Bài làm :

\(a\text{)}3x^2+4x=0\Leftrightarrow x\left(3x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\3x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{4}{3}\end{cases}}\)

\(b\text{)}25x^2-0,64=0\Leftrightarrow\left(5x-0,8\right)\left(5x+0,8\right)=0\Leftrightarrow\orbr{\begin{cases}5x-0,8=0\\5x+0,8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0,16\\-0,16\end{cases}}\)

\(c\text{)}x^4-16x^2=0\Leftrightarrow\left(x^2-4x\right)\left(x^2+4x\right)=0\Leftrightarrow\orbr{\begin{cases}x^2-4x=0\\x^2+4x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x-4\right)=0\\x\left(x+4\right)=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)

\(d\text{)}x^2+x=6\Leftrightarrow x^2+x-6=0\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

Bài làm :

\(a)3x^2+4x=0\)

\(\Rightarrow x\left(3x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\3x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-4}{3}\end{cases}}\)

Vậy x = 0 hoặc \(x=\frac{-4}{3}\) .

\(b)25x^2-0,64=0\)

\(\Rightarrow\left(5x\right)^2=\frac{16}{25}\)

\(\Rightarrow\left(5x\right)^2=\left(\frac{4}{5}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}5x=\frac{4}{5}\\5x=\frac{-4}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{4}{25}\\x=\frac{-4}{25}\end{cases}}\)

Vậy \(x=\frac{4}{25}\) hoặc \(x=\frac{-4}{25}\) .

\(c)x^4-16x^2=0\)

\(\Rightarrow x^2\left(x^2-16\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2=0\\x^2-16=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=4^2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)

Vậy x = 0 hoặc \(x=\pm4\) .

a² + b²

= a² + 2ab + b² - 2ab

= ( a + b )² - 2ab

= 4² - 2.2

= 16 - 4 = 12

           Bài làm :

ab = 2 => 2ab=4 ; a+b=4 => (a+b)² = 16

Ta có :

\(\left(a+b\right)^2-2ab=a^2+2ab+b^2-2ab=a^2+b^2\)

\(\Rightarrow a^2+b^2=\left(a+b\right)^2-2ab=16-4=12\)

Vậy a² + b² = 12

27x⁶ + 125y⁶ = ( 3x² )³ + ( 5y² )³ = ( 3x² + 5y² )( 9x⁴ - 15x²y² + 25y⁴ )

8a⁶ - 8b⁶ = ( 2a² )³ - ( 2b² )³ = ( 2a - 2b )( 4a² + 4ab + 4b² ) = 2( a - b )4( a² + ab + b² ) = 8( a - b )( a² + ab + b² )

x⁴ + 64y⁴ = x⁴ + 16x²y² + 64y⁴ - 16x²y² 

                = ( x⁴ + 16x²y² + 64y⁴ ) - 16x²y²

                = ( x² + 8y² )² - ( 4xy )²

                = ( x² + 8y² - 4xy )( x² + 8y² + 4xy )

x⁴ + x³ + 2x² + x + 1 = x⁴ + x³ + x² + x² + x + 1

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

                                  = x²( x² + x + 1 ) + ( x² + x + 1 )

                                  = ( x² + x + 1 )( x² + 1 )

\(27x^6+125y^6=\left(3x^2\right)^3+\left(5y^2\right)^3=\left(3x^2+5y^2\right)\left(9x^4-15x^2.y^2+25y^4\right)\)

\(8a^6-8b^6=8\left(a^6-b^6\right)=8\left(\left(a^3\right)^2-\left(b^3\right)^2\right)=8\left(a^3-b^3\right)\left(a^3+b^3\right)\)

                                                       \(=8\left(a-b\right)\left(a^2+ab+b^2\right)\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(x^{\text{4}}+64y^4=x^4+64y^4+16x^2y^2-16x^2y^2\)

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

\(x^4+x^3+2x^2+x+1=\left(x^4+2x^2+1\right)+\left(x^3+x\right)\)

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

 (x+2)^2= 9 
=> (x+2)^2= 3^2=(-3)^2
TH1: x+2=3
=> x=3-2=1
TH2: x+2=-3
=> x=(-3)-2=-5

Bài làm :

\(a,\left(x+2\right)^2-9=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+2=3\\x+2=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

Vậy x = 1 hoặc x = -5 .

\(b,\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-\left(25x^2-3^2\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)

\(\Leftrightarrow\left(25x^2-25x^2\right)+10x=30-9-1\)

\(\Leftrightarrow10x=20\)

\(\Leftrightarrow x=2\)

Vậy x = 2 .

\(c,\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\)

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

\(\Leftrightarrow x^3-1+2x^2-x^3+4x-2x^2=5\)

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

\(\Leftrightarrow4x=6\)

\(\Leftrightarrow x=\frac{3}{2}\)

Vậy x = 3/2 .

Học tốt nhé

a) \(x^2-6x-y^2-4y+5=x^2-6x+9-y^2-4y-4\)

\(=\left(x^2-6x+9\right)-\left(y^2+4y+4\right)=\left(x-3\right)^2-\left(y+2\right)^2\)

b) \(4a^2-12a-b^2+2b+8=4a^2-12a+9-b^2+2b-1\)

\(=\left(4a^2-12a+9\right)-\left(b^2-2b+1\right)=\left(2a-3\right)^2-\left(b-1\right)^2\)

x² - 6x - y² - 4y + 5

= ( x² - 6x + 9 ) - ( y² + 4y + 4 )

= ( x - 3 )² - ( y + 2 )²

4a² - 12a - b² + 2b + 8

= ( 4a² - 12a + 9 ) - ( b² - 2b + 1 )

= ( 2a - 3 )² - ( b - 1 )²

M = x³( x² - y² ) + y²( x³ - y³ )

= x⁵ - x³y² + x³y² - y⁵

= x⁵ - y⁵

| y | = 1 => y = ±1

Rồi bạn xét hai trường hợp x = 2 ; y = 1 và x = 2 ; y = -1 nhé

b) N = AB

= ( -2x² + 3x + 5 )( x² - x + 3 )

= -2x⁴ + 2x³ - 6x² + 3x³ - 3x² + 9x + 5x² - 5x + 15

= -2x⁴ + 5x³ - 4x² + 4x + 15

| x | = 2 => x = ±2

Rồi bạn thế vô

Good luck

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

     \(=x^5-x^3y^2+x^3y^2-y^5\)

       \(=x^5-y^5\)

\(|y|=1\Rightarrow y=1\text{hoặc}y=-1\)

TH1: x=2;y=-1Ta có M=1 +1=2

 TH2: tại x=2;y=1 ta có: M= 1-1=0

b)\(N=\left(-2x^2+3x+5\right)\left(x^2-x+3\right)\)

         \(=-2^4+5x^3-4x^2+4x+15\)

\(|x|=2\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

\(\text{Tại x=2 thì }M=-16+40-16+8+15=31\)

\(\text{ Tại x=-2 thì }M=-16-40-16-8+15=-65\)                   

A = ( x + 1 )( x² - 3x - 2 ) + ( x + 1 )( x² - x + 1 )

= ( x + 1 )[ ( x² - 3x - 2 ) + ( x² - x + 1 ) ]

= ( x + 1 )( x² - 3x - 2 + x² - x + 1 )

= ( x + 1 )( 2x² - 4x - 1 )

= x( 2x² - 4x - 1 ) + 2x² - 4x - 1

= 2x³ - 4x² - x + 2x² - 4x - 1

= 2x³ - 2x² - 5x - 1 

B = ( x - y )( x² + xy + y² ) - ( x + y )( x² + 2x + y² )

= x³ - y³ - ( x³ + 2x² + xy² + x²y + 2xy + y³ )

= x³ - y³ - x³ - 2x² - xy² - x²y - 2xy - y³

= -2y³ - 2x² - xy² - x²y - 2xy

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

\(=x.x^2-x.3x-x.2+1.x^2-1.3x-1.2+x.x^2-x.x+x.1+1.x^2-1.x+1.1\)

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

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

\(=2x^3-2x^2-5x-1\)

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

\(=x.x^2+x.xy+x.y^2-y.x^2-y.xy-y.y^2-x.x^2-x.2x-x.y^2+y.x^2+y.2x+y.y^2\)

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

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

\(=-2x^2+2xy\)

Vì ABCD là hình thang cân nên \(AD=BC,\widehat{ADC}=\widehat{BCD}\)

Xét 2 tam giác ADC và BCD có: DC chung, \(\widehat{ADC}=\widehat{BCD}\), AD=BC

\(\Rightarrow\Delta ADC=\Delta BCD\left(c.g.c\right)\Rightarrow\widehat{DAC}=\widehat{CBD}=90^0\Rightarrow AC\perp AD\)