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

bn kiểm tra giúp mk đề 2 câu cuối , mk làm ko ra

Khó quá,hay quá !!!

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) 6x² - 5x + 3 = 2x - 3x(2 - x)

<=> 6x² - 5x + 3 = 2x - 6x + 3x²

<=> 6x² - 5x + 3 = -4x + 3x²

<=> 6x² - 5x + 3 + 4x - 3x² = 0

<=> 3x² - x + 3 = 0

=> Pt vô nghiệm

b) 25x² - 9 = (5x + 3)(2x + 1)

<=> 25x² - 9 = 10x² + 5x + 6x + 3

<=> 25x² - 9 = 10x² + 11x + 3

<=> 25x² - 9 - 10x² - 11x - 3 = 0

<=> 15x² - 12 - 11x = 0

<=> 15x² + 9x - 20x - 12 = 0

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

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

<=> 5x + 3 = 0 hoặc 3x - 4 = 0

<=> x = -3/5 hoặc x = 4/3

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

3x+5 =0 hoặc 4-3x=0

3x=-5 hoặc 3x=-4

x=-5/3 hoặc x=-4/3

9(3x-2)=x(2-3x)

9(3x-2)-x(3x-2)=0

(3x-2)(9-x)=0

3x-2=0 hoặc 9-x=0

3x=2 hoặc x= -9

x =2/3 hoặc x=-9 

vậy x =2/3 ; x= -9

25x^2 - 2=0

(5x)^2 -√2^2=0

(5x-√2)(5x+√2)=0

5x=√2 hoặc 5x = -√2

x=√2/5 hoặc x= -√2/5

vậy x=√2/5 ; x=-√2/5

1) \(\left(5x-4\right)\left(4x+6\right)=0\)

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

Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{4}{5};\dfrac{3}{2}\right\}\)

2) \(\left(4x-10\right)\left(24+5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-24}{5}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{5}{2};\dfrac{-24}{5}\right\}\)

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

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

Vậy phương trình có tập nghiệm S = \(\left\{3;\dfrac{-1}{2}\right\}\)

a) Gần giống cho nó giống luôn.

cần thêm (-x^3+2x^2-x) là giống

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

\(\left(x-1\right)^2\left[\left(x-1\right)^2+x\right]\)

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

Nghiệm duy nhất: x=1

câu d

Với dạng bài này ta chỉ việc chia hoocne là ra nhé!

\(C1:x^4+x^3-8x^2-9x-9=0\\ \Leftrightarrow\left(x-3\right)\left(x^3+4x^2+4x+3\right)\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x^2+x+1\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\\x^2+x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x^2+x+1=0\left(VN\right)\end{matrix}\right.\)

\(C2:x^4+2x^3-3x^2-8x-4=0\\ \Leftrightarrow\left(x+1\right)\left(x^3+x^2-4x-4\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+1\right)\left(x^2-4\right)=0\\ \Leftrightarrow\left(x+1\right)^2\left(x^2-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)

\(a,\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

\(\left(x-1\right)\left(5x+3-3x+8\right)=0\)

\(\left(x-1\right)\left(2x+11\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\2x=-11\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-\frac{11}{2}\end{cases}}}\)

\(b,3x\left(25x+15\right)-35\left(5x+3\right)=0\)

\(15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\left(5x+3\right).5\left(3x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5\left(3x-7\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-3\\3x-7=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{7}{3}\end{cases}}}\)

\(c,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\left(3x-2\right)\left(2-5x\right)+\left(3x-2\right)\left(x+11\right)=0\)

\(\left(3x-2\right)\left(2-5x+x+11\right)=0\)

\(\left(3x-2\right)\left(13-4x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\13-4x=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=2\\4x=13\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{13}{4}\end{cases}}}\)

còn đâu tự lm lười :_#