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1) \(-4x^5\left(x^3-4x^2+7x-3\right)\)
\(=-4x^8+16x^7-28x^6+12x^5\)
2) \(3x^4\left(-2x^3+5x^2-\dfrac{2}{3}x+\dfrac{1}{3}\right)\)
\(=-6x^7+15x^6-2x^5+x^4\)
3) \(-5x^2y^4\left(3x^2y^3-2x^3y^2-xy\right)\)
\(=-15x^4y^7+10x^5y^6+5x^3y^5\)
4) \(4x^3y^2\left(-2x^2y+4x^4-3y^2\right)\)
\(=-8x^5y^3+16x^7y^2-12x^3y^4\)
Bài 2:
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}\)
\(=\dfrac{x}{x-3}+\dfrac{9-6x}{x\left(x-3\right)}\)
\(=\dfrac{x^2-6x+9}{x\left(x-3\right)}\)
\(=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}\)
\(=\dfrac{x-3}{x}\)
b) \(\dfrac{6x-3}{x}:\dfrac{4x^2-1}{3x^2}\)
\(=\dfrac{6x-3}{x}.\dfrac{3x^2}{4x^2-1}\)
\(=\dfrac{3\left(2x-1\right).3x^2}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{9x}{2x+1}\)
c) \(\dfrac{x+2}{3x}+\dfrac{x-5}{5x}-\dfrac{x+8}{4x}\)
\(=\dfrac{20x\left(x+2\right)+12x\left(x-5\right)-15x\left(x+8\right)}{60x}\)
\(=\dfrac{20x^2+40x+12x^2-60x-15x^2-120x}{60x}\)
\(=\dfrac{17x^2-140x}{60x}\)
d) \(\dfrac{x^2-x+1}{x^2+x}.\dfrac{x+1}{3x-2}.\dfrac{9x-6}{x^2-x+1}\)
\(=\dfrac{x^2-x+1}{x\left(x+1\right)}.\dfrac{x+1}{3x-2}.\dfrac{3\left(3x-2\right)}{x^2-x+1}\)
\(=\dfrac{3\left(x^2-x+1\right)\left(x+1\right)\left(3x-2\right)}{x\left(x+1\right)\left(3x-2\right)\left(x^2-x+1\right)}\)
\(=\dfrac{3}{x}\).
a/ \(3x^2\left(4x^3-2x+\dfrac{1}{3}\right)=12x^5-6x^3+x^2\)
b/ \(\left(4x^2+8xy-3xy^2\right)\left(-\dfrac{3}{4}x^2y\right)\)
\(=-3x^4y-6x^3y^2+\dfrac{9}{4}x^3y^3\)
c/ \(4x^3\left(2x^2-x+5\right)5x=20x^4\left(2x^2-x+5\right)\)
\(=40x^6-20x^5+100x^4\)
a, \(3x^2\left(4x^3-2x+\dfrac{1}{3}\right)\)
\(=12x^5-6x^3+x^2\)
b, \(\left(4x^2+8xy-3xy^2\right).\left(\dfrac{-3}{4}x^2y\right)\)
\(=-3x^4y-6x^3y^2+\dfrac{9}{4}x^3y^3\)
c, \(4x^3\left(2x^2-x+5\right)5x\)
\(=\left(8x^5-4x^4+20x^3\right)5x\)
\(=40x^6-20x^5+100x^4=20x^4.\left(2x^2-x+5\right)\)
Chúc bạn học tốt!!! Mình không chắc đâu !
a: =>5-x+6=12-8x
=>-x+11=12-8x
=>7x=1
hay x=1/7
b: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
=>12x+10=6x+5
=>6x=-5
hay x=-5/6
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
a.
\(\left(2x-1\right)^3+6\left(3x-1\right)^3=2\left(x+1\right)^3+6\left(x+2\right)^3\)
\(\Leftrightarrow\left(2x\right)^3-3.\left(2x\right)^2.1+3.2x.1+1^3+6.\left[\left(3x\right)^3-3.\left(3x\right)^2.1+3.3x.1+1^3\right]=2\left(x^3+3x^2+3x+1\right)+6\left(x^2+3.x^2.2+3.x.2^2+2^3\right)\)
nhiều quá bạn ạ
hay bạn tìm hiểu cách thức chung làm dạng bài tìm GTNN chứ như thế này thì làm lâu lắm
a: \(=49x^2-28x+4+21=\left(7x-2\right)^2+21>=21\)
Dấu '=' xảy ra khi x=2/7
b: \(=8\left(x^2-\dfrac{7}{2}x-\dfrac{1}{8}\right)\)
\(=8\left(x^2-2\cdot x\cdot\dfrac{7}{4}+\dfrac{49}{16}-\dfrac{51}{16}\right)\)
\(=8\left(x-\dfrac{7}{4}\right)^2-\dfrac{51}{2}>=-\dfrac{51}{2}\)
Dấu '=' xảy ra khi x=7/4
c: \(C=\left(2x^2+5\right)^2+10>=25+10=35\)
Dấu '=' xảy ra khi x=0
Bài 2:
\(=\dfrac{x^2\left(x^2+4\right)-2x\left(x^2+4\right)}{x^2+4}=x^2-2x\)
Bài 1:
a: \(=\left(\dfrac{2}{3}:\dfrac{-1}{9}\right)\cdot x^4y^2z^6=-6x^4y^2z^6\)
b: \(=-12x^8-21x^5\)
c: =x^3+8
d: \(=125x^3-75x^2+15x-1\)