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f: B-A=21/4xy^3-7/6x^3y-5xy^3+5/8x^3y=1/4xy^3-13/24x^3y
=>A-B=13/24x^3y-1/4xy^3
a: =>A-B=3x^2y-4xy^2+x^2y-2xy^2=4x^2y-6xy^2
b: =>B-A=-7xy^2+8x^2y-5xy^2+6x^2y=-12xy^2+14x^2y
=>A-B=12xy^2-14x^2y
c: =>B-A=8x^2y^3-4x^3y-3x^2y^3+5x^3y^2=5x^2y^3+x^3y^2
=>A-B=-5x^2y^3-x^3y^2
d: =>A-B=2x^2y^3-7x^3y+6x^2y^3+3x^3y^2=8x^2y^3-7x^3y+3x^3y^2
a/
\(\Leftrightarrow A=\dfrac{3}{8}xy^2+B-\dfrac{5}{6}x^2y+\dfrac{3}{4}x^2y-\dfrac{5}{8}xy^2\\ \Leftrightarrow A-B=-\dfrac{1}{12}x^2y-\dfrac{1}{4}xy^2\)
b/
\(\Leftrightarrow A-B=5xy^3-\dfrac{5}{8}yx^3-\dfrac{21}{4}xy^3+\dfrac{3}{7}x^3y\\ \Leftrightarrow A-B=-\dfrac{1}{4}xy^3-\dfrac{11}{56}x^3y\)
a: A+2xy^2-x^2y-B=3x^2y-4xy^2
=>A-B=3x^2y-4xy^2-2xy^2+x^2y=4x^2y-6xy^2
=>A=4x^2y; B=6xy^2
b: 5xy^2-A-6x^2y+B=-7xy^2+8x^2y
=>-A+B=-7xy^2+8x^2y-5xy^2+6x^2y=14x^2y-12xy^2
=>A=12xy^2; B=14x^2y
c: 5xy^3-A-5/8x^3y+B=2+1/4xy^3-7/6x^3y
=>-A+B=2+1/4xy^3-7/6x^3y-5xy^3+5/8x^3y
=>B-A=-19/4xy^3-13/24x^3y+2
=>B=-19/4xy^3; A=13/24x^3y-2
Bài 1:
a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)
\(=\left(2x+1\right)\left(3-2x+5\right)\)
\(=\left(2x+1\right)\left(8-2x\right)\)
\(=2\left(4-x\right)\left(2x+1\right)\)
b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)
\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\left(3x-2\right)\left(3x-6\right)\)
\(=3\left(3x-2\right)\left(x-2\right)\)
Bài 2:
a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)
\(=\left(a-b\right)\left(2a-4b\right)\)
\(=2\left(a-b\right)\left(a-2b\right)\)
f: Ta có: \(x^2-6xy+9y^2+4x-12y\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-3y+4\right)\)
a: a+b=5
=>(a+b)^2=25
=>a^2+b^2+2ab=25
=>2ab=12
=>ab=6
mà a+b=5
nên a,b là các nghiệm của phương trình:
x^2-5x+6=0
=>x=2 hoặc x=3
=>(a,b)=(2;3) hoặc (a,b)=(3;2)
b: a^2-b^2=34
=>(a+b)(a-b)=34
=>a+b=17
mà a-b=2
nên a=19/2 và b=19/2-2=15/2
f: B-A=21/4xy^3-7/6x^3y-5xy^3+5/8x^3y=1/4xy^3-13/24x^3y
=>A-B=13/24x^3y-1/4xy^3