HOC24
Lớp học
Môn học
Chủ đề / Chương
Bài học
\(-x^2+2x-6\le5\)
\(\Leftrightarrow\) \(x^2-2x+6\ge5\)
VT \(=\)\(x^2-2x+6\)
\(=x^2-2x+1+5\)
\(=\left(x-1\right)^2+5\)
Với \(\left(x-1\right)^2\ge0\) nên \(\left(x-1\right)^2+5\) \(\ge5\) hay \(x^2-2x+6\ge5\)
Vậy \(-x^2+2x-6\le-5\)
\(-6\left(x+7\right)+\left(x-5\right)=-52\)
\(\Leftrightarrow-6x-42+x-5+52=0\)
\(\Leftrightarrow-5x+5=0\)
\(\Leftrightarrow-5x=-5\)
\(\Rightarrow x=1\)
A= ( 2x+3)(x-1) - (x+1)(2x-5) -2
= \(2x^2-2x+3x-3-\left(2x^2-5x+2x-5\right)-2\)
= \(2x^2-2x+3x-3-2x^2+5x-2x+5-2\)
= \(4x\)
B= \(\left(x-4\right)\left(x-2\right)-\left(3x+1\right)\left(\frac{1}{3}x-2\right)+2\frac{1}{3}x-10\)
= \(x^2-2x-4x+8-\left(x^2-6x+\frac{1}{3}x-2\right)+\frac{7}{3}x-10\)
= \(x^2-2x-4x+8-x^2+6x-\frac{1}{3}x+2+\frac{7}{3}x-10\)
= \(2x\)
Ta được: \(\frac{A}{B}=\frac{4x}{2x}=2\)
.The children / laugh / talk / noisily / when teacher is / explain / lesson.
\(\Rightarrow\) The children are laughing soisily when teacher is explaining the lesson.
They/now/have breakfast/school canteen.
\(\Rightarrow\) They are having breakfast in the school canteen now.
C3H8