\(a-11b+3c⋮17\Rightarrow2a-22b+6c⋮17\)

\(2a-22b+6c=\left(2a-5b+6c\right)-17b⋮17\)

\(17b⋮17\Rightarrow2a-5b+6c⋮17\)

Trả lời:

a, \(x+5x^2=0\)

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

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

Vậy x = 0; x = - 1/5 là nghiệm của pt.

b, \(x^2-10x=-25\)

\(\Leftrightarrow x^2-10x+25=0\)

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

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

Vậy x = 5 là nghiệm của pt.

a) Áp dụng hằng đẳng thức thứ hai ta có

\(x^2 - 2xy + y^2\) \(<=>\) \((x +y )^2\)

b) Ta có :

\(x^2 - 2xy - 4z^2 + y^2 <=> (x^2 - 2xy + y^2) - (2z)^2 <=> ( x-y)^2 - (2z)^2 <=> ( x-y+2z) (x-y-2z)\)

a) \(73^2-27^2=\left(73+27\right)\left(73-27\right)=100.46=4600\)

b) \(55^2+20^2-25^2+40.45=\left(55^2-25^2\right)+\left(20^2+40.45\right)\)

\(=\left(55-25\right)\left(55+25\right)+\left(40.10+40.45\right)=30.80+40.55\)
\(=40\left(60+55\right)=40.115=4600\)

x²( x + 1 ) + 2x( x + 1 ) = 0 <=> x( x + 1 )( x + 2 ) = 0 <=> x = 0 hoặc x = -1 hoặc x = -2

x( 3x - 1 ) - 5( 1 - 3x ) = 0 <=> x( 3x - 1 ) + 5( 3x - 1 ) = 0 <=> ( 3x - 1 )( x + 5 ) = 0 <=> x = 1/3 hoặc x = -5

Trả lời:

1, \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)

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

\(\Leftrightarrow x=0;x=-1;x=-2\)

Vậy x = 0; x = - 1; x = - 2 là nghiệm của pt.

2, \(x\left(3x-1\right)-5\left(1-3x\right)=0\)

\(\Leftrightarrow x\left(3x-1\right)+5\left(3x-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-5\end{cases}}}\)

Vậy x = 1/3; x = - 5 là nghiệm của pt.

Trả lời:

a) x² + 4y² + 4xy = x² + 2.x.2y + (2y)² = ( x + 2y )²

b) \(\frac{1}{64}-27x^3=\left(\frac{1}{4}\right)^3-\left(3x\right)^3=\left(\frac{1}{4}-3x\right)\left(\frac{1}{16}+\frac{3}{4}x+9x^2\right)\)

c) x³ - 6x² + 12x - 8 = x³ - 3.x².2 + 3.x.2² - 2³ = ( x - 2 )³ 

d) x² -  x - y² - y = ( x² - y² ) - ( x + y ) = ( x - y )( x + y ) - ( x + y ) = ( x + y )( x - y - 1 )

e) 5x - 5y + ax  - ay = ( 5x - 5y ) + ( ax - ay ) = 5 ( x - y ) + a ( x - y ) = ( x - y )( 5 + a )

Trả lời:

a) \(\frac{1}{4}x^2y+5x^3-x^2y^2=x^2\left(\frac{1}{4}y+5x-y^2\right)\)

 b) 5x ( x - 1 ) - 3y ( 1 - x ) = 5x ( x - 1 ) + 3y ( x - 1 ) = ( x - 1 )( 5x + 3y )

 c) 4x² - 25 = ( 2x )² - 5² = ( 2x - 5 )( 2x + 5 )

 d) 6x- 9x² = 3x ( 2 - 3x )

a, sửa đề : \(25x^2+4y^2-10x+12y+10=0\)

\(\Leftrightarrow25x^2-10x+1+4y^2+12y+9=0\)

\(\Leftrightarrow\left(5x-1\right)^2+\left(2y+3\right)^2=0\)

Đẳng thức xảy ra khi x = 1/5 ; y = -3/2 

b, \(3x^2+2y^2-12x+12y+30=0\)

\(\Leftrightarrow3\left(x^2-4x+4\right)+2\left(y^2+6y+9\right)=0\)

\(\Leftrightarrow3\left(x-2\right)^2+2\left(y+3\right)^2=0\)

Đẳng thức xảy ra khi x = 2 ; y = -3 

\(a)\)

\(25x^2+4y^2-10x+12x+10=0\)

\(\Leftrightarrow\left(5x\right)^2-10x+1+\left(2y\right)^2+12y+9=0\)

\(\Leftrightarrow[\left(5x\right)^2-10x+1+\left(2y\right)^2+12y+9=0\)

\(\Leftrightarrow[\left(5x\right)^2-2.5x.1-1^2]+[\left(2y\right)^2+2.2y.3+3^{20}]=0\)

\(\Leftrightarrow\left(5x-1\right)^2+\left(2y+3\right)^2=0\)

\(\Leftrightarrow\left(5x-1\right)^2=0\Leftrightarrow5x-1=0\Leftrightarrow x=\frac{1}{5}\)

\(\Leftrightarrow\left(2y+3\right)^2=0\Leftrightarrow2y+3=0\Leftrightarrow2y=-3\Leftrightarrow y=\frac{-3}{2}\)

\(b)\)

\(3x^2+2y^2-12x+12y+30=0\)

\(\Leftrightarrow3x^2-12x+12+2y^2+12y+18=0\)

\(\Leftrightarrow3\left(x-2\right)^2+2\left(y+3\right)^2=0\)

Mà: \(3\left(x-2\right)^2\ge0\forall x;2\left(y+3\right)^2\ge0\forall y\)

\(\Leftrightarrow3\left(x-2\right)^2+2\left(y+3\right)^2=0\)chỉ khi: \(x-2=y+3=0\)

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

Trả lời:

a, \(A=x^2-6x+15=\left(x^2-6x+9\right)+6=\left(x-3\right)^2+6\ge6\forall x\)\(6\forall x\)

Dấu "=" xảy ra khi x - 3 = 0 <=> x = 3

Vậy GTNN của A = 6 khi x = 3

b, \(B=x^2+5x+7=\left(x^2+2.x.\frac{5}{2}+\frac{25}{4}\right)+\frac{3}{4}=\left(x+\frac{5}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

Dấu "=" xảy ra khi \(x+\frac{5}{2}=0\Leftrightarrow x=-\frac{5}{2}\)

Vậy GTNN của B = 3/4 khi x = - 5/2

c, \(C=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+10\)

\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+10\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+10\)

Đặt \(x^2+5x+4=t\)

\(\Rightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)+10=t\left(t+2\right)+10\)

\(=t^2+2t+10=\left(t^2+2t+1\right)+9=\left(t+1\right)^2+9\ge9\forall t\)

Dấu "=" xảy ra khi \(t+1=0\Leftrightarrow t=-1\)

hay \(x^2+5x+4=-1\)

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

\(\Leftrightarrow4\left(x^2+5x+5\right)=0\)

\(\Leftrightarrow4x^2+20x+20=0\)

\(\Leftrightarrow\left(4x^2+20x+25\right)-5=0\)

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

\(\Leftrightarrow\left(2x+5-\sqrt{5}\right)\left(2x+5+\sqrt{5}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+5-\sqrt{5}=0\\2x+5+\sqrt{5}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-5+\sqrt{5}}{2}\\x=\frac{-5-\sqrt{5}}{2}\end{cases}}}\)

Vậy GTNN của C = 9 khi \(\orbr{\begin{cases}x=\frac{-5+\sqrt{5}}{2}\\x=\frac{-5-\sqrt{5}}{2}\end{cases}}\)