đề là luôn dương 

\(A=x^2-4x+6\)

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

\(A=\left(x-2\right)^2+2\ge2>0\)dấu "=" xảy ra khi và chỉ khi \(x=2\)

\(do:\left(x-2\right)^2+2>0=>x^2-4x+6>0\)vậy biểu thức luôn dương

Trả lời:

\(A=x^2-4x+6=x^2-2.x.2+2^2+2=\left(x-2\right)^2+2\)

Ta có: \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-2\right)^2+2\ge2>0\forall x\)

Dấu "=" xảy ra khi x - 2 = 0 <=> x = 2

Vậy biểu thức A luôn dương.

em hok lớp 7 :(

đây là có 2 ý à bn 

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

                                  \(=x\left(x+2y\right)\)

b) \(\left(x^2+y^2\right)^2-\left(2xy\right)^2=\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\)

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

c) \(\left(x+y\right)^3=x^3+3x^2y+3xy^2+y^3\)

                       \(=\left(x^3-6x^2y+9xy^2\right)+\left(y^3-6xy^2+9x^2y\right)\)

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

\(2\left(3x-1\right)\left(2x+5\right)-6\left(2x-1\right)\left(x+2\right)=-6\)

\(\left(6x-2\right)\left(2x+5\right)-\left(12x-6\right)\left(x+2\right)=-6\)

\(12x^2-4x+30x-10-12x^2+6x-24x+12=-6\)

\(8x=-8\)

\(x=-1\)

\(x=\frac{13}{7}\)

a) x² - 7xy - 18y²

= x² + 2xy - 9xy - 18y²

= x(x + 2y) - 9y(x + 2y) 

= (x - 9y)(x + 2y) 

b) 4x² + 8x - 5

= 4x² - 2x + 10x - 5

= 2x(2x - 1) + 5(2x - 1) 

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

c) 4x⁴ - 21x²y² + y⁴

= (4x⁴ + 4x²y² + y⁴) -25x²y²

= (2x² + y²) - (5xy)²

= (2x² + 5xy + y²)(2x² - 5xy + y²) 

= \(2\left(x^2+\frac{5}{2}xy+\frac{y^2}{2}\right)2\left(x^2-\frac{5}{2}xy+\frac{y^2}{2}\right)\)

= \(4\left[\left(x+\frac{5}{4}y\right)^2-\frac{25}{16}y^2+\frac{y^2}{2}\right]\left[\left(x-\frac{5}{4}\right)y^2-\frac{25}{16}y^2+\frac{y^2}{2}\right]\)

\(=4\left(x+\frac{5}{4}y-\frac{\sqrt{17}}{4}y\right)\left(x+\frac{5}{4}y+\frac{\sqrt{17}}{4}y\right)\left(x-\frac{5}{4}y-\frac{\sqrt{17}y}{4}\right)\left(x-\frac{5}{4}y+\frac{\sqrt{17y}}{4}\right)\)

Trả lời:

a, x² - 7xy - 18y² 

= x² - 9xy + 2xy - 18y²

= ( x² - 9xy ) + ( 2xy - 18y² )

= x ( x - 9y ) + 2y ( x - 9y )

= ( x + 2y ) ( x - 9y )

b, 4x² + 8x - 5

= 4x² + 10x - 2x - 5

= ( 4x² + 10x ) - ( 2x + 5 )

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

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

Ta có x + y = 3

=> (x + y)² = 9

<=> x²  + y² + 2xy = 9

<=> 2xy = 4

<=> xy = 2

Khi đó x³ + y³  = (x +  y)(x² - xy + y²) = 3.(5 - 2) = 9

b) Ta có x - y = 5 

<=> (x - y)² = 25

<=> x² - 2xy + y² = 25

<=> -2xy = 10

<=> xy = -5

Khi đó x³ - y³ = (x - y)(x² - xy + y²) = 5.(15 + 5) = 100 

a) Xét tam giác \(ABC\)có \(H\)là giao điểm của hai đường cao \(BD\)và \(CE\)nên \(H\)là trực tâm của tam giác \(ABC\).

Suy ra \(AH\perp BC\).

\(\Delta ADB=\Delta AEC\)(cạnh huyền - góc nhọn)

suy ra \(\frac{AD}{AB}=\frac{AE}{AC}\)

\(\Rightarrow DE//BC\)

suy ra \(AH\perp DE\).

b) Xét tam giác \(ADH\)vuông tại \(D\)có \(DI\)là đường trung tuyến ứng với cạnh huyền \(AH\): 

\(ID=IH\Rightarrow\widehat{IDH}=\widehat{IHD}\)

Xét tam giác \(BCD\)vuông tại \(D\)có \(DK\)là đường trung tuyến ứng với cạnh huyền \(BC\): 

\(KB=KD\Rightarrow\widehat{KDB}=\widehat{KBD}\)

suy ra \(\widehat{IDK}=\widehat{IDH}+\widehat{KDH}=\widehat{AHD}+\widehat{DBC}=\widehat{AHD}+\widehat{HAD}=90^o\)

suy ra \(DI\perp DK\).

a) = 9x² - ( y² - 10y + 25y² ) = ( 3x )² - ( y - 5 )² = ( 3x - y + 5 )( 3x + y - 5 )

b) = ( x³ - 8 ) - ( x² - 4x + 4 ) = ( x - 2 )( x² + 2x + 4 ) - ( x - 2 )² = ( x - 2 )( x² + x + 6 ) 

c) = ( 4a² - 4a + 1 ) - ( b² - 2bc + c² ) = ( 2a - 1 )² - ( b - c )² = ( 2a - b + c - 1 )( 2a + b - c - 1 )

d) = ( a³ + 3a² + 3a + 1 ) - 27b³ = ( a + 1 )³ - ( 3b )³ = ( a - 3b + 1 )( a² + 9b² + 3ab + 3b )

a. \(9x^2-\left(y^2-10y+25\right)=9x^2-\left(y-5\right)^2=\left(3x-y+5\right)\left(3x+y-5\right)\)

b.\(x^3-8-x^2+4x-4=\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)^2=\left(x-2\right)\left(x^2+x+6\right)\)

c.\(\left(4a^2-4a+1\right)-\left(b^2-2bc+c^2\right)=\left(2a-1\right)^2-\left(b-c\right)^2=\left(2a-1+b-c\right)\left(2a-1-b+c\right)\)

d.\(\left(a^3+3a^2+3a+1\right)-27b^3=\left(a+1\right)^3-\left(3b\right)^3=\left(a+1-3b\right)\left[\left(a+1\right)^2+3b\left(a+1\right)+9b^2\right]\)

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

\(\Leftrightarrow\left(-x-6\right)3x=0\Leftrightarrow x=-6;x=0\)

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

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

\(\Leftrightarrow\left(-x-6\right).3x=0\)

\(\Leftrightarrow-3x\left(x+6\right)=0\)

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

Vậy \(x\in\left\{0;-6\right\}\)