x+1\right)^2\left(x-3\right)}>0\end{matrix}\right.\) \(\Rightarrow x-3>0\Rightarrow x>3\)
\(C=\frac{x^2+5x+8}{x^2+2x+1}=\frac{x^2+2x+1+3x+3+4}{x^2+2x+1}\)
\(=\frac{\left(x+1\right)^2+3\left(x+1\right)+4}{\left(x+1\right)^2}=1+\frac{3}{x+1}+\frac{4}{\left(x+1\right)^2}\)
Đặt \(\frac{1}{x+1}=a\)\(\Rightarrow C=1+3a+4a^2\)
\(\Rightarrow C=4\left(a^2+\frac{3}{4}a+\frac{1}{4}\right)=4\left(a^2+2.\frac{3}{8}+\frac{9}{64}-\frac{9}{64}+\frac{1}{4}\right)\)
\(=4\left(a+\frac{3}{8}\right)^2+\frac{7}{16}\)
\(\Rightarrow C_{min}=\frac{7}{16}\Leftrightarrow\)\(a=-\frac{3}{8}\Leftrightarrow\frac{1}{x+1}=-\frac{3}{8}\)
\(\Rightarrow3\left(x+1\right)=-8\Rightarrow x=-\frac{11}{3}\)
Vậy \(C_{min}=\frac{16}{7}\Leftrightarrow x=-\frac{11}{3}\)
giải câu b trc nha
= ((x-1)^2+2009]/x^2=(x-1)^2/x^2+2009
vậy min=2009 khi x=1
https://olm.vn//hoi-dap/question/57101.html
Tham khảo đây nhá bạn
Ta có: \(A=\frac{5x^2-8x+8}{2x^2}=\frac{3x^2+2x^2-8x+8}{2x^2}=\frac{3x^2+2\left(x-2\right)^2}{2x^2}=\frac{3}{2}+\frac{\left(x-2\right)^2}{x^2}\ge\frac{3}{2}\)
Vậy GTNN của A là \(\frac{3}{2}\) khi \(\frac{\left(x-2\right)^2}{x^2}=0\Rightarrow\left(x-2\right)^2=0\Rightarrow x=2\)
GTNN là 3/2 tại x=2 mình xài casio ra vx