\(A=\dfrac{1}{2}-\left(3x-1\right)^2\)

\(Mà\) \(\left(3x-1\right)^2\ge0\forall x\)

\(\Rightarrow A=\dfrac{1}{2}-\left(3x-1\right)^2\le\dfrac{1}{2}\forall x\)

\(Dấu\) \("="\) \(xảy \) \(ra:\)

\(\Leftrightarrow\left(3x-1\right)^2=0\)

\(\Leftrightarrow3x-1=0\)

\(\Leftrightarrow3x=1\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

\(Vậy\) \(A_{\left(max\right)}=\dfrac{1}{2}\) \(\Leftrightarrow x=\dfrac{1}{3}\forall x\)

\(B=3+\dfrac{4}{\left(x+1\right)^2+1}\)

\(Ta \) \(có:\) \(\) \(\left(x+1\right)^2\ge0\forall x\)

\(\Leftrightarrow\left(x+1\right)^2+1\ge1\forall x\)

\(\Leftrightarrow\dfrac{4}{\left(x+1\right)^2+1}\le4\forall x\)

\(\Leftrightarrow B=3+\dfrac{4}{\left(x+1\right)^2+1}\le7\forall x\)

\(Dấu \) \("="\) \(xảy \) \(ra:\)

\(\Leftrightarrow\left(x+1\right)^2=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

\(Vậy\) \(B_{\left(max\right)}=7\) \(\Leftrightarrow x=-1\forall x\)

1 giờ cả 2 người làm được:

1/4+1/5= 9/20 công việc

hai người cùng làm thì xong sau số giờ là

1: 9/20= 20/9  giờ 

Chúc thầy cô luôn vui tính

Tìm các giới hạn sau  lim x → 1 +   2 x   -   7 x   - 1

\(Ta \) \(có:\)

\(\left(4x-3\right)-\left(x-5\right)=3\left(10-x\right)\)

\(\Leftrightarrow4x-3-x+5\)

\(\Leftrightarrow\left(4x-x\right)+\left(-3+5\right)=30-3x\)

\(\Leftrightarrow3x+2=30-x\)

\(\Leftrightarrow\left(3x+2\right)+x=30\)

\(\Leftrightarrow3x+2+x=30\)

\(\Leftrightarrow\left(3x+x\right)+2=30\)

\(\Leftrightarrow4x+2=30\)

\(\Leftrightarrow4x=28\)

\(\Leftrightarrow x=7\)

\(Vậy \) \(x=7\\\)

\(Ta\) \(có\) \(nghiệm\) \(của\) \(đa\) \(thức\) \(thỏa\) \(mãn:\)

\(x^2+7x+10=0\)

\(\Leftrightarrow\left(x^2+2x\right)+\left(5x+10\right)=0\)

\(\Leftrightarrow x\left(x+2\right)+5\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-2\end{matrix}\right.\)

\(Vậy\) \(nghiệm\) \(của\) \(đa\) \(thức\) \(là\) \(x=-5\) \(hoặc\) \(x=-2\)