Bài làm:

Đặt \(\hept{\begin{cases}x-1=a\\x+2=b\end{cases}}\Rightarrow a+b=2x+1\)

\(Pt\Leftrightarrow a^3+b^3=\left(a+b\right)^3\)

\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-a^3-b^3=0\)

\(\Leftrightarrow3ab\left(a+b\right)=0\)

\(\Leftrightarrow3\left(x-1\right)\left(x+2\right)\left(2x+1\right)=0\)

Đến đây giải PT tích ra ta được: \(x\in\left\{-2;-\frac{1}{2};1\right\}\)

a) 4x³ - 14x² + 6x

= 2x( 2x² - 7x + 3 )

= 2x( 2x² - 6x - x + 3 )

= 2x[ 2x( x - 3 ) - ( x - 3 ) ]

= 2x( 2x - 1 )( x - 3 )

b) x⁴ + 4

= x⁴ + 4 + 4x² - 4x²

= ( x⁴ + 4x² + 4 ) - 4x²

= ( x² + 2 )² - ( 2x )²

= ( x² + 2 + 2x )( x² + 2 - 2x )  

a) 

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

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

b) 

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

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

a) \(A⋮B\Leftrightarrow n\ge2\)\(\left(n\in Z\right)\)

b) \(A⋮B\Leftrightarrow2n\ge2\Leftrightarrow n\ge1\)\(\left(n\in Z\right)\)

Bg

x³ - 3x - x(x² + 5x + 2) = 0

(x² - 3).x - x(x² + 5x + 2) = 0

x.[x² - 3 - (x² + 5x + 2)] = 0

=> x = 0 hoặc x² - 3 - (x² + 5x + 2) = 0

Xét x² - 3 - (x² + 5x + 2) = 0:

=> x² - 3 - x² - 5x - 2 = 0

=> x² - x² - (3 + 2) - 5x = 0

=> -5 - 5x = 0

=> 5x = -5

=> x = -5 : 5

=> x = -1

Vậy x = 0 hoặc x = -1

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

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

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

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

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

\(P=1:\left(\frac{x+2}{\left(\sqrt{x}\right)^3-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

\(P'=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)    (​     \(P=\frac{1}{P'}\))

\(P'=\frac{x+2-\left(x-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(P'=\frac{-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right) \left(x+\sqrt{x}+1\right)}=\frac{\left(1-\sqrt{x}\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\)

=> \(P=-\frac{x+\sqrt{x}+1}{\sqrt{x}+2}\)

Hermit ĐKXĐ đâu bạn eiii ??

pt <=> \(x^2+21x+90=136\)

<=> \(x^2+21x-46=0\)

<=> \(x^2-2x+23x-46=0\)

<=> \(\left(x-2\right)\left(x+23\right)=0\)

<=> \(\orbr{\begin{cases}x=2\\x=-23\end{cases}}\)

Vậy tập hợp nghiệm của pt là S={2;-23}.

\(\left(15+x\right)\left(6+x\right)=136\)

\(\Leftrightarrow90+15x+6x+x^2=136\)

\(\Leftrightarrow-46+21x+x^2=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+23\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-23\end{cases}}\)

pt <=> \(16x^2-8x+1-\left(4x^2-7x-2\right)=12\)

<=>  \(12x^2-x+3=12\)

<=> \(12x^2-x-9=0\)

=> Bạn bấm máy tính tìm nốt x nha

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

<=> 16x² - 8x + 1 - ( 4x² - 7x - 2 ) = 12

<=> 16x² - 8x + 1 - 4x² + 7x + 2 = 12

<=> 12x² - x + 3 = 12

<=> 12x² - x + 3 - 12 = 0

<=> 12x² - x - 9 = 0

\(\Delta=b^2-4ac=\left(-1\right)^2-4\cdot12\cdot\left(-9\right)=1+432=433\)( lại không muốn xài Delta đâu nhưng bí quá ;-; )

\(\Delta>0\)nên phương trình đã cho có hai nghiệm phân biệt :

\(\hept{\begin{cases}x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{1+\sqrt{433}}{2\cdot12}=\frac{1+\sqrt{433}}{24}\\x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{1-\sqrt{433}}{2\cdot12}=\frac{1-\sqrt{433}}{24}\end{cases}}\)

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

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

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

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 )

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 \(S=\left\{\frac{1-\sqrt{433}}{24};\frac{\sqrt{433}+1}{24}\right\}\)