a: \(\left(x^2+x\right)^2+2\left(x^2+x\right)-8=0\)

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

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

hay \(x\in\left\{-2;1\right\}\)

b: \(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x+2\right)\left(x+4\right)+24=0\)

\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x-12\right)+24=0\)

\(\Leftrightarrow\left(x^2+x\right)^2-14\left(x^2+x\right)+48=0\)

\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x-8\right)=0\)

hay \(x\in\left\{-3;2;\dfrac{-1+\sqrt{33}}{2};\dfrac{-1-\sqrt{33}}{2}\right\}\)

\(a,\Leftrightarrow4x^2-24x+36-4x^2+1=10\\ \Leftrightarrow-24x=-27\Leftrightarrow x=\dfrac{9}{8}\\ b,\Leftrightarrow x\left(x^2-25\right)=0\\ \Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

\(a,4.\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)

\(\Leftrightarrow4.\left(x^2-6x+9\right)-\left(2x^2\right)-1^2=10\)

\(\Leftrightarrow4x^2-24x+36-4x^2+1=10\)

\(\Leftrightarrow-24x+27=10\)

\(\Leftrightarrow-24x=-27\)

\(\Leftrightarrow x=\dfrac{27}{24}\)

Vậy \(x=\dfrac{27}{24}\)

\(b,x^3-25x=0\)

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

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

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

Vậy \(x\in\left\{0;\pm5\right\}\)

a: \(6x^4+25x^3+12x^2-25x+6=0\)

\(\Leftrightarrow6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)

\(\Leftrightarrow\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(6x^3+18x^2-5x^2-15x+x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(6x^2-5x+1\right)=0\)

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

hay \(x\in\left\{-2;-3;\dfrac{1}{3};\dfrac{1}{2}\right\}\)

b: \(x^5+2x^4+3x^3+3x^2+2x+1=0\)

\(\Leftrightarrow x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0\)

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

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

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

=>x+1=0

hay x=-1

c: \(x^2\left(x^2+2\right)-x^2-2=0\)

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

=>x=1 hoặc x=-1

b. sửa đề

\(6x^4+25x^3+12x-25x^2+6=0\)

\(\Leftrightarrow6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)

\(\Leftrightarrow6x^3\left(x+2\right)+13x^2\left(x+2\right)-14x\left(x+2\right)+3\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)

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

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

Vậy........

Bài 1 : Giải phương trình

a) (x + 3)⁴ + (x + 5)⁴ = 16

Đặt : x + 3 = t

=> x + 5 = x + 3 + 2 = t + 2

Thay x + 3 = t và x + 5 = t + 2 vào phương trình, ta có :

t⁴ + (t + 2)⁴ = 16

<=> 2t⁴ + 8t³ + 24t² + 32t + 16 = 16

<=> 2(t⁴ + 4t³ + 12t² + 16t) = 0

<=> t⁴ + 4t³ + 12t² + 16t = 0

<=> (t + 2) . t . (t² + 2y + 4) = 0

TH1 : t = 0

TH2 : t + 2 = 0 <=> t = -2

TH3 : t² + 2y + 4 = 0 (vô nghiệm => loại)

Nên t = 0 hoặc t = -2

hay x + 3 = -2 hoặc x + 3 = 0

<=> x = -5 hoặc x = -3

\(S=\left\{-5;-3\right\}\)

b) 6x⁴ + 25x³ + 12x² - 25x + 6 = 0

<=> 6x⁴ + 12x³ + 13x³ + 26x² - 14x² - 28x + 3x + 6 = 0

<=> 6x³ (x + 2) + 13x² (x + 2) - 14x (x + 2) + 3(x + 2) = 0

<=> (x + 2)(6x³ + 13x² - 14x + 3) = 0

<=> (x + 2)(6x³ + 18x² - 5x² - 15x + x + 3) = 0

\(\Leftrightarrow\left(x+2\right)[6x^2\left(x+3\right)-5x\left(x+3\right)+\left(x+3\right)]=0\)

<=> (x + 2)(x + 3) (6x² - 5x + 1) = 0

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

TH1 : x + 2 = 0 <=> x = -2

TH2 : x + 3 = 0 <=> x = -3

TH3 : 2x - 1 = 0 <=> 2x = 1 <=> x = \(\dfrac{1}{2}\)

TH4 : 3x - 1 = 0 <=> 3x = 1 <=> 3x = \(\dfrac{1}{3}\)

\(S=\left\{-2;-3;\dfrac{1}{2};\dfrac{1}{3}\right\}\)

\(\text{a) }\left(x+3\right)^4+\left(x+5\right)^4=16\\ \Leftrightarrow\left(x^2+6x+9\right)^2+\left(x^2+10x+25\right)^2=16\\ \Leftrightarrow x^4+36x^2+81+12x^3+18x^2+108x+x^4+100x^2+625+20x^3+50x^2+500x=16\\ \Leftrightarrow2x^4+32x^3+204x^2+608x+690=0\\ \Leftrightarrow x^4+16x^3+102x^2+304x+345=0\\ \Leftrightarrow x^4+5x^3+11x^3+55x^2+47x^2+235x+373x+69x+345=0\\ \Leftrightarrow\left(x^4+5x^3\right)+\left(11x^3+55x^2\right)+\left(47x^2+235x\right)+\left(69x+345\right)=0\\ \Leftrightarrow x^3\left(x+5\right)+11x^2\left(x+5\right)+47x\left(x+5\right)+69\left(x+5\right)=0\\ \Leftrightarrow\left(x^3+11x^2+47x+69\right)\left(x+5\right)=0\\ \Leftrightarrow\left(x^3+3x^2+8x^2+24x+23x+69\right)\left(x+5\right)=0\\ \Leftrightarrow\left[\left(x^3+3x^2\right)+\left(8x^2+24x\right)+\left(23x+69\right)\right]\left(x+5\right)=0\\ \Leftrightarrow\left[x^2\left(x+3\right)+8x\left(x+3\right)+23\left(x+3\right)\right]\left(x+5\right)=0\\ \Leftrightarrow\left(x^2+8x+23\right)\left(x+3\right)\left(x+5\right)=0\)\(\Leftrightarrow\left(x^2+8x+16+7\right)\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left[\left(x+4\right)^2+7\right]\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x+5\right)=0\left(\text{Vì }\left(x+4\right)^2+7\ne0\right)\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

Vậy tập nghiệm phương trình là \(S=\left\{-3;-5\right\}\)

giúp tôi với

1) 2x⁴ - 9x³ + 14x² - 9x + 2 = 0

<=> (2x⁴ - 4x³) - (5x³ - 10x²) + (4x² - 8x) - (x - 2) = 0

<=> 2x³(x - 2) - 5x²(x - 2) + 4x(x - 2) - (x - 2) = 0

<=> (2x³ - 5x² + 4x - 1)(x - 2) = 0

<=> [(2x³ - 2x²) - (3x² - 3x) + (x - 1)](x - 2) = 0

<=> [2x²(x - 1) - 3x(x - 1) + (x - 1)](x - 2) = 0

<=> (2x² - 2x - x + 1)(x - 1)(x - 2) = 0

<=> (2x - 1)(x - 1)²(x - 2) = 0

<=> 2x - 1=0

hoặc x - 1 = 0

hoặc x - 2 = 0

<=> x = 1/2

hoặc x = 1

hoặc x = 2

Vậy S = {1/2; 1; 2}

1) \(2x^4-9x^3+14x^2-9x+2=0\)

\(\Leftrightarrow2x^4-2x^3-7x^3+7x^2+7x^2-7x-2x+2=0\)

\(\Leftrightarrow2x^3\left(x-1\right)-7x^2\left(x-1\right)+7x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x^3-7x^2+7x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[2\left(x^3-1\right)-7x\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[2\left(x-1\right)\left(x^2+x+1\right)-7x\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(2x^2+2x+2-7x\right)=0\)

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

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

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

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

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

hoặc \(2x-1=0\)

hoặc \(x-2=0\)

\(\Leftrightarrow\)\(x=1\)hoặc \(x=\frac{1}{2}\)hoặc \(x=2\)

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

2) \(6x^4+25x^3+12x^2-25x+6=0\)

\(\Leftrightarrow6x^4-3x^3+28x^3-14x^2+26x^2-13x-12x+6=0\)

\(\Leftrightarrow3x^3\left(2x-1\right)+14x^2\left(2x-1\right)+13x\left(2x-1\right)-6\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x^3+14x^2+13x-6\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x^3-x^2+15^2-5x+18x-6\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left[x^2\left(3x-1\right)+5x\left(3x-1\right)+6\left(3x-1\right)\right]=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x-1\right)\left(x^2+5x+6\right)=0\)

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

\(\Leftrightarrow\)\(2x-1=0\)

hoặc \(3x-1=0\)

hoặc \(x+2=0\)

hoặc \(x+3=0\)

\(\Leftrightarrow\)\(x=\frac{1}{2}\)hoặc \(x=\frac{1}{3}\)hoặc \(x=-2\)hoặc \(x=-3\)

Vậy tập nghiệm của phương trình là \(S=\left\{\frac{1}{2};\frac{1}{3};-2;-3\right\}\)

3) Ktra lại đề nhé :D

4) \(x^3-3x^2+3x+8=0\)

\(\Leftrightarrow2x^3+2x^2-5x^2-5x+8x+8=0\)

\(\Leftrightarrow2x^2\left(x+1\right)-5x\left(x+1\right)+8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2-5x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\2x^2-5x+8=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\left(TM\right)\\2\left(x-\frac{5}{4}\right)^2+\frac{39}{8}=0\left(L\right)\end{cases}}\)

Vậy x = -1

5) \(x^4+2x^3+x^2=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{0;-1\right\}\)

a) 9-64x^2=0

=>  64x^2  = 8

=>  \(x^2=\frac{8}{64}=\frac{1}{8}\)

=> \(x=\frac{1}{\sqrt{8}}\)

 b )   25x^2  -  3  =  0

=>  25x^2  =  3 

=>  \(x^2=\frac{3}{25}\)    

=>  \(x=\frac{\sqrt{3}}{5}\)           

C)  7  -  16x^2  =0

=>  16x^2   =  7

=>  \(x^2=\frac{7}{16}\)       

=>   \(x=\frac{\sqrt{7}}{4}\)    

d)  4x^2  -  (x-4)^2 = 0

=>  4x^2  - x^2 + 8x - 16 =0

=>  3x^2 + 8x -16  =  0 

=> ( 3x^2 + 12x ) - ( 4x  +16 ) =  0 

=>  3x( x + 4 ) - 4( x + 4 ) =  0 

=>( x + 4 )( 3x - 4 ) =  0 

=>   \(\orbr{\begin{cases}x+4=0\\3x-4=0\end{cases}}\)    

=>  \(\orbr{\begin{cases}x=-4\\x=\frac{4}{3}\end{cases}}\)         

e)  ( 3x + 4 )^2 - ( 2x - 5 )^2 = 0

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

=>   ( 5x -1 ) ( x + 9 )  = 0 

=>  \(\orbr{\begin{cases}5x-1=0\\x+9=0\end{cases}}\)     

=> \(\orbr{\begin{cases}x=\frac{1}{5}\\x=-9\end{cases}}\)            

Trả lời:

a, \(9-64x^2=0\)

\(\Leftrightarrow\left(3-8x\right)\left(3+8x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3-8x=0\\3+8x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{8}\\x=-\frac{3}{8}\end{cases}}}\)

Vậy x = 3/8; x = - 3/8 là nghiệm của pt.

b, \(25x^2-3=0\)

\(\Leftrightarrow\left(5x-\sqrt{3}\right)\left(5x+\sqrt{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x-\sqrt{3}=0\\5x+\sqrt{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{3}}{5}\\x=-\frac{\sqrt{3}}{5}\end{cases}}}\)

Vậy \(x=\pm\frac{\sqrt{3}}{5}\)

c, \(7-16x^2=0\)

\(\Leftrightarrow\left(\sqrt{7}-4x\right)\left(\sqrt{7}+4x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{7}-4x=0\\\sqrt{7}+4x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{7}}{4}\\x=-\frac{\sqrt{7}}{4}\end{cases}}}\)

Vậy \(x=\pm\frac{\sqrt{7}}{4}\)

d, \(4x^2-\left(x-4\right)^2=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\3x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=\frac{4}{3}\end{cases}}}\)

Vậy x = - 4; x = 4/3 là nghiệm của pt.

e, \(\left(3x+4\right)^2-\left(2x-5\right)^2=0\)

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

\(\Leftrightarrow\left(x+9\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+9=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-9\\x=\frac{1}{5}\end{cases}}}\)

Vậy x = - 9; x = 1/5 là nghiệm của pt.

\(6x^4+25x^3+12x^2-25x+6=0.\)

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

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

x²  - 1 + 3(6x + 6)=0

\(\left(3x+5\right)\left(7-2x\right)+6x\left(x+4\right)=0\)

\(\Leftrightarrow 21x-6x^2+35-10x+6x^2+24x=0\)

\(\Leftrightarrow35x+35=0\)

\(\Leftrightarrow35\left(x+1\right)=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

\(x^3-25x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-25=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)

Vậy ......

Toán mik ghi nhầm ko phải ta

em ghi nhầm môn em có thể đăng lại ko em

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

Vậy \(x\in\left\{-2;3\right\}\) là nghiệm của pt.

z) \(3x^2-5x-8=0\\ \Leftrightarrow\left(3x-8\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-1\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{8}{3};-1\right\}\) là nghiệm của pt.

j) \(25x^2-4=0\\ \Leftrightarrow\left(5x-2\right)\left(5x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{-2}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{2}{5};\dfrac{-2}{5}\right\}\) là nghiệm của pt.

r) \(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Vậy \(x\in\left\{-3;2\right\}\) là nghiệm của pt.

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

Vậy \(x\in\left\{-1;1;3\right\}\) là nghiệm của pt.