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

Ta có x² ≥ 0 ∀ x => -x² ≤ 0

=> A = -x² + 2x + 7 ≤ 2x + 7

Dấu "=" xảy ra <=> -x² = 0 <=> x² = 0 <=> x = 0

Khi đó A = -0² + 2 . 0 + 7 = 7

`x^2-y^2-2x+2y`

`=(x^2-y^2)-(2x-2y)`

=(x−y)(x+y)−2(x−y)

1+1=2 4567890

( 3x - 1 )² - 9 = 0

=> ( 3x - 1 )² = 9

=> ( 3x - 1 )² = ( ±3 )²

\(\Rightarrow\orbr{\begin{cases}3x-1=3\\3x-1=-3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=4\\3x=-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{-2}{3}\end{cases}}\)

\(\left(3x-1\right)^2-9=0\)

\(\left(3x-1\right)^2=9\)

\(\Rightarrow\left(3x-1\right)^2=3^2\text{ hoặc }\left(3x-1\right)^2=\left(-3\right)^2\)

\(\Rightarrow3x-1=3\text{ hoặc }3x-1=\left(-3\right)\)

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

                  \(x=\frac{4}{3}\)                 \(x=-\frac{2}{3}\)

Vậy ...

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

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

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

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

\(\left(3x-1\right)^2-9=0\)

\(\Leftrightarrow\left(3x-1-9\right)\left(3x-1+9\right)=0\)

\(\Leftrightarrow\left(3x-10\right)\left(3x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-10=0\\3x+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\\x=-\frac{8}{3}\end{cases}}}\).

\(x^4-9x^3-x^2+9x=0\)

\(\Leftrightarrow\left(x^4-x^2\right)+\left(-9x^3+9x\right)=0\)

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

\(\Leftrightarrow x\left(x-9\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\)\(x=0\) hoặc \(x-9=0\) hoặc \(x-1=0\) hoặc \(x+1=0\).

1.\(x=0\).

2. \(x-9=0\Leftrightarrow x=-9\).

3. \(x-1=0\Leftrightarrow x=1\).

4. \(x+1=0\Leftrightarrow x=-1\).

Vậy \(x\in\left\{-9;-1;0;1\right\}\).

(x2−25)2−(x−5)2=0(x2−25)2−(x−5)2=0

⇔[(x−5)(x+5)]2−(x−5)2=0⇔[(x−5)(x+5)]2−(x−5)2=0

⇔(x−5)2⋅(x+5)2−(x−5)2=0⇔(x−5)2⋅(x+5)2−(x−5)2=0

⇔(x−5)2⋅[(x+5)2−1]=0⇔(x−5)2⋅[(x+5)2−1]=0

⇔[(x−5)2=0(x+5)2−1=0⇒(x+5)2=1⇔[(x−5)2=0(x+5)2−1=0⇒(x+5)2=1

⇔⎡⎢⎣x−5=0x+5=1x+5=−1⇔⎡⎢⎣x=5x=−4x=−6⇔[x−5=0x+5=1x+5=−1⇔[x=5x=−4x=−6

Vậy: x= 5; x=-4 hoặc x=-6

sai r bn ơi