ai ủng hộ vài li-ke lên 90 điểm hỏi đáp đi

Trên tia Ox ta có: \(OA=3cm\) ; \(OB=7cm\)

Vì \(3cm< 7cm\) nên \(OA< OB\) => A nằm giữa O và B

Ta có A nằm giữa O và B nên:

\(OA+AB=OB\)

\(\Rightarrow AB=OB-OA=7-3=4\left(cm\right)\)

Vì M là trung điểm của AB

\(\Rightarrow AM=MB=\dfrac{AB}{2}=\dfrac{4}{2}=2\left(cm\right)\)

Vậy \(BM=2cm\)

1)

a) \(\left(x-2\right)\left(x^2+3x+4\right)\)

\(\Leftrightarrow x^3+3x^2+4x-2x^2-6x-8\)

\(\Leftrightarrow x^3+x^2-2x-8\)

b) \(\left(x-2\right)\left(x-x^2+4\right)\)

\(=x^2-x^3+4x-2x+2x^2-8\)

\(=3x^2-x^3+2x-8\)

c) \(\left(x^2-1\right)\left(x^2+2x\right)\)

\(=x^4+2x^3-x^2-2x\)

d) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)

\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)

\(=18x^2+12x-9x-6-6x^3-4x^2+3x^2+2x\)

\(=17x^2+5x-6-6x^3\)

2)

a) \(3x^3-3x=0\)

\(\Leftrightarrow3x\left(x^2-1\right)=0\)

\(\Leftrightarrow3x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-1=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

Vậy x=0 ; x=-1 ; x=1

b) \(x^2-x+\dfrac{1}{4}=0\)

\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)

\(\Leftrightarrow x-\dfrac{1}{2}=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

Gọi x là cân nặng của \(7m^3\) sắt

Đổi \(14dm^3=0,014m^3\)

Vì số khối sắt và cân nặng là hai đại lượng tỉ lệ thuận nên ta có.

\(\dfrac{0,014}{109,2}=\dfrac{7}{x}\)

\(\Rightarrow x=\dfrac{109,2.7}{0,014}=\dfrac{764,4}{0,014}=54600\)

Vậy \(7m^3\) sắt nặng 54600 kg

\(1+2+3+4+5+.....+100+101\)

\(=\dfrac{\left(1+101\right).\left[\left(101-1\right).1+1\right]}{2}\)

\(=\dfrac{102.\left(100+1\right)}{2}\)

\(=\dfrac{102.101}{2}\)

\(=\dfrac{102.101}{2}\)

\(=\dfrac{10302}{2}\)

\(=5151\)

\(k\left(x\right)=\dfrac{5x^2-22x+25}{x^2-4x+4}\)

\(\Leftrightarrow k\left(x\right)=\dfrac{5x^2-20x+20-x+2-x+2+1}{x^2-4x+4}\)

\(\Leftrightarrow k\left(x\right)=\dfrac{\left(5x^2-20x+20\right)-\left(x-2\right)-\left(x-2\right)+1}{x^2-4x+4}\)

\(\Leftrightarrow k\left(x\right)=\dfrac{5\left(x^2-4x+4\right)-\left(x-2\right)-\left(x-2\right)+1}{x^2-4x+4}\)

\(\Leftrightarrow k\left(x\right)=\dfrac{5\left(x-2\right)^2-\left(x-2\right)-\left(x-2\right)+1}{\left(x-2\right)^2}\)

\(\Leftrightarrow k\left(x\right)=\dfrac{5\left(x-2\right)^2}{\left(x-2\right)^2}-\dfrac{x-2}{\left(x-2\right)^2}-\dfrac{x-2}{\left(x-2\right)^2}+\dfrac{1}{\left(x-2\right)^2}\)

\(\Leftrightarrow k\left(x\right)=5-\dfrac{1}{x-2}-\dfrac{1}{x-2}+\dfrac{1}{\left(x-2\right)^2}\)

Đặt \(y=\dfrac{1}{x-2}\)

\(\Rightarrow k\left(x\right)=5-y-y+y^2=y^2-2y+1+4=\left(y-1\right)^2+4\ge4\)

Vậy GTNN của \(k\left(x\right)=4\) khi \(y=1\Rightarrow\dfrac{1}{x-2}=1\Leftrightarrow x=3\)

\(h\left(x\right)=\dfrac{x^2-x+1}{\left(x-1\right)^2}\)

\(\Leftrightarrow h\left(x\right)=\dfrac{x^2-2x+1+x-1+1}{\left(x-1\right)^2}\)

\(\Leftrightarrow h\left(x\right)=\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2}+\dfrac{x-1}{\left(x-1\right)^2}+\dfrac{1}{\left(x-1\right)^2}\)

\(\Leftrightarrow h\left(x\right)=1+\dfrac{1}{x-1}+\dfrac{1}{\left(x-1\right)^2}\)

Đặt \(y=\dfrac{1}{x-1}\)

\(\Rightarrow h\left(x\right)=1+y+y^2\)

\(\Rightarrow h\left(x\right)=y^2+y+1\)

\(\Rightarrow h\left(x\right)=y^2+2.y.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(\Rightarrow h\left(x\right)=\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

=> GTNN của \(h\left(x\right)=\dfrac{3}{4}\) khi \(y+\dfrac{1}{2}=0\Leftrightarrow y=\dfrac{-1}{2}\)

\(\Leftrightarrow\dfrac{1}{x-1}=\dfrac{-1}{2}\)

\(\Leftrightarrow x=-1\)