Ta có: \(A=x^2+5y^2-2xy+4y+3\)

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

\(=\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\)

Dấu " = " khi \(\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(2y+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow x=y=\dfrac{-1}{2}\)

Vậy \(MIN_A=2\) khi \(x=y=\dfrac{-1}{2}\)

Ta có: \(\dfrac{n^2+2n+1}{n+23}\in Z\Rightarrow n^2+2n+1⋮n+23\)

\(\Rightarrow n^2+23n-\left(21n-1\right)⋮n+23\)

\(\Rightarrow n\left(n+23\right)-\left(21n-1\right)⋮n+23\)

\(\Rightarrow21n-1⋮n+23\)

\(\Rightarrow21n+483-484⋮n+23\)

\(\Rightarrow21\left(n+23\right)-484⋮n+23\)

\(\Rightarrow484⋮n+23\)

Để n lớn nhất thì n + 23 = 484

\(\Rightarrow n=461\)

Vậy n = 461

a, \(x^3-1+5x^2-5+3x-3\)

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

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

\(=\left(x-1\right)\left[x^2+x+1+5x+5+3\right]\)

\(=\left(x-1\right)\left(x^2+6x+9\right)\)

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

b, \(a^5+a^4+a^3+a^2+a+1\)

\(=a^3\left(a^2+a+1\right)+a^2+a+1\)

\(=\left(a^3+1\right)\left(a^2+a+1\right)\)

c, \(ab\left(x^2+y^2\right)+xy\left(a^2+b^2\right)\)

\(=abx^2+aby^2+xya^2+xyb^2\)

\(=ax\left(bx+ay\right)+by\left(ay+bx\right)\)

\(=\left(ax+by\right)\left(ay+bx\right)\)

Turn into passive voise .

1. We have told him not to be late again .

=> He have been told not to be late again.

2. They paid me a lot of money to do the job .

=> I was paid a lot of money to do the job.

3. About thirty milion people are watching this new programme.

=> This new programme is watched by a about thirty million people.

4. They should help Jane with the sewing

=> Jane should be helped with the sewing.

5. The students will need the registration in order to take advantage of the good studying materials .

=> The registration will be need in order to take advantage of the good studying materials .

a, \(x^2-x-30=0\)

\(\Leftrightarrow x^2-6x+5x-30=0\)

\(\Leftrightarrow x\left(x-6\right)+5\left(x-6\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)

Vậy x = -5 hoặc x = 6

a, \(E=-4x^2+4x-3\)

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

\(=-\left[\left(2x+1\right)^2-4\right]=-\left(2x+1\right)^2+4\le4\)

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

Vậy \(MAX_E=4\) khi \(x=\dfrac{-1}{2}\)

b, \(F=13-2x^2+4y+4xy-3y^2\)

\(=17-\left(2x^2-4xy+2y^2\right)-\left(y^2-4y+4\right)\)

\(=17-2\left(x-y\right)^2-\left(y-2\right)^2\le17\)

Dấu " = " khi \(\left\{{}\begin{matrix}2\left(x-y\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=2\end{matrix}\right.\Leftrightarrow x=y=2\)

Vậy \(MAX_F=17\) khi x = y = 2

\(\dfrac{x+4}{2016}+\dfrac{x+2}{2018}\ge\dfrac{x+14}{2006}+\dfrac{x+83}{1937}\)

\(\Leftrightarrow\dfrac{x+4}{2016}+1+\dfrac{x+2}{2018}+1\ge\dfrac{x+14}{2006}+1+\dfrac{x+83}{1937}+1\)

\(\Leftrightarrow\dfrac{x+2020}{2016}+\dfrac{x+2020}{2018}-\dfrac{x+2020}{2006}-\dfrac{x+2020}{1937}\ge0\)

\(\Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{2016}+\dfrac{1}{2018}-\dfrac{1}{2006}-\dfrac{1}{1937}\right)\ge0\)

\(\Leftrightarrow x+2020\ge0\Leftrightarrow x\ge-2020\)

Vậy \(x\ge-2020\)

a, \(x^2-9=\left(x-3\right)\left(x+3\right)\)

b, \(4x^2-25=\left(2x-5\right)\left(2x+5\right)\)

c, \(x^6-y^6=\left(x^3-y^3\right)\left(x^3+y^3\right)\)

Câu hỏi của NT Ánh - Toán lớp 9 | Học trực tuyến

Sorry bạn.Mình không biết làm.

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