a chia 5 dư 4 thì a có dạng 5k + 4

\(\Rightarrow a^2=\left(5k+4\right)^2=25k^2+40k+16\)

\(=5\left(5k^2+8k+3\right)+1\)

Vậy a² chia 5 dư 1

\(ab\left(b-a\right)-bc\left(b-c\right)-ac\left(c-a\right)\)

\(=ab\left(b-a\right)-\left(b^2c-bc^2\right)-\left(ac^2-a^2c\right)\)

\(=ab\left(b-a\right)-b^2c+bc^2-ac^2+a^2c\)

\(=ab\left(b-a\right)-\left(b^2c-a^2c\right)+\left(bc^2-ac^2\right)\)

\(=ab\left(b-a\right)-c\left(b^2-a^2\right)+c^2\left(b-a\right)\)

\(=ab\left(b-a\right)-c\left(b-a\right)\left(b+a\right)+c^2\left(b-a\right)\)

\(=\left(b-a\right)\left[ab-c\left(b+a\right)+c^2\right]=\left(b-a\right)\left[ab-\left(bc+ac\right)+c^2\right]\)

\(=\left(b-a\right)\left(ab-bc-ac+c^2\right)=\left(b-a\right)\left[\left(ab-bc\right)-\left(ac-c^2\right)\right]\)

\(=\left(b-a\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]=\left(b-a\right)\left(b-c\right)\left(a-c\right)\)

Ơ tưởng là GTNN chứ nhỉ :D

Từ đa thức, ta suy ra:

\(A=-2\cdot\left(-4x+x^2\right)-5\)

\(A=-2\left(x^2-4x+4\right)+8-5\)

\(A=-2\cdot\left(x-2\right)^2-3\)

\(\)Vì 2(x-2)²\(\le\)0 \(\forall x\)nên minA=-3

Vậy...

\(A=-2x^2+8x-5=-2\left(x^2-4x+4\right)+8-5\)

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

Có : \(-2\left(x-2\right)^2\le0\)

=> \(A=-2\left(x-2\right)^2+3\le0+3=3\)

Dấu "=" xảy ra <=> x - 2 = 0 <=> x = 2

Vậy max A = 3 tại x = 2.

\(ab\left(b-a\right)-bc\left(b-c\right)-ac\left(c-a\right)\)

\(=ab\left(b-a\right)-b^2c+bc^2-ac^2+a^2c\)

\(=ab\left(b-a\right)+c^2\left(b-a\right)-c\left(b^2-a^2\right)\)

\(=\left(b-a\right)\left(ab+c^2-bc-ca\right)\)

\(=\left(b-a\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)

\(=\left(b-a\right)\left(a-c\right)\left(b-c\right)\)

\(\left(5x-4\right)^2+3\left(16-25x^2\right)=0\)

\(\Leftrightarrow\left(5x-4\right)^2-3\left(25x^2-16\right)=0\)

\(\Leftrightarrow\left(5x-4\right)^2-3\left(5x-4\right)\left(5x+4\right)=0\)

\(\Leftrightarrow\left(5x-4\right)\left[5x-4-3\left(5x+4\right)\right]=0\)

\(\Leftrightarrow\left(5x-4\right)\left(5x-4-15x-12\right)=0\)

\(\Leftrightarrow\left(5x-4\right)\left(-10x-16\right)=0\)

\(\Leftrightarrow5x-4=0\)hoặc \(-10x-16=0\)

\(\Leftrightarrow5x=4\)         hoặc \(-2\left(5x+8\right)=0\)

\(\Leftrightarrow x=\frac{4}{5}\)         hoặc \(5x+8=0\)

\(\Leftrightarrow x=\frac{4}{5}\)hoặc \(x=\frac{-8}{5}\)

Vậy tập nghiệm của phương trình là : \(S=\left\{\frac{-8}{5};\frac{4}{5}\right\}\)

Ta có: \(\left(5x-4\right)^2-3.\left(5x-4\right).\left(5x+4\right)=0\)

    \(\Leftrightarrow\left(5x-4\right).\left[\left(5x-4\right)-3\left(5x+4\right)\right]=0\)

    \(\Leftrightarrow\left(5x-4\right).\left(5x-4-15x-12\right)=0\)

    \(\Leftrightarrow-2.\left(5x-4\right).\left(5x+8\right)=0\)

    \(\Leftrightarrow\orbr{\begin{cases}5x-4=0\\5x+8=0\end{cases}}\)

    \(\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{5}\\x=\frac{-8}{5}\end{cases}}\)

 Vậy \(S=\left\{\frac{4}{5};\frac{-8}{5}\right\}\)

\(ab\left(b-a\right)-bc\left(b-c\right)-ac\left(c-a\right)\)

\(=ab\left[\left(b-c\right)+\left(c-a\right)\right]-bc\left(b-c\right)-ac\left(c-a\right)\)

\(=ab\left(b-c\right)+ab\left(c-a\right)-bc\left(b-c\right)-ac\left(c-a\right)\)

\(=\left[ab\left(b-c\right)-bc\left(b-c\right)\right]+\left[ab\left(c-a\right)-ac\left(c-a\right)\right]\)

\(=\left(b-c\right)\left(ab-bc\right)+\left(c-a\right)\left(ab-ac\right)\)

\(=-b\left(b-c\right)\left(c-a\right)+a\left(c-a\right)\left(b-c\right)\)

\(=\left(b-c\right)\left(c-a\right)\left(a-b\right)\)

Bài làm

x³ - 8 = 2x² - 4x

<=> x³ - 2x² + 4x - 8 = 0

<=> x²( x - 2 ) + 4( x - 2 ) = 0

<=> ( x - 2 )( x² + 4 ) = 0

<=> x - 2 = 0 hoặc x² + 4 = 0

<=> x = 2 hoặc x² = -4 ( vô lí )

Vậy x = 2 là nghiệp phương trình.

"nghiệp" :v

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

\(=>x.\left(x-2\right)=-4+3\)

\(=>x.\left(x-2\right)=-1\)

\(TH1:\orbr{\begin{cases}x=-1\\x-2=1\end{cases}=>\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)

\(TH2:\orbr{\begin{cases}x=1\\x-2=-1\end{cases}=>\orbr{\begin{cases}x=1\\x=1\end{cases}}}\)

Vậy ...

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

<=> \(x^2-2x+1=0\)

<=> \(\left(x-1\right)^2=0\)

<=> x - 1 = 0 

<=> x = 1

Vậy S = { 1 }