a) (5x - 1)(2x + 1) = (5x -1)(x + 3)

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

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

<=> (5x - 1)(x - 2) = 0

<=> \(\orbr{\begin{cases}5x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0,2\\x=2\end{cases}}\)

Vậy x = 0,2 ; x = 2 là nghiệm phương trình

b) x³ - 5x² - 3x + 15 = 0

<=> x²(x - 5) - 3(x - 5) = 0

<=> (x² - 3)(x - 5) = 0

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

<=> \(x-\sqrt{3}=0\text{ hoặc }x+\sqrt{3}=0\text{ hoặc }x-5=0\)

<=> \(x=\sqrt{3}\text{hoặc }x=-\sqrt{3}\text{hoặc }x=5\)

Vậy \(x\in\left\{\sqrt{3};\sqrt{-3};5\right\}\)là giá trị cần tìm

c) (x - 3)² - (5 - 2x)² = 0

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

<=> (-x + 2)(3x - 8) = 0

<=> \(\orbr{\begin{cases}-x+2=0\\3x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{8}{3}\end{cases}}\)

Vậy tập nghiệm phương trình \(S=\left\{2;\frac{8}{3}\right\}\)

d) x³ + 4x² + 4x = 0

<=> x(x² + 4x + 4) = 0

<=> x(x + 2)² = 0

<=> \(\orbr{\begin{cases}x=0\\\left(x+2\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)

Vậy tập nghiệm phương trình S = \(\left\{0;-2\right\}\)

Ví dụ cho bạn một bài, còn lại tương tự.

a)Ta có: \(3x^4-5x^3+8x^2-5x+3\)

\(=3x^2\left(x-\frac{5}{6}\right)^2+\frac{71}{12}\left(x-\frac{30}{71}\right)^2+\frac{138}{71}>0\)

Vậy phương trình vô nghiệm.

tth_new bạn làm hết ra đc ko. mình đọc không hiểu đc

a, ( 8x + 5 )( 4x + 3 )( 2x + 1 ) = 9

<=> ( 8x + 5 )[ 2( 4x+3)] [ 4 ( 2x+1 )] = 9* 2 * 4

<=> (8x+5)(8x+6)(8x+4) = 72

Đặt 8x+5 = y ta có phương trình tương đương :

y ( y -1 ) ( y+1) = 72

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

b, Tương tự phần a nhé

c, x^3 + 5x^2 + 5x + 2=0 

<=> x^3 + 1 + 5x^2 + 5x + 1 = 0

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

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

1. \(\left(x-4\right)^2-25=0\)

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

<=> (x+1)(x-9) = 0

<=> \(\left[\begin{matrix}x+1=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-1\\x=9\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = {-1;9}

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

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

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

<=> \(\left[\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}2x=1\\x=-1\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0.5\\x=-1\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = {-1 ; 0,5}

3. \(x^2+6x+9=4x^2\)

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

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

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

<=> \(\left[\begin{matrix}3x+3=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}3x=-3\\x=3\end{matrix}\right.\Leftrightarrow}\left[\begin{matrix}x=-1\\x=3\end{matrix}\right.\) Vậy phương trình có tập nghiệm S = {-1 ; 3}

4. (2x-5)(x+11) = (5-2x)(2x+1)

<=> (2x-5)(x+11) = - (2x-5)(2x+1)

<=> x + 11 = -2x - 1

<=> x+2x = -12

<=> 3x = -12

<=> x = -4

Vậy phương trình có một nghiệm duy nhất là x = -4

5. \(2x^2+5x+3=0\)

<=> \(2x^2+2x+3x+3=0\)

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

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

<=> \(\left[\begin{matrix}x+1=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-1\\2x=-3\end{matrix}\right.\Leftrightarrow}\left[\begin{matrix}x=-1\\x=\frac{-3}{2}\end{matrix}\right.\) Vậy phương trình có tập nghiệm S = { -1 ; -3/2 }

1) (x-4)^2-25=0

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

\(\Leftrightarrow\left[\begin{matrix}x=-1\\x=9\end{matrix}\right.\)

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

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

<=> \(\left[\begin{matrix}x=\frac{1}{2}\\x=4\end{matrix}\right.\)

3) x^2+6x+9=4x^2

<=> 3x^2 -6x-9=0

<=> x^2 -2x -3=0

<=> x^2 -3x+x-3=0

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

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

=>\(\left[\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

4) (2x-5)(x+11)=(5-2x)(2x+1)

-(5-2x)(x+11)-(5-2x)(2x+1)=0

(5-2x)(x+11+2x+1)=0

=>\(\left[\begin{matrix}x=\frac{5}{2}\\x=-4\end{matrix}\right.\)

5)2x^2+5x+3=0

2x^2+2x+3x+3=0

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

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

=>\(\left[\begin{matrix}x=-1\\x=\frac{-3}{2}\end{matrix}\right.\)

a) Ta có: \(x^2-3x+2=0\)

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

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

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

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

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

Vậy: \(x\in\left\{1;2\right\}\)

b) Ta có: \(-x^2+5x-6=0\)

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

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

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

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

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

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

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

Vậy: x∈{2;3}

c) Ta có: \(4x^2-12x+5=0\)

\(\Leftrightarrow4x^2-10x-2x+5=0\)

⇔(4x²-10x)-(2x-5)=0

\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{1}{2};\frac{5}{2}\right\}\)

d) Ta có: \(2x^2+5x+3=0\)

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

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

\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)

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

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

Vậy: \(x\in\left\{-1;\frac{-3}{2}\right\}\)

e) Ta có: \(x^3+2x^2-x-2=0\)

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

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

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

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

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

Vậy: \(x\in\left\{-2;1;-1\right\}\)

g) Ta có: \(\left(3x-1\right)^2-5\left(2x+1\right)^2+\left(6x-3\right)\left(2x+1\right)=\left(x-1\right)^2\)

\(\Leftrightarrow9x^2-6x+1-20x^2-20x-5+12x^2-3-x^2+2x-1=0\)

\(\Leftrightarrow-24x-8=0\)

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

⇔3x+1=0

\(\Leftrightarrow3x=-1\)

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

Vậy: \(x=-\frac{1}{3}\)

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

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

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

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

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

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

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

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

Vậy S = {2; 1; \(\frac{1}{2}\)}

i) \(x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)

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

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

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

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

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

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\\left(x+\frac{1}{2}\right)^2=\frac{-23}{4}\left(loai\right)\end{matrix}\right.\)

Vậy S = {1;-2}

1) \(x^4-8x^3+11x^2+8x-12=0\)

\(\Leftrightarrow x^4-x^3-7x^3+7x^2+4x^2-4x+12x-12=0\)

\(\Leftrightarrow x^3\left(x-1\right)-7x^2\left(x-1\right)+4x\left(x-1\right)+12\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-7x^2+4x+12\right)=0\)

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

\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+1\right)-8x\left(x+1\right)+12\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2-8x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2-2x-6x+12\right)=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\\x=6\end{matrix}\right.\)

Vậy ...

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

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\pm\sqrt{3}\\\end{matrix}\right.\)

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

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

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

c/ \(\Leftrightarrow3x^3-15x^2+9x+x^2-5x+3=0\)

\(\Leftrightarrow3x\left(x^2-5x+3\right)+\left(x^2-5x+3\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(x^2-5x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{3}\\x=\frac{5\pm\sqrt{13}}{2}\end{matrix}\right.\)

d/ \(x\left(x^2+6x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\pm\sqrt{14}\end{matrix}\right.\)

1) \(x^4-6x^3-x^2+54x-72=0\)

\(\Leftrightarrow x^3\left(x-2\right)-4x^2\left(x-2\right)-9x\left(x-2\right)+36\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3-4x^2-9x+36\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-4\right)-9\left(x-4\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x^2-9\right)=0\)

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

Tự làm nốt...

2) \(x^4-5x^2+4=0\)

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

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

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

Tự làm nốt...

\(x^4-2x^3-6x^2+8x+8=0\)

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

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

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

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

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left[\left(x-1\right)^2-\left(\sqrt{3}\right)^2\right]=0\)

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

...

\(2x^4-13x^3+20x^2-3x-2=0\)

\(\Leftrightarrow2x^3\left(x-2\right)-9x^2\left(x-2\right)+2x\left(x-2\right)+\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x^3-9x^2+2x+1\right)=0\)

Bí

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