a) x = 2 7                         b) x = 2.

c) x = 2                          d) x = 1.

a. (3x - 1)² - (x + 3)² = 0

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

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

\(\Leftrightarrow4x+2=0\)  hoặc  \(2x-4=0\)

1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)

2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)

S=\(\left\{-\dfrac{1}{2};2\right\}\)

b. \(x^3=\dfrac{x}{49}\)

\(\Leftrightarrow49x^3=x\)

\(\Leftrightarrow49x^3-x=0\)

\(\Leftrightarrow x\left(49x^2-1\right)=0\)

\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)

\(\Leftrightarrow x=0\) hoặc  \(7x+1=0\) hoặc \(7x-1=0\)

1. x=0

2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)

3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)

*Cách khác:

a) Ta có: \(\left(3x-1\right)^2-\left(x+3\right)^2=0\)

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

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

Vậy: \(S=\left\{-\dfrac{1}{2};2\right\}\)

e: =>x(x^3-4x^2-8x+8)=0

=>x[(x^3+8)-4x(x+2)]=0

=>x(x+2)(x^2-2x+4-4x)=0

=>x(x+2)(x^2-6x+4)=0

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

g: =>2x^4+5x^3-6x^3-15x^2+6x^2+15x-2x-5=0

=>(2x+5)(x^3-3x^2+3x-1)=0

=>(2x+5)(x-1)^3=0

=>x=1 hoặc x=-5/2

h: =>(x^2+8x+7)(x^2+8x+15)+15=0

=>(x^2+8x)^2+22(x^2+8x)+120=0

=>(x^2+8x+10)(x^2+8x+12)=0

=>(x^2+8x+10)(x+2)(x+6)=0

=>\(x\in\left\{-2;-6;-4+\sqrt{6};-4-\sqrt{6}\right\}\)

Bài 1

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

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

b/

\(\Leftrightarrow x^3-6x^2+9x+5x^2-30x+45=0\)

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

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

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

1.

c/ \(\Leftrightarrow x^3+2x^2+2x+x^2+2x+2=0\)

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

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

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

d/

\(\Leftrightarrow x^4+x^3-2x^2-x^3-x^2+2x+4x^2+4x-8=0\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Bài 1:

e/ \(\Leftrightarrow x^4+2x^2-8x+5=0\)

\(\Leftrightarrow x^4-2x^3+x^2+2x^3-4x^2+2x+5x^2-10x+5=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+5=0\left(vn\right)\\x=1\end{matrix}\right.\)

Bài 2:

a/ Đặt \(x^2-5x=t\)

\(t^2+10t+24=0\Rightarrow\left[{}\begin{matrix}t=-4\\t=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x=-4\\x^2-5x=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x+4=0\\x^2-5x+6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=4\\x=2\\x=3\end{matrix}\right.\)

\(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)

\(\Leftrightarrow\left(x+1\right)\left(x+5\right)\left(x+2\right)\left(x+4\right)=40\)

\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)

\(\Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+5+3\right)=40\)

\(\Leftrightarrow p\left(p+3\right)=40\) (khi đặt \(\left(x^2+6x+5\right)=p\)

\(\Leftrightarrow p^2+3p=40\)

\(\Leftrightarrow p^2\cdot2\cdot p\cdot\frac{3}{2}+\left(\frac{3}{2}\right)^2=\frac{169}{4}\)

\(\Leftrightarrow\left(p+\frac{3}{2}\right)^2-\left(\frac{13}{2}\right)^2=0\)

\(\Leftrightarrow\left(p+\frac{3}{2}-\frac{13}{2}\right)\left(p+\frac{3}{2}+\frac{13}{2}\right)=0\)

\(\Leftrightarrow\left(p-5\right)\left(p+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}p=5\\p=-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+6x+5=5\\x^2+6x+5=-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+6x=0\\x^2+2\cdot x\cdot3+9-9+5=-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\left(x+6\right)=0\\\left(x+3\right)^2=-4\left(\text{vôlí}\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-6\end{cases}}\)

\(\left(x-2\right)\left(x^2+5x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x^2+5x-7=0\end{cases}}\)

Ta có: \(\Delta=25-4\cdot\left(-7\right)=25+28=53\)

\(\Rightarrow\Delta>0\)

\(\Rightarrow\text{pt có 2 nghiệm pb}\)

\(\Rightarrow\hept{\begin{cases}x_1=\frac{-5-\sqrt{53}}{2}\\x_2=\frac{-5+\sqrt{53}}{2}\end{cases}}\)

\(\text{Vậy pt trên có nghiệm là x=2; x=}\frac{-5\pm\sqrt{53}}{2}\)

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

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

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

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

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

\(\text{Vậy pt có nghiệm là x=2;x=-3}\)

bạn đăng tách cho mn cùng giúp nhé 

Bài 1 : 

a, \(\Leftrightarrow11-x=12-8x\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)

b, \(\Leftrightarrow2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)

\(\Leftrightarrow2x^3+8x^2+8x-8x^2=2x^3-16\Leftrightarrow x=-2\)

c, \(\Leftrightarrow3-2x=-x-4\Leftrightarrow x=7\)

d, \(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1\)

\(\Leftrightarrow3x^2+12x-9=3x^2+3x+1\Leftrightarrow x=\dfrac{10}{9}\)

e, \(\Leftrightarrow2x^2-x-3=2x^2+9x-5\Leftrightarrow x=5\)

f, \(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x=10x-5x^2-11x-22\)

\(\Leftrightarrow-5x^2+2x-1=-5x^2-x-22\Leftrightarrow3x=-21\Leftrightarrow x=-7\)

h) \(PT\Leftrightarrow x^2+4x-3x-12-6x+4=x^2-8x+16\)

\(\Leftrightarrow3x=24\)

\(\Leftrightarrow x=8\)

Vậy: \(S=\left\{8\right\}\)

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

\(\Leftrightarrow x=1\)

Vậy: \(S=\left\{1\right\}\)

Lời giải:

\(\frac{4}{9}\times \frac{3}{7}\times \frac{7}{4}=\frac{1}{3}\)

\(\frac{6}{5}\times \frac{4}{5}\times \frac{25}{16}=\frac{3}{2}\)

\(\frac{7}{8}\times \frac{16}{9}\times \frac{3}{14}=\frac{1}{3}\)

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

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

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

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

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