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

\(\Leftrightarrow2x^4-6x^3+3x^2-2x^2+6x-3=0\)

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

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+1=0\\2x^2-6x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\\Delta_{2x^2-6x+3}=\left(-6\right)^2-4\left(2.3\right)=12\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\x_{1,2}=\frac{6\pm\sqrt{12}}{4}\end{array}\right.\)

b)\(x^3+9x^2+26x+24=0\)

\(\Leftrightarrow x^3+5x^2+6x+4x^2+20x+24=0\)

\(\Leftrightarrow x\left(x^2+5x+6\right)+4\left(x^2+5x+6\right)=0\)

\(\Leftrightarrow\left(x^2+5x+6\right)\left(x+4\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\\x=-4\end{array}\right.\)

de qua

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

A) x³-6x²+12x-8=0

<=>(x-2)³=0

<=>x-2=0

<=>x=2

B)4(x-3)² -(2x-1)(2x+1)=13

<=>4(x²-6x+9)-4x²+1=13

<=>4x²-24x+36-4x²+1=13

<=>-24x+37=13

<=>24x=37-13

<=>24x=24

<=>x=1

C)25x²-6(x+1)²=0

<=>(5x-\(\sqrt{6}\left(x+1\right)\))(5x+\(\sqrt{6}\left(x+1\right)\))=0

<=>5x-\(\sqrt{6}\left(x+1\right)\)=0 hoặc 5x+\(\sqrt{6}\left(x+1\right)\))=0

<=>5x-\(\sqrt{6}x-\sqrt{6}\)=0         <=>5x+\(\sqrt{6}x+\sqrt{6}\)=0

<=>x(5-\(\sqrt{6}\))=\(\sqrt{6}\)               <=>x(5+\(\sqrt{6}\))=\(-\sqrt{6}\)

<=>x=\(\frac{\sqrt{6}}{5-\sqrt{6}}\)                           <=>x=\(\frac{-\sqrt{6}}{5+\sqrt{6}}\)

Rút gọn C=(4+2A+A^2).(4-A^2).(4-2a+a^2) GIẢI GIÚP MIK ĐI

a) \(x^3-6x^2+12x-8=0\)

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

\(\left[\left(x-2\right)\left(x^2+2x+4\right)\right]-6x\left(x-2\right)=0\)

\(\left(x-2\right)\left[x^2+2x+4-6x\right]=0\)

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

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

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

\(\Rightarrow x-2=0\)

\(\Rightarrow x=2\)

b) \(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=13\)

\(4\left(x^2-6x+9\right)-\left(4x^2-1\right)-13=0\)

\(4x^2-24x+36-4x^2+1-13=0\)

\(-24x+24=0\)

\(-24x=-24\)

\(x=1\)

c) \(25x^2-6\left(x+1\right)^2=0\)

\(25x^2-6\left(x^2+2x+1\right)=0\)

\(25x^2-6x^2-12x-6=0\)

\(19x^2-12x-6=0\)

câu này có vẻ kq lẻ, xem lại đề em nhé

Chị cũng là fan của BTS à

Chị hâm mộ V đúng không

Em hâm mộ JungKook

d) (x - 2)^2 = 1 

= x = 2 + 1 = 3  

c) (x^2 + 1). (x + 2011) = 0

Tim x:

a) x^2 + 2x = 0 

= \(x^2+2x=0\)

= \(x^2=0:2=0\)

b) (x - 3) + 2x^2 - 6x = 0

Rút gọn thừa số chung : 

\(2x^2-5x-3=0\)

x = \(\frac{-1}{2}\)x = 3

=\(x^2=0\)

=> x = 0 

Bạn Nguyễn Phương Trung làm đúng rồi, các bn tk cho bn ấy nha !

a) = (3x +1)² =0 

3x+1 =0

x = -1/3

b) = (5x)² -2² =0

(5x+2)(5x-2) = 0

5x+2 =0

x = -2/5

5x -2  =0

x= 2/5

xem đi rui lam tip

a) 9x² + 6x + 1 = 0  => (3x)² + 2 x 3x + 1 = 0  => (3x + 1)²  = 0  => 3x + 1 = 0  => x = \(\frac{-1}{3}\)

b) 25x² = 4  => x² = 4 : 25  => x² = 0,16  => x = 0,4 hoặc x = -0,4

c) 8 - 125x³ = 0  => 125x³ = 8  => x³ = 8 : 125  => x³ = \(\frac{8}{125}\)=> x = \(\frac{2}{5}\)

c) = 2³ - (5x)³ =0

(2-5x)(4 +10x +25x²) =0

2-5x=0

x = 2/5

4 + 10x +25x²  = 0 (máy tính giải dc)

d) = (( 2x+1) - 5(x+2))² =0

2x+1 -5x -10=0

3x= -9

x = -3

(hoàn toàn ad hđt đáng nhớ thui,bn à)

a: =>|7x-9|=5x-3

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(7x-9-5x+3\right)\left(7x-9+5x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(2x-6\right)\left(12x-12\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{3;1\right\}\)

b: =>|17x-5|=|17x+5|

=>17x-5=17x+5(vô lý) hoặc 17x-5=-17x-5

=>34x=0

hay x=0

c: =>|3x+4|=|4x-18|

=>4x-18=3x+4 hoặc 4x-18=-3x-4

=>x=22 hoặc 7x=14

=>x=22 hoặc x=2

*vn:vô nghiệm.

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

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

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

\(\Leftrightarrow x=\pm\sqrt{2}\)

-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).

b. \(16x^2-8x+5=0\)

\(\Leftrightarrow16x^2-8x+1+4=0\)

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

-Vậy S=∅.

c. \(2x^3-x^2-8x+4=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)

-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).

d. \(3x^3+6x^2-75x-150=0\)

\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)

\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)

-Vậy \(S=\left\{-2;\pm5\right\}\)

x4−3x3−2x2+6x+4=0x4−3x3−2x2+6x+4=0

⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0

⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0

⇔(x2−x−2)(x2−2x−2)=0⇔(x2−x−2)(x2−2x−2)=0

⇔(x+1)(x−2)(x−1−√3)(x−1+√3)=0⇔(x+1)(x−2)(x−1−3)(x−1+3)=0

⇔⎡⎢ ⎢ ⎢ ⎢⎣x=−1x=2x=1+√3x=1−√3

tl

x4−3x3−2x2+6x+4=0x4−3x3−2x2+6x+4=0

⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0⇔x4−2x3−2x2−x3+2x2+2x−2x2+4x+4=0

⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0⇔x2(x2−2x−2)−x(x2−2x−2)−2(x2−2x−2)=0

⇔(x2−x−2)(x2−2x−2)=0⇔(x2−x−2)(x2−2x−2)=0

⇔(x+1)(x−2)(x−1−√3)(x−1+√3)=0⇔(x+1)(x−2)(x−1−3)(x−1+3)=0

⇔⎡⎢ ⎢ ⎢ ⎢⎣x=−1x=2x=1+√3x=1−√3

^HT^

Ta thấy x=0 không là nghiệm của phương trình

chia cả 2 vế cho x^2 ta được:

PT <=> x^2-3x-6+3/x+1/(x^2)=0

       <=> (x^2-2+1/(x^2))-3(x-1/x)-4=0

      <=> (x-1/x)^2-3(x-1/x)-4=0

Đặt x-1/x=y

PT <=> y^2-3y-4=0

     <=> y=-4 hoặc y=1

Tại y=-4 , ta có x+1/x+4=0

                       <=> x^2+4x+1=0

                       <=> x=-2+ √3 hoăc x=-2-  √ 3

Tại y=1 ta có x^2-x-1=0

                 <=> x=(1- √  5)/2 hoặc x=(1+  √5)/2