a) Ta có: \(x^2-2x+1=0\)

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

\(\Leftrightarrow x-1=0\)hay x=1

Vậy: S={1}

c) Ta có: \(x+x^4=0\)

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

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

mà \(x^2-x+1>0\forall x\)

nên x(x+1)=0

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

Vậy: S={0;-1}

a) Ta có: \(x^2+3x-10=0\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

Vậy: S={-5;2}

b) Ta có: \(3x^2-7x+1=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)

c) Ta có: \(3x^2-7x+8=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)

Vậy: \(x\in\varnothing\)

ko bt

a) \(x\left(x+2\right)-3x-6=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

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

\(x^3+1=0\)

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

\(x=-1\)

c) \(4x^2-25=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

\(a,x\left(x+2\right)-3x-6=0\)

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

\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)\(b,\left(x^3+3x^2+3x+1\right)-3x^2-3x=0\)

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

\(\Leftrightarrow x^3+1=0\)

\(\Rightarrow x^3=1\Rightarrow x=1\)

\(c,4x^2-25=0\)

\(\Leftrightarrow\left(2x+5\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}2x+5=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

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)

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

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

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

b)

\(\Rightarrow3x\left(x-2\right)-2\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(3x-2\right)=0\)

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

c)

\(\Rightarrow\left(3x-1\right)\left(5x+x-2\right)=0\)

\(\Rightarrow\left(3x-2\right)^2.2=0\)

\(\Rightarrow3x-2=0\)

\(\Rightarrow x=\frac{2}{3}\)

\(\left(4-3x\right)\left(10x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)

\(\left(7-2x\right)\left(4+8x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)

rồi thực hiện đến hết ... 

Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>

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

\(2x^2-7x+3=4x^2+4x-3\)

\(2x^2-7x+3-4x^2-4x+3=0\)

\(-2x^2-11x+6=0\)

\(2x^2+11x-6=0\)

\(2x^2+12x-x-6=0\)

\(2x\left(x+6\right)-\left(x+6\right)=0\)

\(\left(x+6\right)\left(2x-1\right)=0\)

\(x+6=0\Leftrightarrow x=-6\)

\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)

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

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

\(x=0\)

\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Tự lm tiếp nha 

chiu lop 3 ma

c(x-1)^2=4

x^2-2x+1=4

x^2-2x+1-4=0

x^2-2x-3=0

x^2-3x+x-3=0

x(x-3)+(x-3)=0

(x-3)(x+1)=0

\(\Rightarrow\hept{\begin{cases}x-3=0\\x+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=-1\end{cases}}}\)

d, x^3+2x^2-x-2=0

x^2(x+2)-(x+2)=0

(x+2)(x^2-1)=0

\(\Rightarrow\hept{\begin{cases}x=-2\\x=+-1\end{cases}}\)

e, (3x+2)^2-(2x-1)^2=0

(3x+2-2x+1)(3x+2+2x-1)=0

(x+3)(5x-1)=0

x+3=0=>x=-3

5x-1=0=>5x=1=>x=1/5

a) \(2\left(x+3\right)-x^2-3x=0\)

\(\Leftrightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\)

\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)

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

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

Vậy tập nghiệm của phương trình là \(S=\left\{-3;2\right\}\)

c) \(3x\left(x-5\right)-x^2+25=0\)

\(\Leftrightarrow3x\left(x-5\right)-\left(x^2-25\right)=0\)

\(\Leftrightarrow3x\left(x-5\right)-\left(x-5\right)\left(x+5\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\2x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\2x=5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{5}{2}\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{5;\frac{5}{2}\right\}\)