\(a.\left(x^2-2x+1\right)-4=0\\\Leftrightarrow \left(x-1\right)^2-2^2=0\\\Leftrightarrow \left(x-1-2\right)\left(x-1+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{3;-1\right\}\)

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

Vậy tập nghiệm của phương trình trên là \(S=\left\{-2;1\right\}\)

\(c.4x^2+4x+1=x^2\\ \Leftrightarrow4\left(x^2+x+\frac{1}{4}\right)-x^2=0\\ \Leftrightarrow4\left(x+\frac{1}{2}\right)^2-x^2=0\\ \Leftrightarrow\left[2\left(x+\frac{1}{2}\right)-x\right]\left[2\left(x-\frac{1}{2}\right)+x\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2\left(x+\frac{1}{2}\right)-x=0\\2\left(x+\frac{1}{2}\right)+x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x+1-x=0\\2x+1+x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{1}{3}\end{matrix}\right.\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{-1;-\frac{1}{3}\right\}\)

bạn cứ tra gg rồi ấn thừa số là ra

kinh nghiệm đó

1000%

\(a,\) PT thứ 2 bị lỗi rồi bạn, dấu '' = '' đou

\(b,\)

\(4x^2-32=0\Leftrightarrow4x^2=32\Leftrightarrow x^2=8\Leftrightarrow x=\pm\sqrt{8}\)

\(3x^2=48\Leftrightarrow x^2=16\Leftrightarrow x=\pm4\)

Vậy 2 pt trên không tường đương

xin lỗi bạn, mình không để ý 

a)6(x²-2x+3)=2(3x²-6x+9) và 3x-6=3(x-2)

\(a,6\left(x^2-2x+3\right)=2\left(3x^2-6x+9\right)\)

\(\Leftrightarrow6x^2-12x+18=6x^2-12x+18\)

\(\Leftrightarrow\) pt vô nghiệm

\(3x-6=3\left(x-2\right)\)

\(\Leftrightarrow3x-6=3x-6\)

\(\Leftrightarrow\) pt vô nghiệm

Vậy 2 pt tương đương

\(b,4x^2-32=0\Leftrightarrow x^2=8\Leftrightarrow x=\pm\sqrt{8}\)

\(3x^2=48\Leftrightarrow x=\pm4\)

Vậy 2 pt ko tương đương

Phương trình b tương đương vì chúng có cùng tập nghiệm là S={4;-4}

a: 6(x^2-2x+3)=2(3x^2-6x+9)

=>6x^2-12x+18=6x^2-12x+18

=>-12x=-12x

=>0x=0(luôn đúng)

3x-6=3(x-2)

=>3x-6=3x-6

=>0x=0(luôn đúng)

=>Hai phương trình tương đương

\(1,\\ a,=3x^3-2x^2+5x\\ b,=2x^3y^2+\dfrac{2}{9}x^4y^2-\dfrac{1}{3}x^2y^3\\ c,=x^2-2x+6x-12=x^2+4x-12\\ 2,\\ a,\Rightarrow6x-9+4-2x=-3\\ \Rightarrow4x=2\Rightarrow x=\dfrac{1}{2}\\ b,\Rightarrow5x-2x^2+2x^2-2x=13\\ \Rightarrow3x=13\Rightarrow x=\dfrac{13}{3}\\ c,\Rightarrow5x^2-5x-5x^2+7x-10x+14=6\\ \Rightarrow-8x=-8\Rightarrow x=1\\ d,\Rightarrow6x^2+9x-6x^2+4x-15x+10=8\\ \Rightarrow-2x=-2\Rightarrow x=1\)

\(3,\\ A=2x^2+x-x^3-2x^2+x^3-x+3=3\\ B=6x^2-10x+33x-55-6x^2-14x-9x-21=-76\)

\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\)

\(< =>\left(1-x\right)\left(5x+3+3x-7\right)=0\)

\(< =>\left(1-x\right)\left(8x-4\right)=0\)

\(< =>\orbr{\begin{cases}1-x=0\\8x-4=0\end{cases}< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)

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

\(< =>\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)

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

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

\(< =>2-x=0< =>x=2\)

1) Ta có: \(x^2-4x+4=0\)

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

\(\Leftrightarrow x-2=0\)

hay x=2

Vậy: S={2}

a: \(\frac{x+1}{2x-6}-\frac{4}{2x-6}\)

\(=\frac{x+1-4}{2\left(x-3\right)}\)

\(=\frac{x-3}{2\left(x-3\right)}=\frac12\)

b: \(\frac{3x-4}{6x+3}-\frac{x-5}{6x+3}\)

\(=\frac{3x-4-x+5}{6x+3}\)

\(=\frac{2x+1}{3\left(2x+1\right)}=\frac13\)

c: \(\frac{x-1}{x-3}-\frac{3x-8}{3-x}+\frac{3-2x}{x-3}\)

\(=\frac{x-1}{x-3}+\frac{3x-8}{x-3}+\frac{3-2x}{x-3}\)

\(=\frac{x-1+3x-8+3-2x}{x-3}=\frac{2x-6}{x-3}=\frac{2\left(x-3\right)}{x-3}\)

=2

d: \(\frac{3}{x+5}-\frac{5}{x-7}\)

\(=\frac{3\left(x-7\right)-5\left(x+5\right)}{\left(x+5\right)\left(x-7\right)}=\frac{3x-21-5x-25}{\left(x+5\right)\left(x-7\right)}\)

\(=\frac{-2x-46}{\left(x+5\right)\left(x-7\right)}\)

e: \(\frac{3}{x+5}-\frac{5}{x-7}\)

\(=\frac{3\left(x-7\right)-5\left(x+5\right)}{\left(x+5\right)\left(x-7\right)}=\frac{3x-21-5x-25}{\left(x+5\right)\left(x-7\right)}\)

\(=\frac{-2x-46}{\left(x+5\right)\left(x-7\right)}\)

f: \(\frac{2}{x-2}+\frac{3}{x+2}+\frac{5x-18}{x^2-4}\)

\(=\frac{2}{x-2}+\frac{3}{x+2}+\frac{5x-18}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{2\left(x+2\right)+3\left(x-2\right)+5x-18}{\left(x-2\right)\left(x+2\right)}=\frac{2x+4+3x-6+5x-18}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{10x-20}{\left(x-2\right)\left(x+2\right)}=\frac{10}{x+2}\)

Bài 1:

a) (3x - 2)(4x + 5) = 0

<=> 3x - 2 = 0 hoặc 4x + 5 = 0

<=> 3x = 2 hoặc 4x = -5

<=> x = 2/3 hoặc x = -5/4

b) (2,3x - 6,9)(0,1x + 2) = 0

<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0

<=> 2,3x = 6,9 hoặc 0,1x = -2

<=> x = 3 hoặc x = -20

c) (4x + 2)(x^2 + 1) = 0

<=> 4x + 2 = 0 hoặc x^2 + 1 # 0

<=> 4x = -2

<=> x = -2/4 = -1/2

d) (2x + 7)(x - 5)(5x + 1) = 0

<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0

<=> 2x = -7 hoặc x = 5 hoặc 5x = -1

<=> x = -7/2 hoặc x = 5 hoặc x = -1/5

bài 2:

a, (3x+2)(x^2-1)=(9x^2-4)(x+1)

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

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

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

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

x(x^2-9)-(x^3+8)=0

x^3-9x-x^3-8=0

-9x-8=0

tự tìm x nha

a: ĐKXĐ: \(x^2-6x+6\ge0\)

=>\(x^2-6x+9-3\ge0\)

=>\(\left(x-3\right)^2-3\ge0\)

=>\(\left(x-3\right)^2\ge3\)

=>\(\left[\begin{array}{l}x-3\ge\sqrt3\\ x-3\le-\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge\sqrt3+3\\ x\le-\sqrt3+3\end{array}\right.\)

Ta có: \(x^2-6x+9=4\sqrt{x^2-6x+6}\)

=>\(x^2-6x+6-4\cdot\sqrt{x^2-6x+6}+3=0\)

=>\(\left(\sqrt{x^2-6x+6}-3\right)\left(\sqrt{x^2-6x+6}-1\right)=0\)

TH1: \(\sqrt{x^2-6x+6}-3=0\)

=>\(\sqrt{x^2-6x+6}=3\)

=>\(x^2-6x+6=9\)

=>\(x^2-6x-3=0\)

=>\(x^2-6x+9-12=0\)

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

=>\(\left[\begin{array}{l}x-3=2\sqrt3\\ x-3=-2\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\sqrt3+3\left(nhận\right)\\ x=3-2\sqrt3\left(nhận\right)\end{array}\right.\)

TH2: \(\sqrt{x^2-6x+6}-1=0\)

=>\(x^2-6x+6=1\)

=>\(x^2-6x+5=0\)

=>(x-1)(x-5)=0

=>\(\left[\begin{array}{l}x=1\left(nhận\right)\\ x=5\left(nhận\right)\end{array}\right.\)

b: ĐKXĐ: x∈R

\(x^2-x+8-4\sqrt{x^2-x+4}=0\)

=>\(x^2-x+4-4\cdot\sqrt{x^2-x+4}+4=0\)

=>\(\left(\sqrt{x^2-x+4}-2\right)^2=0\)

=>\(\sqrt{x^2-x+4}-2=0\)

=>\(\sqrt{x^2-x+4}=2\)

=>\(x^2-x+4=4\)

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

=>x(x-1)=0

=>x=0 hoặc x=1

c: \(x^2+\sqrt{4x^2-12x+44}=3x+4\)

=>\(x^2-3x-4+2\sqrt{x^2-3x+11}=0\)

=>\(x^2-3x+11+2\sqrt{x^2-3x+11}-15=0\)

=>\(\left(\sqrt{x^2-3x+11}+5\right)\left(\sqrt{x^2-3x+11}-3\right)=0\)

=>\(\sqrt{x^2-3x+11}-3=0\)

=>\(\sqrt{x^2-3x+11}=3\)

=>\(x^2-3x+11=9\)

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

=>(x-1)(x-2)=0

=>x=1(nhận) hoặc x=2(nhận)