a, \(\frac{1}{25}x^2-\frac{1}{9}y^2=\left(\frac{1}{5}x\right)^2-\left(\frac{1}{3}y\right)^2=\left(\frac{1}{5}x-\frac{1}{3}y\right)\left(\frac{1}{5}x+\frac{1}{3}y\right)\)

b, \(-\left(4x^2-20x+25\right)=-\left[\left(2x\right)^2-2.2x.5+5^2\right]=-\left(2x-5\right)^2\)

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

e, \(x^6-y^6=\left(x^3-y^3\right)\left(x^3+y^3\right)=\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\)

3x³ + 10x² + 2 = 3x³ + x² + 9x² + 3x - 3x - 1 + 3

= x²( 3x + 1 ) + 3x( 3x + 1 ) - ( 3x + 1 ) + 3

= ( 3x + 1 )( x² + 3x - 1 ) + 3

Vì ( 3x + 1 )( x² + 3x - 1 ) ⋮ ( 3x + 1 )

=> 3 ⋮ ( 3x + 1 ) <=> ( 3x + 1 ) ∈ Ư(3) ( đến đây bạn tự xét giá trị nhé )

Trả lời:

a, \(x^2=6x\)

\(\Leftrightarrow x^2-6x=0\)

\(\Leftrightarrow x\left(x-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)

Vậy x = 0; x = 6 là nghiệm của pt.

b, \(x^2-4x+3=0\)

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

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

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

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

Vậy x = 1; x = 3 là nghiệm của pt.

Bài 3:

a) \(x^2=6x\)

\(\Leftrightarrow x^2-6x=0\)

\(\Leftrightarrow x\left(x-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}\)

Vậy \(S=\left\{0;6\right\}\)

b) \(x^2-4x+3=0\)

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

Vậy \(S=\left\{3;1\right\}\)

x² + x + 1/4 = ( x + 1/2 )²

Trả lời:

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

...

1) =\(x^7-x+x^2+x\)+1

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

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

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

=[(x^4+x)(x-1)+1](x^2+x+1)

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

Trả lời:

1, x⁷ + x² + 1 

= x⁷ + x² + 1 + x⁶ - x⁶ + x⁵ - x⁵ + x⁴ - x⁴ + x³ - x³ + x² - x² + x - x 

= ( x⁷ + x⁶ + x⁵ ) - ( x⁶ + x⁵ + x⁴ ) + ( x⁴ + x³ + x² ) - ( x³ + x² + x ) + ( x² + x + 1 )

= x⁵ ( x² + x + 1 ) - x⁴ ( x² + x + 1 ) + x² ( x² + x + 1 ) - x ( x² + x + 1 ) + ( x² + x + 1 )

= ( x² + x + 1 )( x⁵ - x⁴ + x² - x + 1 )

b, x⁸ + x⁷ + 1 

= x⁸ + x⁷ + 1 + x⁶ - x⁶ + x⁵ - x⁵ + x⁴ - x⁴ + x³ - x³ + x² - x² + x - x 

= ( x⁸ + x⁷ + x⁶ ) - ( x⁶ + x⁵ + x⁴ ) + ( x⁵ + x⁴ + x³ ) - ( x³ + x² + x ) + ( x² + x + 1 ) 

= x⁶ ( x² + x + 1 ) - x⁴ ( x² + x + 1 ) + x³ ( x² + x + 1 ) - x ( x² + x + 1 ) + ( x² + x + 1 )

= ( x² + x + 1 )( x⁶ - x⁴ + x³ - x + 1 )

Trả lời:

a, 3x²y - 6xy = 3xy ( x - 2 )

b, x² - y² - 9x + 9y 

= ( x² - y² ) - ( 9x - 9y )

= ( x - y )( x + y ) - 9 ( x - y )

= ( x - y )( x + y - 9 )

c, x³ - 6x² - y²x + 9x 

= x ( x² - 6x - y² + 9 )

= x [ ( x² - 6x + 9 ) - y² ]

= x [ ( x - 3 )² - y² ]

= x ( x - 3 - y )( x - 3 + y )

3x²y - 6xy = 3xy( x - 2 )

x² - y² - 9x + 9y = ( x - y )( x + y ) - 9( x - y ) = ( x - y )( x + y - 9 )

x³ - 6x² - y²x + 9x = x( x² - 6x - y² + 9 ) = x[ ( x - 3 )² - y² ] = x( x - y - 3 )( x + y - 3 )

Trả lời:

a, \(-xy.\left(x^2+2xy-3\right)=-x^3y-2x^2y^2+3xy\)

b, \(\left(12x^6y^5-3x^3y^4+4x^2y\right):6x^2y\)

\(=12x^6y^5:6x^2y^2-3x^3y^4:6x^2y+4x^2y+6x^2y\)

\(=2x^4y^3-\frac{1}{2}xy^3+\frac{2}{3}\)

a.\(\left(-xy\right)\left(x^2+2xy-3\right)=-x^3y-2x^2y^2+6xy\)

b.\(\left(12x^6y^5-3x^3y^4+4x^2y\right):6x^2y=2x^4y^4-\frac{1}{2}xy^3+\frac{2}{3}\)

Mn giúp mk vs, mk cần gấp á

Thanks!

Giải phương trình: x^4-5x^3+6x^2-5x+1=0

x=2-căn bậc hai(3),

x=căn bậc hai(3)+2;

x = -(căn bậc hai(3)*i-1)/2;

x = (căn bậc hai(3)*i+1)/2;

Trả lời:

2ab⁴ - 8ab³ + 8ab² 

= 2ab² ( b² - 4b + 4 )

= 2ab² ( b - 2 )² 

\(\frac{1}{a+2b+3c}+\frac{1}{2a+3b+c}+\frac{1}{3a+b+2c}\)

\(\le\frac{1}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{b+c}+\frac{1}{a+b}+\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{a+c}\right)\)

\(\le\frac{1}{36}\left(\frac{1}{a}+\frac{1}{c}+\frac{1}{b}+\frac{1}{c}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{a}+\frac{1}{c}+\frac{1}{a}+\frac{1}{c}\right)\)

\(=\frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(=\frac{1}{6}\left(\frac{a+b+c}{abc}\right)=\frac{1}{6}\)

Dấu \(=\)khi \(a=b=c=3\).

bài này áp dụng bđt phụ \(\frac{1}{x_1+x_2+x_3+x_4+x_5+x_6}\le\frac{1}{36}\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}+\frac{1}{x_4}+\frac{1}{x_5}+\frac{1}{x_6}\right)\)

là oke nhé =)) 

về cách chứng minh thì bạn dùng svacxo cho vp là xong