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

Trả lời:

a, \(\left(3x+1\right)\left(x-3\right)-x\left(3x-14\right)=15\)

\(\Leftrightarrow3x^2-9x+x-3-3x^2+14x=15\)

\(\Leftrightarrow6x-3=15\)

\(\Leftrightarrow6x=18\)

\(\Leftrightarrow x=3\)

Vậy x = 3 là nghiệm của pt.

b, \(\left(x-3\right)^2=9-x^2\)

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

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

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

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

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

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

Vậy x = 3; x = 0 là nghiệm của pt.

c, \(\left(2x-\frac{1}{2}\right)^2-\left(1-2x\right)^2=2\)

\(\Leftrightarrow4x^2-2x+\frac{1}{4}-\left(1-4x+4x^2\right)=2\)

\(\Leftrightarrow4x^2-2x+\frac{1}{4}-1+4x-4x^2=2\)

\(\Leftrightarrow2x-\frac{3}{4}=2\)

\(\Leftrightarrow2x=\frac{11}{4}\)

\(\Leftrightarrow x=\frac{11}{8}\)

Vậy x = 11/8 là nghiệm của pt.

d, \(4x^2+4x-3=0\)

\(\Leftrightarrow4x^2-2x+6x-3=0\)

\(\Leftrightarrow2x\left(2x-1\right)+3\left(2x-1\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{3}{2}\end{cases}}}\)

Vậy x = 1/2; x = - 3/2 là nghiệm của pt.

Trả lời:

a, x² + 2022x - xy - 2022y 

= ( x² - xy ) + ( 2022x - 2022y )

= x ( x - y ) + 2022 ( x - y )

= ( x - y )( x + 2022 )

b, sai đề

c, 4a² - 12ab + 9b² - 25 

= ( 9b² - 12ab + 4a² ) - 25

= ( 3b - 2a )² - 25

= ( 3b - 2a - 5 )( 3b - 2a + 5 )

d, x³ + 8 - ( 2 + x )( 5 - 2x )

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

= ( x + 2 )( x² - 2x + 4 - 5 + 2x )

= ( x + 2 )( x² - 1 )

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

mình đánh máy vội nên sai mn đừng trả lời câu này nha !!!!