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Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
a: \(N=\dfrac{3x^5-4x^4+6x^3}{-2x^2}=-\dfrac{3}{2}x^3+2x^2-3x\)
b: \(N=\dfrac{\left(6x^4y^5-3x^3y^4+\dfrac{1}{2}x^4y^3z\right)}{-\dfrac{1}{3}x^2y^3}=-18x^2y^2+9xy-\dfrac{3}{2}x^2z\)
c: \(\Leftrightarrow N\cdot\left(y-x\right)=\left(x-y\right)^3\)
\(\Leftrightarrow N=\dfrac{\left(x-y\right)^3}{y-x}=-\left(y-x\right)^2\)
d: \(\Leftrightarrow N\cdot\left(y^2-x^2\right)=\left(y^2-x^2\right)^2\)
hay \(N=y^2-x^2\)
a: \(=3\left(x^2-\dfrac{2}{3}x+\dfrac{4}{3}\right)\)
\(=3\left(x^2-2\cdot x\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{11}{9}\right)\)
\(=3\left(x-\dfrac{1}{3}\right)^2+\dfrac{11}{3}>=\dfrac{11}{3}\)
Dấu '=' xảy ra khi x=1/3
b: \(=2\left(x^2+\dfrac{3}{2}x\right)\)
\(=2\left(x^2+2\cdot x\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}\right)\)
\(=2\left(x+\dfrac{3}{4}\right)^2-\dfrac{9}{8}>=-\dfrac{9}{8}\)
Dấu '=' xảy ra khi x=-3/4
d: \(=3\left(x^2-2x+\dfrac{2}{3}\right)\)
\(=3\left(x^2-2x+1-\dfrac{1}{3}\right)\)
\(=3\left(x-1\right)^2-1>=-1\)
Dấu '=' xảy ra khi x=1
bài 1:
b, x(2x-1)(x+5)-(2x\(^2\)+1)(x+4,5)=3,5
(2x\(^2\)-x)(x+5)-(2x\(^3\)+9x\(^2\)+x+4,5)=3,5
2x\(^3\)+10x\(^2\)-x\(^2\)-5x-2x\(^3\)-9x\(^2\)-x-4,5=3,5
-6x-4,5=3,5
-6x=8
x=\(\frac{-4}{3}\)
vậy x=\(\frac{-4}{3}\)
a,6x\(^2\)-(2x+5)(3x-2)=7
6x\(^2\)-6x\(^2\)+4x-15x+10=7
4x-15x+10=7
-11x+10=7
-11x=-3
x=\(\frac{3}{11}\)
vậy x=\(\frac{3}{11}\)
câu 2 mk đọc k hiểu!
CHÚC BẠN HỌC TỐT
a: =>3M+2x^4y^4=x^4y^4
=>3M=-x^4y^4
=>M=-1/3*x^4y^4
b: x^2-2M=3x^2
=>2M=-2x^2
=>M=-x^2
c: =>M=-x^2y^3-3x^2y^3=-4x^2y^3
d: =>M=7x^2y^2-3x^2y^2=4x^2y^2