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6 tháng 8 2020

a/ \(x^2-4y^2-3x+6y=\left(x^2-4y^2\right)-\left(3x-6y\right)=\left(x-2y\right)\left(x+2y\right)-3\left(x-2y\right)=\left(x-2y\right)\left(x+2y-3\right)\)

b/ \(a^2+2ab+b^2-ac-bc=\left(a^2+2ab+b^2\right)-\left(ac+bc\right)=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b\right)\left(a+b-c\right)\)

c/ \(25x^2-10x-3x=25x^2-13x=x\left(25x-13\right)\)

d/ \(16x^2+24x-7=16x^2-4x+28x-7=4x\left(4x-1\right)+7\left(4x-1\right)=\left(4x-1\right)\left(4x+7\right)\)

NV
15 tháng 10 2019

\(A=3\left(x-\frac{5}{6}\right)^2+\frac{11}{12}\)

\(B=2\left(x-\frac{3}{4}\right)^2+\frac{23}{8}\)

\(C=\left(x+\frac{3}{2}\right)^2+\frac{11}{4}\)

\(D=\left(x-5\right)^2+\left(3y+1\right)^2+4\)

\(E=\left(4x+1\right)^2+\left(y-2\right)^2+1\)

\(M=-\left(x+\frac{7}{2}\right)^2-\frac{11}{4}\)

\(N=-5\left(x-\frac{3}{5}\right)^2-\frac{41}{5}\)

\(C\) đề sai ví dụ \(x=3\Rightarrow C=2>0\)

\(D=-5\left(x-\frac{7}{10}\right)^2-\frac{131}{20}\)

AH
Akai Haruma
Giáo viên
28 tháng 12 2018

a)

\(a^2+b^2+2ab+2a+2b+1\)

\(=(a^2+2ab+b^2)+(2a+2b)+1\)

\(=(a+b)^2+2(a+b)+1^2=(a+b+1)^2\)

b)

\(3x(x-2y)+6y(2y-x)\)

\(=3x(x-2y)-6y(x-2y)=(3x-6y)(x-2y)=3(x-2y)(x-2y)\)

\(=3(x-2y)^2\)

AH
Akai Haruma
Giáo viên
28 tháng 12 2018

c)

\(16xy+4y^2-9+16x^2\)

\(=(16x^2+16xy+4y^2)-9\)

\(=(4x+2y)^2-3^2=(4x+2y-3)(4x+2y+3)\)

d)

\(x^4+64y^8=(x^2)^2+(8y^4)^2=(x^2)^2+(8y^4)^2+2.x^2.8y^4-2x^2.8y^4\)

\(=(x^2+8y^4)^2-16x^2y^4=(x^2+8y^4)^2-(4xy^2)^2\)

\(=(x^2+8y^4-4xy^2)(x^2+8y^4+4xy^2)\)

e)

\(3x^2-7x+2=3x^2-6x-x+2=(3x^2-6x)-(x-2)\)

\(=3x(x-2)-(x-2)=(3x-1)(x-2)\)

28 tháng 12 2018

a, a2+b2+2ab+2a+2b+1=(a+b+1)2

b,3x(x-2y)+6y(2y-x)=3x(x-2y)-6y(x-2y)

=3(x-2y)(x-2y)=3(x-2y)2

c, 16xy +4y2-9 +16x2=(16x2+16xy+4y2)-32

=(4x-2y)2-32=(4x-2y+3)(4x-2y-3)

y: Ta có: \(x^2-x-6=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

z: Ta có: \(3x^2-5x-8=0\)

\(\Leftrightarrow\left(3x-8\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-1\end{matrix}\right.\)

j: Ta có: \(25x^2-4=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

28 tháng 8 2020

x+ y2 + 10x + 6y + 34 = 0

=> (x2 + 10x + 25) + (y2 + 6y + 9) = 0

=> (x + 5)2 + (y + 3)2 = 0

=> \(\hept{\begin{cases}x+5=0\\y+3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-5\\y=-3\end{cases}}\)

Vậy x = - 5 ; y = -3

b) 25x2 + 4y2 + 10x + 4y + 2 = 0

=> (25x2 + 10x + 1) + (4y2 + 4y + 1) = 0

=> (5x + 1)2 + (2y + 1)2 = 0

=> \(\hept{\begin{cases}5x+1=0\\2y+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-0,2\\y=-0,5\end{cases}}\)

Vậy x = -0,2 ; y = -0,5

28 tháng 8 2020

a) 

\(x^2+10x+25+y^2+6y+9=0\)    

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

Ta có : 

\(\left(x+5\right)^2\ge0\forall x\) 

\(\left(y+3\right)^2\ge0\forall y\) 

\(\left(1\right)=0\Leftrightarrow\hept{\begin{cases}\left(x+5\right)^2=0\\\left(y+3\right)^2=0\end{cases}}\)   

\(\hept{\begin{cases}x+5=0\\y+3=0\end{cases}}\)         

\(\hept{\begin{cases}x=-5\\y=-3\end{cases}}\)   

b) 

\(25x^2+10x+1+4y^2+4y+1=0\)     

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

Ta có : 

\(\left(5x+1\right)^2\ge0\forall x\)      

\(\left(2y+1\right)^2\ge0\forall y\)  

\(\left(1\right)=0\Leftrightarrow\hept{\begin{cases}\left(5x+1\right)^2=0\\\left(2y+1\right)^2=0\end{cases}}\)   

\(\hept{\begin{cases}5x+1=0\\2y+1=0\end{cases}}\)    

\(\hept{\begin{cases}x=\frac{-1}{5}\\y=\frac{-1}{2}\end{cases}}\)

25 tháng 10 2020

d, \(2x^3-12x^2+24x-16\)

= 2(\(x^3-6x^2+12x-8\))

=2(x-2)\(^3\)

e, \(x^3-10x^2+25x-9xy^2\)

=\(x\left(x-10x+25-9y^2\right)\)

=\(x\left[\left(x-5\right)^2-\left(3y\right)^2\right]\)

=\(x\left[\left(x-5-3y\right)\left(x-5+3y\right)\right]\)

25 tháng 10 2020

a, \(x^3-8x^2+16x\)

=\(x^3-4x^2-4x^2+16x\)

= (\(x^3-4x^2\))-\(\left(4x^2-16x\right)\)

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

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

b, \(3x^2-27\)

=3(\(x^2-9\))

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

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

c,\(3x^2-5xy+6x-10y\)

=\(\left(3x^2+6x\right)-\left(5xy+10y\right)\)

=3x(x+2)-5y(x+2)

=(x+2)(3x-5y)