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

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

\(\Rightarrow\left(2x+2\right)\left(-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+2=0\\-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=-2\\-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\left(tm\right)\\-4=0\left(ktm\right)\end{cases}}\)

Vậy chọn đáp án A và C: \(x=-1\)

${\left(2x+3\right)}^2-{\left(x-1\right)}^2=0$

$\iff \left(2x+3+x-1\right)\left(2x+3-x+1\right)=0$

$\iff \left(3x+2\right)\left(x+4\right)=0$

$\iff [\begin{array}{c}3x+2=0\\ x+4=0\end{array}\Rightarrow \{\begin{array}{c}x=-\frac{2}{3}\\ x=-\end{array}$

Bài 1:

a,27x3+27x2+9x+1a,27x3+27x2+9x+1

=(3x)3+3.(3x)2.1+3.3x.12+13=(3x)3+3.(3x)2.1+3.3x.12+13

=(3x+1)3=(3x+1)3

b,x3+3√2x2y+6xy2+2√2y3b,x3+32x2y+6xy2+22y3

=x3+3.x2.√2y+3.x.(√2y)2+(√2y)3=x3+3.x2.2y+3.x.(2y)2+(2y)3

=(x+√2y)3=(x+2y)3

Bài 2:

a,x3+9x2+27x+27=0a,x3+9x2+27x+27=0

⇔(x+3)3=0⇔(x+3)3=0

⇔x+3=0⇔x=−3⇔x+3=0⇔x=−3

b,(x+1)3−x(x−2)2+x−1=0b,(x+1)3−x(x−2)2+x−1=0

⇔x3+3x2+3x+1−x3−4x2+4x+x−1=0⇔x3+3x2+3x+1−x3−4x2+4x+x−1=0

⇔−x2+8x=0⇔−x2+8x=0

⇔−x(x−8)=0

27x³ - 27x² + 3x + 1

= 27x³ - 9x² - 18x² - 3x + 6x + 1

= (27x³ - 9x²) - (18x² - 6x) - (3x - 1)

= 9x² (3x - 1) - 6x (3x - 1) - (3x - 1)

= (3x - 1) (9x² - 6x - 1)

em xem lại xem nào, sao lại 4y dc

MAX =1

\(\left(3x-5\right)\left(5-3x\right)+9\left(x+1\right)^2=30\)

\(\Leftrightarrow-\left(3x-5\right)^2+9\left(x+1\right)^2=30\)

\(\Leftrightarrow-\left(9x^2-30x+25\right)+9\left(x^2+2x+1\right)=30\)

\(\Leftrightarrow-9x^2+9x^2+30x+18x-25+9-30=0\)

\(\Leftrightarrow48x=46\Leftrightarrow x=\frac{23}{24}\)