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\(\left(2x+1\right)^2-\left(x-1\right)^2=\left(2x+1-x+1\right)\left(2x+1+x-1\right)=\left(x+2\right)3x\)
TL:
\(\left(2x+1\right)^2-\left(x-1\right)^2\)
\(=\left(2x+1+x-1\right)\left(2x+1-x+1\right)\)
\(=3x.\left(x+2\right)\)
\(64-27x^3=4^3-\left(3x\right)^3=\left(4-3x\right)\left(16+12x+9x^2\right)\)
\(A=4x^2+6x=2x\left(2x+3\right)\)
\(B=\left(2x+3\right)^2-x\left(2x+3\right)=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\)
\(C=\left(9x^2-1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1-3x+1\right)=2\left(3x+1\right)\)
\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)
\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)
\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)
\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)
a) 16x2 - ( x2 + 4 )2
= ( 4x )2 - ( x2 + 4 )2
= [ 4x - ( x2 + 4 ) ][ 4x + ( x2 + 4 ) ]
= ( -x2 + 4x - 4 )( x2 + 4x + 4 )
= [ -( x2 - 4x + 4 ) ]( x + 2 )2
= [ -( x - 2 )2 ]( x + 2 )2
b) ( x + y )3 + ( x - y )3
= [ ( x + y ) + ( x - y ) ][ ( x + y )2 - ( x + y )( x - y ) + ( x - y )2 ]
= ( x + y + x - y )[ x2 + 2xy + y2 - ( x2 - y2 ) + x2 - 2xy + y2 ]
= 2x( 2x2 + 2y2 - x2 + y2
= 2x( x2 + 3y2 )
Bài 1:
a. $2x^3+3x^2-2x=2x(x^2+3x-2)=2x[(x^2-2x)+(x-2)]$
$=2x[x(x-2)+(x-2)]=2x(x-2)(x+1)$
b.
$(x+1)(x+2)(x+3)(x+4)-24$
$=[(x+1)(x+4)][(x+2)(x+3)]-24$
$=(x^2+5x+4)(x^2+5x+6)-24$
$=a(a+2)-24$ (đặt $x^2+5x+4=a$)
$=a^2+2a-24=(a^2-4a)+(6a-24)$
$=a(a-4)+6(a-4)=(a-4)(a+6)=(x^2+5x)(x^2+5x+10)$
$=x(x+5)(x^2+5x+10)$
Bài 2:
a. ĐKXĐ: $x\neq 3; 4$
\(A=\frac{2x+1-(x+3)(x-3)+(2x-1)(x-4)}{(x-3)(x-4)}\\ =\frac{2x+1-(x^2-9)+(2x^2-9x+4)}{(x-3)(x-4)}\\ =\frac{x^2-7x+14}{(x-3)(x-4)}\)
b. $x^2+20=9x$
$\Leftrightarrow x^2-9x+20=0$
$\Leftrightarrow (x-4)(x-5)=0$
$\Rightarrow x=5$ (do $x\neq 4$)
Khi đó: $A=\frac{5^2-7.5+14}{(5-4)(5-3)}=2$
Bài `1:`
`a)3x^3+6x^2=3x^2(x+2)`
`b)x^2-y^2-2x+2y=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)`
Bài `2:`
`a)(2x-1)^2-25=0`
`<=>(2x-1-5)(2x-1+5)=0`
`<=>(2x-6)(2x+4)=0`
`<=>[(x=3),(x=-2):}`
`b)Q.(x^2+3x+1)=x^3+2x^2-2x-1`
`<=>Q=[x^3+2x^2-2x-1]/[x^2+3x+1]`
`<=>Q=[x^3-x^2+3x^2-3x+x-1]/[x^2+3x+1]`
`<=>Q=[(x-1)(x^2+3x+1)]/[x^2+3x+1]=x-1`
Đặt \(x^2+x+1=t\)
Ta có: \(\left(x^2+x+1\right)^2+3x\left(x^2+x+1\right)+2x^2\)
\(=t^2+3xt+2x^2\)
\(=t^2+xt+2xt+2x\)
\(=t\left(t+x\right)+2x\left(t+x\right)\)
\(=\left(t+x\right)\left(t+2x\right)\)
\(=\left(x^2+x+1+x\right)\left(x^2+x+1+2x\right)\)
\(=\left(x^2+2x+1\right)\left(x^2+3x+1\right)\)
\(=\left(x+1\right)^2\left(x^2+3x+1\right)\)
Chúc bạn học tốt.
a) 9 -(x-y)2
= 32 - (x-y)2
= (3-x+y).(3+x-y)
b) (x2 +4)2 - 16x2
= (x2+4)2 - (4x)2
= (x2 + 4 -4x).(x2 + 4 +4x)
\(9-\left(x-y\right)^2\)
\(=3^2-\left(x-y\right)^2\)
\(=\left(3-x+y\right)\left(3+x-y\right)\)
\(\left(x^2+4\right)^2-16x^2\)
\(=\left(x^2+4\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+4\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)^2\left(x+2\right)^2\)
Ta có:\(TH1:\left(3x+1\right)^2-\left(1-2x\right)^2=\left(3x+1+1-2x\right)\left(3x+1-1+2x\right)=\left(x+2\right)\left(5x\right)\)
Còn ra hằng đẳng thức thì mk chịu