a, \(14x^2-21x^2+28x^2y^2\)

\(=-7x^2+28x^2y^2=-7x^2\left(1-4y^2\right)\)

\(=-7x\left(1-2y\right)\left(1+2y\right)\)

b, \(8x^3+4x^2-y^3-y^2\)

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

\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)+\left(2x-y\right)\left(2x+y\right)\)

\(=\left(2x-y\right)\left(4x^2+2xy+y^2+2x+y\right)\)

\(x^3+6x^2+11x+6\)

\(=x^3+3x^2+3x^2+9x+2x+6\)

\(=x^2\left(x+3\right)+3x\left(x+3\right)+2\left(x+3\right)\)

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

\(\dfrac{x+2}{x+3}-\dfrac{1}{x-2}-\dfrac{5}{x^2+x-6}\)

\(=\dfrac{\left(x+2\right)\left(x-2\right)-x-3}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{x^2+x-6}\)

\(=\dfrac{x^2-4-x-3}{x^2+x-6}-\dfrac{5}{x^2+x-6}\)

\(=\dfrac{x^2-x-7-5}{x^2+x-6}=\dfrac{x^2-x-12}{x^2+x-6}\)

Vậy...

Sửa đề: \(\dfrac{100+\dfrac{99}{2}+\dfrac{98}{3}+...+\dfrac{1}{100}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{101}}-2\)

\(=\dfrac{\left(\dfrac{99}{2}+1\right)+\left(\dfrac{98}{3}+1\right)+...+\left(\dfrac{1}{100}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{101}}-2\)

\(=\dfrac{\dfrac{101}{2}+\dfrac{101}{3}+...+\dfrac{101}{100}+\dfrac{101}{101}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}+\dfrac{1}{101}}-2\)

\(=\dfrac{101\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}+\dfrac{1}{101}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}+\dfrac{1}{101}}-2\)

\(=101-2=99\)

Vậy...

b, Câu hỏi của Cry... - Toán lớp 8 | Học trực tuyến

a, \(9x^2+y^2+2z^2-18x-6y+4z+20=0\)

\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\)

\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)

Vì \(\left\{{}\begin{matrix}9\left(x-1\right)^2\ge0\\\left(y-3\right)^2\ge0\\2\left(z+1\right)^2\ge0\end{matrix}\right.\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)

Mà \(9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\\2\left(z+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

Vậy...

Ta có: \(\left(a+b+c\right)^2=9^2=81\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=81\)

\(\Leftrightarrow A=2\left(ab+bc+ac\right)=28\)

Vậy A = 28

a, \(\sqrt{\left(x-3\right)^2}=9\)

\(\Leftrightarrow\left|x-3\right|=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)

Vậy...

b, \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy...

Ta có:

  264-24=240

 363-43=320

Do đó ƯC(240;320) ;a>3

=>ƯCLN(240;320)=80 ƯC>43=80.Vậy a=80