Bài 2:

1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)

=>(2x-1)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

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

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

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

=>(2x-3)(2x-3-2)=0

=>(2x-3)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

=>\(x\left(2x-5\right)=0\)

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

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

\(=3xy\left(x^2y-2\right)\)

2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)

\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)

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

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)

\(=(x-2y)(3x-1+5x)\)

\(=\left(x-2y\right)\left(8x-1\right)\)

4: \(x^2-y^2-6y-9\)

\(=x^2-\left(y^2+6y+9\right)\)

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

\(=\left(x-y-3\right)\left(x+y+3\right)\)

5: \(\left(3x-y\right)^2-4y^2\)

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

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

\(=\left(3x-3y\right)\left(3x+y\right)\)

\(=3\left(x-y\right)\left(3x+y\right)\)

6: \(4x^2-9y^2-4x+1\)

\(=\left(4x^2-4x+1\right)-9y^2\)

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

\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)

8: \(x^2y-xy^2-2x+2y\)

\(=xy\left(x-y\right)-2\left(x-y\right)\)

\(=\left(x-y\right)\left(xy-2\right)\)

9: \(x^2-y^2-2x+2y\)

\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)

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

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

Vì \(\dfrac{1}{2}\ne\dfrac{-2}{3}\)

nên hệ luôn có nghiệm duy nhất

a: \(\left\{{}\begin{matrix}x-2y=-3m-4\\2x+3y=8m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-4y=-6m-8\\2x+3y=8m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-4y-2x-3y=-6m-8-8m+1\\2x+3y=8m-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-7y=-14m-7\\2x=8m-1-3y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2m+1\\2x=8m-1-6m-3=2m-4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2m+1\\x=m-2\end{matrix}\right.\)

Đặt \(A=y^2+3x-1\)

\(=\left(2m+1\right)^2+3\left(m-2\right)-1\)

\(=4m^2+4m+1+3m-6-1\)

\(=4m^2+7m-6\)

\(=4\left(m^2+\dfrac{7}{4}m-\dfrac{3}{2}\right)\)

\(=4\left(m^2+2\cdot m\cdot\dfrac{7}{8}+\dfrac{49}{64}-\dfrac{145}{64}\right)\)

\(=4\left(m+\dfrac{7}{8}\right)^2-\dfrac{145}{16}>=-\dfrac{145}{16}\)
Dấu '=' xảy ra khi m=-7/8

b: Đặt B=x^2-y^2

\(=\left(m-2\right)^2-\left(2m+1\right)^2\)

\(=m^2-4m+4-4m^2-4m-1\)

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

\(=-3\left(m^2+\dfrac{8}{3}m-1\right)\)

\(=-3\left(m^2+2\cdot m\cdot\dfrac{4}{3}+\dfrac{16}{9}-\dfrac{25}{9}\right)\)

\(=-3\left(m+\dfrac{4}{3}\right)^2+\dfrac{25}{3}< =\dfrac{25}{3}\)

Dấu '=' xảy ra khi m=-4/3

\(1,\left(2x+1\right)^2+2\left(2x+1\right)+1\\ =\left(2x+1\right)^2+2.\left(2x+1\right).1+1^2\\ =\left[\left(2x+1\right)+1\right]^2\\ b,\left(3x-2y\right)^2+4\left(3x-2y\right)+4\\ =\left(3x-2y\right)^2+2.\left(3x-2y\right).2+2^2\\ =\left[\left(3x-2y\right)+2\right]^2\)

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

\(=\left(2x+1\right)^2+2\left(2x+1\right)\cdot1+1^2\)

\(=\left[\left(2x+1\right)+1\right]^2\)

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

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

\(=\left(3x+2y\right)^2+2\cdot\left(3x+2y\right)\cdot2+2^2\)

\(=\left[\left(3x+2y\right)+2\right]^2\)

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

c: \(=\dfrac{3x\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}=\dfrac{3x}{x^2+1}\)

C gửi nha tích giúp c 

\(\left(x^2-3\right)\left(x^4+3x^2+9\right)=x^6-27\)

a: \(=3y^2-5x^2y^3-2y^2+3x^2y^3=y^2-2x^2y^3\)

b: \(=6x-y+2x^2+3y^2-2x^2+x=7x-y+3y^2\)

c: \(=x-y+4y^2-6xy+\dfrac{10x^2}{y}\)

\(a.\left(9x^2y^3-15x^4y^4\right):3x^2y-\left(2-3x^2y\right)y^2\)

\(=3y^2-5x^2y^3-2y^2+3x^2y^3\)

\(=y^2-2x^2y^3\)

\(b.\left(6x^2-xy\right):x+\left(2x^3y+3xy^2\right):xy-\left(2x-1\right)x\)

\(=6x-y+2x^2+3y-2+x\)

\(=2x^2+7x+2y-2\)

\(c.\left(x^2-xy\right):x+\left(6x^2y^5-9x^3y^4+15x^4y^3\right):\dfrac{3}{2}x^2y^3\)

\(=x-y+4y^2-6xy+10x^2\)

Cho \(0< x< y\le z\le1\) và \(3x+2y+z\le4\). Tìm Max \(S=3x^2+2y^2+z^2\) - Hoc24

Tham khảo

Khai triển Abel ta có:

\(S=\left(z-y\right)z+\left(y-x\right)\left(z+2y\right)+x\left(3x+2y+z\right)\)

\(\le\left(z-y\right).1+\left(y-x\right).3+4x=x+2y+z\)

\(=\left(1-1\right)z+\left(1-\dfrac{1}{3}\right)\left(2y+z\right)+\dfrac{1}{3}\left(3x+2y+z\right)\)

\(\le\dfrac{2}{3}.3+\dfrac{1}{3}.4=\dfrac{10}{3}\)

Dấu = xảy ra khi \(x=\dfrac{1}{3},y=z=1\)