2x³ - 3x² + 3x - 1 

= x³ + (x³ - 3x² + 3x - 1)

= x³ + (x - 1)³

= (x + x - 1)[x² - x(x - 1) + (x - 1)²]

= (2x - 1)(x² - x + 1) 

à_à mình chép sai đề 

sửa nhé : \(2-x=2\left(x-2\right)^3\Leftrightarrow2\left(x-2\right)^3+x-2=0\)

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

\(2-x=2\left(x-3\right)^3\Leftrightarrow2-x=2\left(x^3-9x^2+27x-27\right)\)

\(\Leftrightarrow2x^3-18x^2+54x-54=2-x\Leftrightarrow2x^3-18x^2+55x-56=0\)

xem lại đề nhé 

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

\(\Leftrightarrow\left(x^2-1\right)\left(x-5\right)=0\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-5\right)=0\Leftrightarrow x\pm1;x=5\)

Trả lời:

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=\pm1\end{cases}}}\)

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

tui cần gấp

\(x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=\left[\left(x+y\right)^2-2xy\right]^2-2x^2y^2\)

Thay vào ta được : 

\(\left(64-2.15\right)^2-2\left(15\right)^2=\left(64-30\right)^2-2\left(15\right)^2\)

\(=34^2-2.225=1156-450=706\)

4(2-x)²+xy-2y

= 4(x-2)²+y(x-2)

=(x-2)(4x-8+y)

* Nguồn : Hoc 24 *

Trả lời:

4 ( 2 - x )² + xy - 2y 

= 4 ( x - 2 )² + ( xy - 2y )

= 4 ( x - 2 )² + y ( x - 2 )

= ( x - 2 )[ ( 4 ( x - 2 ) + y ]

= ( x - 2 )( 4x - 8 + y )

Trả lời :

3( x - y ) - 5x( x - y )

= ( 3 - 5x ) . ( x - y )

4x^2-6x = 2x ( 2x - 3 )

4x² - 6x

= 2x . 2x - 2x . 3

= 2x( 2x - 3)

c, \(C=3\sqrt{a^2+1}-4\sqrt{4a^2+4}+2\sqrt{9a^2+9}\)

\(=3\sqrt{a^2+1}-8\sqrt{a^2+1}+6\sqrt{a^2+1}=\sqrt{a^2+1}\)

Trả lời:

\(C=3\sqrt{a^2+1}-4\sqrt{4a^2+4}+2\sqrt{9a^2+9}\)

\(=3\sqrt{a^2+1}-4\sqrt{4\left(a^2+1\right)}+2\sqrt{9\left(a^2+1\right)}\)

\(=3\sqrt{a^2+1}-4\sqrt{2^2\left(a^2+1\right)}+2\sqrt{3^2\left(a^2+1\right)}\)

\(=3\sqrt{a^2+1}-4.2\sqrt{a^2+1}+2.3\sqrt{a^2+1}\)

\(=3\sqrt{a^2+1}-8\sqrt{a^2+1}+6\sqrt{a^2+1}\)

\(=\left(3-8+6\right).\sqrt{a^2+1}\)

\(=\sqrt{a^2+1}\)

Ta có: (a-b)^2=a^2-2ab+b^2=9

=>ab=1

Mặt khác, a^3-b^3= (a-b)(a^2-ab+b^2)= 3.6=18