x³-4x²+7x-6=0

=>x³-2x²-2x²+3x+4x-6=0

=>x³-2x²+3x-2x²+4x-6=0

=>x(x²-2x+3)-2(x²-2x+3)=0

=>(x-2)(x²-2x+3)=0

=>x-2=0 hoặc x²-2x+3=0

Vậy pt trên có nghiệm là x=2

bậc 3 thì dùng PP nhẩm nghiệm đi.......

hoặc có thể nhóm như này 

x³-4x²+7x-6=0

<=>x³-4x²+4x+3x-6=0

<=>x.(x-2)²+3.(x-2)=0

<=>(x-2)(x²-2x+3)=0

<=>x=2 ( vì x²-2x+3 >0)

1:

a: =>3x=6

=>x=2

b: =>4x=16

=>x=4

c: =>4x-6=9-x

=>5x=15

=>x=3

d: =>7x-12=x+6

=>6x=18

=>x=3

2:

a: =>2x<=-8

=>x<=-4

b: =>x+5<0

=>x<-5

c: =>2x>8

=>x>4

Bài 1: ĐKXĐ: $2\leq x\leq 4$
PT $\Leftrightarrow (\sqrt{x-2}+\sqrt{4-x})^2=2$

$\Leftrightarrow 2+2\sqrt{(x-2)(4-x)}=2$
$\Leftrightarrow (x-2)(4-x)=0$

$\Leftrightarrow x-2=0$ hoặc $4-x=0$

$\Leftrightarrow x=2$ hoặc $x=4$ (tm)

Bài 2:
PT $\Leftrightarrow 4x^3(x-1)-3x^2(x-1)+6x(x-1)-4(x-1)=0$

$\Leftrightarrow (x-1)(4x^3-3x^2+6x-4)=0$
$\Leftrightarrow x=1$ hoặc $4x^3-3x^2+6x-4=0$

Với $4x^3-3x^2+6x-4=0(*)$

Đặt $x=t+\frac{1}{4}$ thì pt $(*)$ trở thành:
$4t^3+\frac{21}{4}t-\frac{21}{8}=0$

Đặt $t=m-\frac{7}{16m}$ thì pt trở thành:

$4m^3-\frac{343}{1024m^3}-\frac{21}{8}=0$
$\Leftrightarrow 4096m^6-2688m^3-343=0$

Coi đây là pt bậc 2 ẩn $m^3$ và giải ta thu được \(m=\frac{\sqrt[3]{49}}{4}\) hoặc \(m=\frac{-\sqrt[3]{7}}{4}\)

Khi đó ta thu được \(x=\frac{1}{4}(1-\sqrt[3]{7}+\sqrt[3]{49})\)

Nãy mình tìm được một cách giải tương tự cho câu 2.

PT \(\Leftrightarrow\left(x-1\right)\left(4x^3-3x^2+6x-4\right)=0\)

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

Vậy pt có 1 nghiệm bằng 1.

\(\left(1\right)\Rightarrow8x^3-6x^2+12x-8=0\)

\(\Leftrightarrow7x^3+x^3-6x^2+12x-8=0\)

\(\Leftrightarrow\left(x-2\right)^3=-7x^3\)

\(\Leftrightarrow x-2=-\sqrt[3]{7}x\)

\(\Leftrightarrow x=\dfrac{2}{1+\sqrt[3]{7}}\)

Vậy pt có nghiệm \(S=\left\{1;\dfrac{2}{1+\sqrt[3]{7}}\right\}\)

Lưu ý: Nghiệm của người kia hoàn toàn tương đồng với nghiệm của mình (\(\dfrac{2}{1+\sqrt[3]{7}}=\dfrac{1}{4}\left(1-\sqrt[3]{7}+\sqrt[3]{49}\right)\))

a: \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)

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

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

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

=>\(\left[\begin{array}{l}3x+2=0\\ x+1=0\\ 2x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac23\\ x=-1\\ x=\frac12\end{array}\right.\)

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

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

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

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

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

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

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

=>\(\left[\begin{array}{l}x-1=0\\ x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=-1\end{array}\right.\)

e: \(x^3-7x+6=0\)

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

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

=>x(x-1)(x+1)-6(x-1)=0

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

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

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

g: \(x^5-5x^3+4x=0\)

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

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

=>\(\left[\begin{array}{l}x=0\\ x^2-1=0\\ x^2-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^2=1\\ x^2=4\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\\ x=-1\\ \left[\begin{array}{l}x=2\\ x=-2\end{array}\right.\end{array}\right.\)

Bài 2:

a: =>2x^2-4x+1=x^2+x+5

=>x^2-5x-4=0

=>\(x=\dfrac{5\pm\sqrt{41}}{2}\)

b: =>11x^2-14x-12=3x^2+4x-7

=>8x^2-18x-5=0

=>x=5/2 hoặc x=-1/4

a) \(\left(4x-3\right)^3+\left(3x-2\right)^3=\left(7x-5\right)^3\)

\(\Leftrightarrow64x^3-144x^2+108x-27+27x^3-54x^2+36x-8=343x^3-735x^2+525x-125\)

\(\Leftrightarrow-252x^3+537x^2-381x+90=0\)

\(\Leftrightarrow-3\left(84x^3-179x^2+127-30\right)=0\)

\(\Leftrightarrow-3\left(7x-5\right)\left(3x-2\right)\left(4x-3\right)=0\)

\(\Leftrightarrow x\in\left\{\frac{5}{7};\frac{2}{3};\frac{3}{4}\right\}\)

b) \(x^3-2x^2-x-6=0\)

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

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

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

Vì \(x^2+x+2>0\)

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

Vậy....

x²-4x+7 = 0 ⇔ x² -4x + 4 + 3 = 0 

⇔ (x-2)²+3=0 ⇔ (x-2)²=-3 (vô lí)

Vậy pt vô nghiệm

*Chứng minh phương trình \(x^2-4x+7=0\) vô nghiệm

Ta có: \(x^2-4x+7=0\)

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

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

mà \(\left(x-2\right)^2+3\ge3>0\forall x\)

nên \(x\in\varnothing\)(đpcm)

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

\(\Leftrightarrow\left(2x^3+4x^2\right)+\left(3x^2+6x\right)+\left(x+2\right)=0\)

\(\Leftrightarrow2x^2\left(x+2\right)+3x\left(x+2\right)+\left(x+2\right)=0\)

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

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

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

.......................................................................................

\(x^3-8x^2-8x+1=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)-8x\left(x+1\right)=0\)

......................................................................................

cảm ơn nha