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a/ \(A=\left(x-y\right)^2+\left(x+y\right)^2.\)
\(A=\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)\)
\(A=x^2-2xy+y^2+x^2+2xy+y^2\)
\(A=2x^2+2y^2\)
b/ \(B=\left(2a+b\right)^2-\left(2a-b\right)^2\)
\(B=\left(4a^2+4ab+b^2\right)-\left(4a^2-4ab+b^2\right)\)
\(B=4a^2+4ab+b^2-4a^2+4ab-b^2\)
\(B=8ab\)
c/ \(C=\left(x+y\right)^2-\left(x-y\right)^2\)
\(C=\left(x^2+2xy+y^2\right)-\left(x^2-2xy+y^2\right)\)
\(C=x^2+2xy+y^2-x^2+2xy-y^2\)
\(C=4xy\)
d/ \(D=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(D=\left(4x^2-4x+1\right)-2\left(4x^2-12x+9\right)+4\)
\(D=4x^2-4x+1-8x^2+24x-18+4\)
\(D=-4x^2+20x-13\)
a. \(8x\left(x-2017\right)-2x+4034=0\)
\(8x\left(x-2017\right)-2\left(x-2017\right)=0\)
\(\left(8x-2\right)\left(x-2017\right)=0\)
\(\Rightarrow TH1:8x-2=0\)
\(8x=2\)
\(x=\frac{1}{4}\)
\(TH2:x-2017=0\)
\(x=2017\)
Vậy \(x\in\left\{\frac{1}{4};2017\right\}\)
Bài 1
a) \(8x\left(x-2017\right)-2x+4034=0\)
\(\Rightarrow8x\left(x-2017\right)-2\left(x-2017\right)=0\)
\(\Rightarrow\left(x-2017\right)\left(4x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=2017\\x=\frac{1}{4}\end{cases}}\)
\(\left(x+5\right)^2-3\left(x+5\right)\)
\(=\left(x+5\right)\left(x+5-3\right)\)
\(=\left(x+5\right)\left(x+2\right)\)
\(2x\left(x-3\right)-\left(x-3\right)^2\)
\(=\left(x-3\right)\left(2x-x+3\right)\)
\(=\left(x-3\right)\left(x+3\right)\)
Bài 6
\(\left(a-b\right)^2=a^2-2ab+b^2\)
\(=\left(a^2+2ab+b^2\right)-4ab\)
\(=\left(a+b\right)^2-4ab\)
Bài 5 :
\(a,16x^2-\left(4x-5\right)^2=15\)
\(16x^2-16x^2+40x-25-15=0\)
\(40x-40=0\)
\(40x=40\)
\(x=1\)
\(b,\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)
\(4x^2+12x+9-4x^2+4=49\)
\(12x=36\)
\(x=3\)
\(c,\left(2x+1\right)\left(2x-1\right)+\left(1-2x\right)^2=18\)
\(4x^2-1+1-4x+4x^2=18\)
\(8x^2-4x-18=0\)
\(2\left(4x^2-2x-9\right)=0\)
\(x=\frac{1-\sqrt{37}}{4}\)
\(d,2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)
\(2x^2+4x+2-x^2+9-x^2+8x-16=0\)
\(12x=4\)
\(x=\frac{1}{3}\)
e sẽ cố gắng !!!
\(3x-15=2x\left(x-5\right)\)
\(3x-15=2x^2-10x\)
\(3x-15-2x^2+10x=0\)
\(13x-15-2x^2=0\)
\(x\left(13-2x\right)-15=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\13-2x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\-2-2x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\2x=-2\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
\(f,x\left(2x-7\right)-4x+14=0\)
\(2x^2-7x-4x+14=0\)
\(2x^2-11x+14=0\)
\(x\left(2x-11\right)=-14\)
\(\Rightarrow\orbr{\begin{cases}x=-14\\2x-11=-14\end{cases}\Rightarrow\orbr{\begin{cases}x=-14\\2x=-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=-14\\x=-\frac{3}{2}\end{cases}}}\)
a) \(4x^2-6x=2x\left(2x-3\right)\)
b) \(9x^4y^3+3x^2y^4=3x^2y^3\left(3x^2+y\right)\)
c) \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(5x+3\right)\left(x-y\right)\)
d) \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)
e) \(5\left(x+3y\right)-15x\left(x+3y\right)=\left(5-15x\right)\left(x+3y\right)\)
\(=5\left(1-3x\right)\left(x+3y\right)\)
f) \(2x^2\left(x+1\right)-4\left(x+1\right)=\left(2x^2-4\right)\left(x+1\right)\)
\(=\left(\sqrt{2}x-2\right)\left(\sqrt{2}x+2\right)\left(x+1\right)\)
\(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x-1\right)\left(x+1\right).\)
\(=x^3-3x^2+3x-1-\left(x^3-2^3\right)+3\left(x^2-1\right)\)
\(=x^3-3x^2+3x-1-x^3+8+3x^2-3\)
\(=3x+4\)
\(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)^3-\left(x^3+8\right)+3\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)^3-x^3-8+3\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1-x\right)+3x\left(x-1\right)\left(x-1-x\right)-8+3\left(x-1\right)\left(x+1\right)\)(1)
\(=-1-3x\left(x-1\right)-8+3\left(x-1\right)\left(x+1\right)\)
\(=3\left(x-1\right)\left(-x+x+1\right)-9=3\left(x-1\right)-9=3\left(x-4\right)=3x-12\)
(1) là hằng đẳng thức \(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)\)
a) ( 3 - x )( x2 + 2x - 7 ) + ( x - 3 )( x2 + x - 5 )
= ( 3 - x )( x2 + 2x - 7 ) - ( 3 - x )( x2 + x - 5 )
= ( 3 - x )( x2 + 2x - 7 - x2 - x + 5 )
= ( 3 - x )( x - 2 )
b) ( x - 5 )2 + 3( 5 - x )
= ( x - 5 )2 - 3( x - 5 )
= ( x - 5 )( x - 5 - 3 ) = ( x - 5 )( x - 8 )
c) 2x( x - 1 )2 - ( 1 - x )3
= 2x( 1 - x )2 - ( 1 - x )3
= ( 1 - x )2( 2x - 1 + x ) = ( 1 - x )2( 3x - 1 )
d) x2 + 8x + 16 = ( x + 4 )2
e) x2 - 4xy + 4y2 = ( x - 2y )2
g) 4x2 - 25y2 = ( 2x )2 - ( 5y )2 = ( 2x - 5y )( 2x + 5y )
h) 25( x + 1 )2 - 4( x - 3 )2
= 52( x + 1 )2 - 22( x - 3 )2
= ( 5x + 5 )2 - ( 2x - 6 )2
= ( 5x + 5 - 2x + 6 )( 5x + 5 + 2x - 6 )
= ( 3x + 11 )( 7x - 1 )
i) x3 + 27 = ( x + 3 )( x2 - 3x + 9 )
k) 8x3 - 125 = ( 2x )3 - 53 = ( 2x - 5 )( 4x2 + 10x + 25 )
l) x3 + 6x2 + 12x + 8 = ( x + 2 )3
m) -x3 + 9x2 - 27x + 27 = -( x3 - 9x2 + 27x - 27 ) = -( x - 3 )3
1) \(x\left(2^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=x-5x^3-x^2+x\)
\(=2x-5x^3-x^2\)
2) \(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=\left(6x^2+23x-55\right)-\left(6x^2+23x+21\right)\)
\(=-76\)
Làm lại câu 1
\(x\left(2^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=x-5x^3-x^2+x^2\)
\(=x-5x^3\)