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\(\left(2a+b\right)^2-\left(2a+a\right)^2\)
\(=\left(2a+b-2a-a\right)\left(2a+b+2a+a\right)\)
\(=\left(b-a\right)\left(5a+b\right)\)
\(\left(2a+b\right)^2-\left(2a+a\right)^2\)
\(=\left(2a+b\right)^2-\left(3a\right)^2\)
\(=\left(2a+b-3a\right)\left(2a+b+3a\right)\)
\(=\left(b-a\right)\left(5a+b\right)\)
c: \(5\left(a+b\right)+x\left(a+b\right)\)
=(a+b)(x+5)
d: \(\left(a-b\right)^2-\left(b-a\right)\)
\(=\left(a-b\right)^2+\left(a-b\right)\)
=(a-b)(a-b+1)
e: \(=\left(12x^2+6x\right)\left(y+z+y-z\right)\)
\(=2y\cdot6x\cdot\left(2x+1\right)=12xy\left(2x+1\right)\)
a)\(\left(a^3-b^3\right)+\left(a-b\right)^2\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)^2\)
\(\left(a-b\right)\left(a^2+ab+b^2+a-b\right)\)
b) \(\left(8a^3-27b^3\right)-2a\left(4a^2-9b^2\right)\)
\(=\left(2a-3b\right)\left(4a^2+6ab+9b^2\right)-2a\left(2a-3b\right)\left(2a+3b\right)\)
\(=\left(2a-3b\right)\left(4a^2+6ab+9b^2-4a^2-6ab\right)\)
\(=\left(2a-3b\right)\cdot9b^2\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)+a^2-2ab+b^2\)
= ...........
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\begin{array}{l} a){\left( {ab - 1} \right)^2} + {\left( {a + b} \right)^2}\\ = {a^2}{b^2} - 2ab + 1 + {a^2} + 2ab + {b^2}\\ = {a^2}{b^2} + 1 + {a^2} + {b^2}\\ = {a^2}\left( {{b^2} + 1} \right) + \left( {{b^2} + 1} \right)\\ = \left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\\ c){x^3} - 4{x^2} + 12x - 27\\ = {x^3} - 27 + \left( { - 4{x^2} + 12x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) - 4x\left( {x - 3} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9 - 4x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} - x + 9} \right)\\ b){x^3} + 2{x^2} + 2x + 1\\ = {x^3} + 2{x^2} + x + x + 1\\ = x\left( {{x^2} + 2x + 1} \right) + \left( {x + 1} \right)\\ = x{\left( {x + 1} \right)^2} + \left( {x + 1} \right)\\ = \left( {x + 1} \right)\left( {x\left( {x + 1} \right) + 1} \right)\\ = \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\ d){x^4} - 2{x^3} + 2x - 1\\ = {x^4} - 2{x^3} + {x^2} - {x^2} + 2x - 1\\ = {x^2}\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)\\ = \left( {{x^2} - 2x + 1} \right)\left( {{x^2} - 1} \right)\\ = {\left( {x - 1} \right)^2}\left( {x - 1} \right)\left( {x + 1} \right)\\ = {\left( {x - 1} \right)^3}\left( {x + 1} \right)\\ e){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\ = {x^4} + 2{x^3} + {x^2} + {x^2} + 2x + 1\\ = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\ = \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right)\\ = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right) \end{array} |
Câu 1: \(a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=\left(a^4+b^4+c^4-2a^2b^2-2c^2a^2+2b^2c^2\right)-4b^2c^2=\left(a^2-b^2-c^2\right)^2-4b^2c^2=\left(a^2-b^2-c^2-2bc\right)\left(a^2-b^2-c^2+2bc\right)=\left[a^2-\left(b+c\right)^2\right]\left[a^2-\left(b-c\right)^2\right]=\left(a-b-c\right)\left(a+b+c\right)\left(a-b+c\right)\left(a+b-c\right)\)Câu 2: \(a^3+a^2-ab^2-b^2=a^2\left(a+1\right)-b^2\left(a+1\right)=\left(a^2-b^2\right)\left(a+1\right)=\left(a+b\right)\left(a-b\right)\left(a+1\right)\)
Câu 3: \(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)=a\left(b^3-c^3\right)-b\left[\left(b^3-c^3\right)+\left(a^3-b^3\right)\right]+c\left(a^3-b^3\right)=\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)-\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)=\left(a-b\right)\left(b-c\right)\left[b\left(c-a\right)+\left(c-a\right)\left(c+a\right)\right]=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)
Câu 1.
a4 + b4 + c4 - 2a2b2 - 2b2c2 - 2a2c2
= [ ( a4 - 2a2b2 + b4 ) - 2a2c2 + 2b2c2 + c4 ] - 4b2c2
= [ ( a2 - b2 )2 - 2( a2 - b2 )c2 + ( c2 )2 ] - ( 2bc )2
= ( a2 - b2 - c2 ) - ( 2bc )2
= ( a2 - b2 - c2 - 2bc )( a2 - b2 - c2 + 2bc )
= [ a2 - ( b2 + 2bc + c2 ) ][ a2 - ( b2 - 2bc + c2 ) ]
= [ a2 - ( b + c )2 ][ a2 - ( b - c )2 ]
= ( a - b - c )( a + b + c )( a - b + c )( a + b - c )
Câu 2.
a3 + a2 - ab2 - b2
= a2( a + 1 ) - b2( a + 1 )
= ( a + 1 )( a2 - b2 )
= ( a + 1 )( a - b )( a + b )
a) x2-y2-5x+5y=(x2-y2)-(5x-5y)
=(x-y)(x+y)-5(x-y)
=(x-y)(x+y-5)
b) 16x2-(x+y)2=(4x)2-(x+y)2
=(4x-x-y)(4x+x+y)
=(3x-y)(5x+y)
c) 5x2+6xy+y2=5x2+5xy+xy+y2
=5x(x+y)+y(x+y)
=(x+y)(5x+y)
d) a2-2a+2b-b2=( a2-b2)-(2a-2b)
= (a-b)(a+b)-2(a-b)
=(a-b)(a+b-2)
a) x2 - y2 - 5x + 5y
= (x2 - y2) - (5x - 5y)
= (x - y)(x + y) - 5(x - y)
= (x - y)(x + y - 5)
b) 16x2 - (x + y)2
= (4x)2 - (x + y)2
= (4x - x - y)(4x + x + y)
= (3x - y)(5x + y)
c) 5x2 + 6xy + y2
= 5x2 + 5xy + xy + y2
= 5x(x + y) + y(x + y)
= (x + y)(5x + y)
d) a2 - 2a + 2b - b2
= (a2 - b2) - (2a - 2b)
= (a - b)(a + b) - 2(a - b)
= (a - b)(a + b - 2)
a) \(P\left(a,b\right)=3a^2-2ab+b^2=3a^2-3ab+ab-b^2\)\(=3a\left(a-b\right)+b\left(a-b\right)=\left(a-b\right)\left(3a+b\right)\)
b) \(P\left(a,b\right)=0\Leftrightarrow\orbr{\begin{cases}a-b=0\\3a+b=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=b\\a=\frac{-b}{3}\end{cases}}}\)
+) \(a=b\Leftrightarrow M=\frac{a^2+a.a+2a^2}{2a^2-a^2}=4\)
+) \(a=\frac{-b}{3}\Rightarrow M=\frac{\left(\frac{-b}{3}\right)^2+\left(\frac{-b}{3}\right).b+2b^2}{2.\left(\frac{-b}{3}\right)^2-b^2}=\frac{\frac{16}{9}b^2}{\frac{-7}{9}b^2}=\frac{-16}{7}\)
`a)x^4+2x^2y+y^2`
`=(x^2+y)^2`
`b)(2a+b)^2-(2b+a)^2`
`=(2a+b-2b-a)(2a+b+2b+a)`
`=(a-b)(3a+3b)`
`=3(a-b)(a+b)`
`c)8a^3-27b^3-2a(4a^2-9b^2)`
`=(2a-3b)(4a^2+6ab+9b^2)-2a(2a-3b)(2a+3b)`
`=(2a-3b)(4a^2+6ab+9b^2-3a^2-6ab)`
`=9b^2(2a-3b)`
a) Ta có: \(x^4+2x^2y+y^2\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot y+y^2\)
\(=\left(x^2+y\right)^2\)
b) Ta có: \(\left(2a+b\right)^2-\left(2b+a\right)^2\)
\(=\left(2a+b-2b-a\right)\left(2a+b+2b+a\right)\)
\(=\left(a-b\right)\left(3a+3b\right)\)
\(=3\left(a+b\right)\left(a-b\right)\)