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

\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)

\(\Leftrightarrow x^2\left(2x+54\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)

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

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)

\(\Leftrightarrow6x^2=-2\)

\(\Leftrightarrow x^2=-3\) ( vô lí)

Vậy pt vô nghiệm

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

\(\Leftrightarrow x^2-3x-x+3=0\)

\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)

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

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

\(d,4x^2+4x+1=0\)

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

\(\Rightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)

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

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

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

\(\Leftrightarrow-2x-5=0\)

\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)

Học tốt nha you <3

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

\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)

\(\Leftrightarrow x^2\left(2x+54\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)

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

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)

\(\Leftrightarrow6x^2=-2\)

\(\Leftrightarrow x^2=-3\) ( vô lí)

Vậy pt vô nghiệm

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

\(\Leftrightarrow x^2-3x-x+3=0\)

\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)

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

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

\(d,4x^2+4x+1=0\)

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

\(\Rightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)

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

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

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

\(\Leftrightarrow-2x-5=0\)

\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)

Học tốt nha you <3

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

\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)

\(\Leftrightarrow x^2\left(2x+54\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)

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

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)

\(\Leftrightarrow6x^2=-2\)

\(\Leftrightarrow x^2=-3\) ( vô lí)

Vậy pt vô nghiệm

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

\(\Leftrightarrow x^2-3x-x+3=0\)

\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)

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

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

\(d,4x^2+4x+1=0\)

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

\(\Rightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)

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

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

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

\(\Leftrightarrow-2x-5=0\)

\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)

Học tốt nha you <3

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

1. \(\sqrt{x^2-4x+3}=x-2\)

<=> x² - 4x + 3 = (x - 2)²

<=> x² - 4x + 3 = x² - 4x + 4

<=> x² - x² - 4x + 4x = 1

<=> 0 = 1 (Vô lí)

vậy PT có nghiệm là S = \(\varnothing\)

2. \(\sqrt{4x^2-4x+1}=x-1\)

<=> \(\sqrt{\left(2x-1\right)^2}=x-1\)

<=> 2x - 1 = x - 1

<=> 2x - x = -1 + 1

<=> x = 0

a: \(x^3-4x^2-x+4=0\)

=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)

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

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

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

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

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)

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

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

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

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

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

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

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

a) Ta có: \(x^2-9x+20=0\)

\(\Leftrightarrow x^2-5x-4x+20=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)

Vậy: x∈{4;5}

b) Ta có: \(x^3-4x^2+5x=0\)

\(\Leftrightarrow x\left(x^2-4x+5\right)=0\)(1)

Ta có: \(x^2-4x+5\)

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

Ta có: \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-2\right)^2+1\ge1>0\forall x\)

hay \(x^2-4x+5>0\forall x\)(2)

Từ (1) và (2) suy ra x=0

Vậy: x=0

c) Sửa đề: \(x^2-2x-15=0\)

Ta có: \(x^2-2x-15=0\)

\(\Leftrightarrow x^2+3x-5x-15=0\)

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

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

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

Vậy: x∈{-3;5}

d) Ta có: \(\left(x^2-1\right)^2=4x+1\)

\(\Leftrightarrow x^4-2x^2+1-4x-1=0\)

\(\Leftrightarrow x^4-2x^2-4x=0\)

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

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

\(\Leftrightarrow x\cdot\left[x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\right]=0\)

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

Ta có: \(x^2+2x+2\)

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

Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+1\right)^2+1\ge1>0\forall x\)

hay \(x^2+2x+2>0\forall x\)(4)

Từ (3) và (4) suy ra

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

Vậy: x∈{0;2}

cảm ơn bạn

a)  \(x^3\)\(-\)\(\frac{1}{4}x\)\(=\)\(0\)

\(x\left(x^2-\frac{1}{4}\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2-\frac{1}{4}=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2=0,5^2\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=+-0,5\end{cases}}\)

Vậy .............................

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

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+2=0\\x-4=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=-2\\x=4\end{cases}}\)\(\orbr{\begin{cases}x=\frac{-2}{3}\\x=4\end{cases}}\)

Vậy ................................

c)  \(x^2\)\(\left(x-3\right)\)\(+\)\(12\)\(-\)\(4x\)\(=\)\(0\)

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

\(\left(x^2-4\right)\left(x-3\right)\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x-3=0\end{cases}-4=0}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x=3\end{cases}=2^2}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=+-2\\x=3\end{cases}}\)

a)\(x^3-\frac{1}{4}x=0\)

\(\Leftrightarrow x\left(x^2-\frac{1}{4}\right)=0\)

\(\Leftrightarrow x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)

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

\(x^2+x+3x+3=0\)

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

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

\(\left[\begin{array}{nghiempt}x+1=0\\x+3=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=-1\\x=-3\end{array}\right.\)

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

\(4x^2-2x+6x-3=0\)

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

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

\(\left[\begin{array}{nghiempt}2x-1=0\\2x+3=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}2x=1\\2x=-3\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=\frac{1}{2}\\x=-\frac{3}{2}\end{array}\right.\)

\(x^2-x-12=0\)

\(x^2-4x+3x-12=0\)

\(x\left(x-4\right)+3\left(x-4\right)=0\)

\(\left(x-4\right)\left(x+3\right)=0\)

\(\left[\begin{array}{nghiempt}x-4=0\\x+3=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=4\\x=-3\end{array}\right.\)

\(x^2-25-\left(x-5\right)=0\)

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

\(\left(x-5\right)\left(x+5-1\right)=0\)

\(\left(x-5\right)\left(x+4\right)=0\)

\(\left[\begin{array}{nghiempt}x-5=0\\x+4=0\end{array}\right.\)

\(\left[\begin{array}{nghiempt}x=5\\x=-4\end{array}\right.\)

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

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

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

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

\(\left[\begin{array}{nghiempt}x-1=0\\x+1=0\end{array}\right.\) (vì \(x^2+1\ge1>0\))

\(\left[\begin{array}{nghiempt}x=1\\x=-1\end{array}\right.\)