Bài làm:

Ta có: \(\left(4x-1\right)^2-\left(4x+1\right)\left(x-2\right)=12\)

\(\Leftrightarrow16x^2-8x+1-4x^2+7x+2-12=0\)

\(\Leftrightarrow12x^2-x-9=0\)

\(\Leftrightarrow12\left(x^2-\frac{1}{12}x+\frac{1}{576}\right)-\frac{433}{48}=0\)

\(\Leftrightarrow\left[2\sqrt{3}\left(x-\frac{1}{24}\right)\right]^2-\left(\frac{\sqrt{433}}{\sqrt{48}}\right)^2=0\)

\(\Leftrightarrow\left[2\sqrt{3}\left(x-\frac{1}{24}\right)-\sqrt{\frac{433}{48}}\right]\left[2\sqrt{3}\left(x-\frac{1}{24}\right)+\sqrt{\frac{433}{48}}\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}2\sqrt{3}\left(x-\frac{1}{24}\right)=\sqrt{\frac{433}{48}}\\2\sqrt{3}\left(x-\frac{1}{24}\right)=-\sqrt{\frac{433}{48}}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{24}=\frac{\sqrt{433}}{24}\\x-\frac{1}{24}=\frac{-\sqrt{433}}{24}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{433}+1}{24}\\x=\frac{1-\sqrt{433}}{24}\end{cases}}\)

Vậy tập nghiệm của PT \(S=\left\{\frac{1-\sqrt{433}}{24};\frac{\sqrt{433}+1}{24}\right\}\)

\(\left(4x-1\right)^2-\left(4x+1\right)\left(x-2\right)=12\) 

\(\Leftrightarrow\left(4x-1\right)\left(4x-1-x+2\right)=12\)

\(\Leftrightarrow\left(4x-1\right)\left(3x+1\right)=12\)

Rồi bạn tự tính tiếp nhớ :3

Học tốt 

\(\left(4x-1\right)^2-\left(4x+1\right)\left(x-2\right)=12\)

\(\Leftrightarrow16x^2-8x+1-4x^2+8x-x+2=12\)

\(\Leftrightarrow12x^2-x-9=0\)( vô nghiệm )

a: \(x^3-4x^2-x+4=0\)

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

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

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

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

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

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

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

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

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

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

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

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

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

=>(x-2)(x-6)=0

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

a.\(\left(4x-1\right)-\left(4x+1\right).\left(x-2\right)=12\)

\(\Leftrightarrow4x-1-\left(4x^2-7x-2\right)-12=0\)

\(\Leftrightarrow4x-1-4x^2+7x+2-12=0\)

\(\Leftrightarrow-4x^2+11x-11=0\)

\(\Rightarrow4x^2-11x+11=0\)

\(\Leftrightarrow\left(2x\right)^2-2.2x.\frac{11}{4}+\frac{11^2}{4^2}-\frac{11^2}{4^2}+11=0\)

\(\Leftrightarrow\left(2x-\frac{11}{4}\right)^2+\frac{55}{16}=0\)( VÔ LÝ )

VẬY KHÔNG CÓ GIÁ TRỊ NÀO CỦA x THỎA MÃN PT ĐÃ CHO

b. \(\left(2x-3\right).\left(2x+1\right)-\left(2x-2\right)^2=15\)
\(\Leftrightarrow4x^2-4x-3-4x^2+8x-4-15=0\)

\(\Leftrightarrow4x-22=0\)\

\(\Leftrightarrow x=\frac{11}{2}\)

VẬY PT CÓ NGHIỆM x= 11/2

a) \(\left(4x-1\right)-\left(4x+1\right)\left(x-2\right)=12\)

\(\Leftrightarrow4x-1-\left(4x^2-7x-2\right)=12\)

\(\Leftrightarrow4x-1-4x^2+7x+2=12\)

\(\Leftrightarrow4x^2-11x+11=0\)( Pt vô nghiệm )

b) \(\left(2x-3\right)\left(2x+1\right)-\left(2x-2\right)^2=15\)

\(\Leftrightarrow\left(4x^2-4x-3\right)-\left(4x^2-8x+4\right)=15\)

\(\Leftrightarrow4x=22\)

\(\Leftrightarrow x=\frac{11}{2}\)

(4x - 1) - (4x + 1)(x - 2) = 12

=> 4x - 1 - 4x² - 7x - 2 = 12

=> (4x - 7x) + (- 1 - 2) - 4x² = 12

=> -3x - 3 - 4x² = 12

=> -3x - 4x² = 15

=> không tồn tại x 

b. (2x - 3)(2x + 1) - (2x - 2)(2x - 2) = 15

=> 2x(2x + 1) - 3(2x + 1) - 2x(2x - 2) + 2(2x - 2) = 15

=> 4x² + 2x - 6x - 3 - 4x² + 4x - 4x - 4 = 15

=> (4x² - 4x²) + (2x - 6x + 4x - 4x) + (-3 - 4) = 15

=> -4x - 7 = 15

=> -4x = 22

=> x = \(-\frac{11}{2}\)

a, \(\left(4x-1\right)-\left(4x+1\right)\left(x-2\right)=12\)

\(\Leftrightarrow4x-1-4x^2+8x-x+2=12\)

\(\Leftrightarrow11x+1-4x^2=12\)

\(\Leftrightarrow11x-11-4x^2=0\)( vô nghiệm )

b, \(\left(2x-3\right)\left(2x+1\right)-\left(2x-2\right)^2=15\)

\(\Leftrightarrow4x^2+2x-6x-3-4x^2+8x-4=15\)

\(\Leftrightarrow4x-7=15\Leftrightarrow4x=22\Leftrightarrow x=\frac{11}{2}\)

Ây da,làm lại câu b nè

b. (2x - 3)(2x + 1) - (2x - 2)(2x - 2) = 15

=> 2x(2x + 1) - 3(2x + 1) - 2x(2x - 2) + 2(2x - 2) = 15

=> 4x² + 2x - 6x - 3 - 4x² + 4x + 4x - 4 = 15

=> (4x² - 4x²) - (2x - 6x + 4x + 4x) + (-3 - 4) = 15

=>  4x - 7 = 15

=> 4x = 22

=> x = 11/2

Bài 1.

1) ( 2x + 1 )³ - ( 2x + 1 )( 4x² - 2x + 1 ) - 3( 2x - 1 ) = 15

<=> 8x³ + 12x² + 6x + 1 - [ ( 2x )³ - 1³ ] - 6x + 3 = 15

<=> 8x³ + 12x² + 4 - 8x³ + 1 = 15

<=> 12x² + 15 = 15

<=> 12x² = 0

<=> x = 0

2) x( x - 4 )( x + 4 ) - ( x - 5 )( x² + 5x + 25 ) = 13

<=> x( x² - 16 ) - ( x³ - 5³ ) = 13

<=> x³ - 16x - x³ + 125 = 13

<=> 125 - 16x = 13

<=> 16x = 112

<=> x = 7

Bài 2.

A = ( x + 5 )( x² - 5x + 25 ) - ( 2x + 1 )³ - 28x³ + 3x( -11x + 5 )

= x³ + 5³ - ( 8x³ + 12x² + 6x + 1 ) - 28x³ - 33x² + 15x

= -27x³ + 125 - 8x³ - 12x² - 6x - 1 - 33x² + 15x

= -33x³ - 45x² + 9x + 124 ( có phụ thuộc vào biến )

B = ( 3x + 2 )³ - 18x( 3x + 2 ) + ( x - 1 )³ - 28x³ + 3x( x - 1 )

= 27x³ + 54x² + 36x + 8 - 54x² - 36x + x³ - 3x² + 3x - 1 - 28x³ + 3x² - 3x

= 7 ( đpcm )

C = ( 4x - 1 )( 16x² + 4x + 1 ) - ( 4x + 1 )³ + 12( 4x + 1 )³ + 12( 4x + 1 ) - 15

= ( 4x )³ - 1³ - [ ( 4x + 1 )³ - 12( 4x + 1 )³ - 12( 4x + 1 ) ] - 15

= 64x³ - 1 - ( 4x + 1 )[ ( 4x + 1 )² - 12( 4x + 1 )² - 12 ] - 15

= 64x³ - 16 - ( 4x + 1 )[ 16x² + 8x + 1 - 12( 16x² + 8x + 1 ) - 12 ]

= 64x³ - 16 - ( 4x + 1 )( 16x² + 8x - 11 - 192x² - 96x - 12 )

= 64x³ - 16 - ( 4x + 1 )( -176x² - 88x - 23 )

= 64x³ - 16 - ( -704x³ - 528x² - 180x - 23 )

= 64x³ - 16 + 704x³ + 528x² + 180x + 23 

= 768x³ + 528x² + 180x + 7 ( có phụ thuộc vào biến )

a)  \(x^3\)\(-\)\(\frac{1}{4}x\)\(=\)\(0\)

\(x\left(x^2-\frac{1}{4}\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2-\frac{1}{4}=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2=0,5^2\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=+-0,5\end{cases}}\)

Vậy .............................

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

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

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

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

Vậy ................................

c)  \(x^2\)\(\left(x-3\right)\)\(+\)\(12\)\(-\)\(4x\)\(=\)\(0\)

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

\(\left(x^2-4\right)\left(x-3\right)\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x-3=0\end{cases}-4=0}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x=3\end{cases}=2^2}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=+-2\\x=3\end{cases}}\)

a)\(x^3-\frac{1}{4}x=0\)

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

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

\(\Leftrightarrow\hept{\begin{cases}x=0\\x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)

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

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

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

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