Tính:
a) \(\dfrac{1}{2}{x^2}.\dfrac{6}{5}{x^3}\); b) \({y^2}(\dfrac{5}{7}{y^3} - 2{y^2} + 0,25)\);
c) \((2{x^2} + x + 4)({x^2} - x - 1)\); d) \((3x - 4)(2x + 1) - (x - 2)(6x + 3)\).
K
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Những câu hỏi liên quan
28 tháng 6 2023
2:
a: =>2/3:x=1,4-2,4=-1
=>x=-2/3
b: =>x/5=25/30-19/30=6/30=1/5
=>x=1
3:
Số học sinh giỏi là 40*1/4=10 bạn
Số học sinh khá là 30*3/5=18 bạn
Số học sinh TB là 30-18=12 bạn
14 tháng 9 2023
a: \(=\dfrac{13}{3}+\dfrac{17}{6}=\dfrac{26}{6}+\dfrac{17}{6}=\dfrac{43}{6}\)
b: \(=7-\dfrac{8}{3}=\dfrac{21-8}{3}=\dfrac{13}{3}\)
c: \(=\dfrac{17}{7}\cdot\dfrac{7}{4}=\dfrac{17}{4}\)
d: \(=\dfrac{16}{3}:\dfrac{16}{5}=\dfrac{16}{3}\cdot\dfrac{5}{16}=\dfrac{5}{3}\)
23 tháng 11 2021
\(a,=\sqrt{5}-4\sqrt{5}-12\sqrt{5}=-15\sqrt{5}\\ b,=2\sqrt{3}-\dfrac{2+\sqrt{3}}{1}=2\sqrt{3}-2-\sqrt{3}=\sqrt{3}-2\\ c,=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2\sqrt{x}}\\ =\dfrac{2\left(x+1\right)}{2\sqrt{x}}=\dfrac{x+1}{\sqrt{x}}\)
29 tháng 12 2021
c: \(=\dfrac{2\left(x+3\right)}{x\left(3x-1\right)}\cdot\dfrac{-\left(3x-1\right)}{x\left(x+3\right)}=\dfrac{-2}{x^2}\)
H
23 tháng 9 2023
\(\dfrac{2x+4}{x^3-1}-\dfrac{2}{x-1}+\dfrac{x+2}{x^2+x+1}\\ =\dfrac{2x+4}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{2}{x-1}+\dfrac{x+2}{x^2+x+1}\\ =\dfrac{2x+4}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{2x+4-2x^2-2x-2+x^2-x+2x-2}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{-x^2+x}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{-x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=-\dfrac{x}{x^2+x+1}\)
H
23 tháng 9 2023
`a, 2/(x+1)` hay `2/(x-1)` cậu nhỉ?
`b,`
\(\dfrac{x-1}{x^2-5x+6}-\dfrac{x-3}{x-2}+\dfrac{x-2}{x-3}\\ =\dfrac{x-1}{\left(x-2\right)\left(x-3\right)}-\dfrac{x-3}{x-2}+\dfrac{x-2}{x-3}\\ =\dfrac{x-1}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)^2}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(x-2\right)^2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x-1-\left(x^2-6x+9\right)+x^2-4x+4}{\left(x-2\right)\left(x-3\right)}\\ =\dfrac{x-1-x^2+6x-9+x^2-4x+4}{\left(x-2\right)\left(x-3\right)}\\ =\dfrac{3x-6}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{3\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\\ =\dfrac{3}{x-3}\)
1 tháng 11 2021
a: \(=\dfrac{x^2-5x+x+4}{x\left(x-2\right)}=\dfrac{x^2-4x+4}{x\left(x-2\right)}=\dfrac{x-2}{x}\)
b: \(=\dfrac{x^2-6x+9+4x^2+8x-4x^2-8x}{\left(x-3\right)\left(x+2\right)}\)
\(=\dfrac{x-3}{x+2}\)
1 tháng 11 2021
a) \(=\dfrac{x\left(x-5\right)+x+4}{x\left(x-2\right)}=\dfrac{x^2-4x+4}{x\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}=\dfrac{x-2}{x}\)
b) \(=\dfrac{\left(x-3\right)^2+4x\left(x+2\right)-8x-4x^2}{\left(x+2\right)\left(x-3\right)}=\dfrac{x^2-6x+9+4x^2+8x-8x-4x^2}{\left(x+2\right)\left(x-3\right)}\)
\(=\dfrac{x^2-6x+9}{\left(x+2\right)\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{\left(x+2\right)\left(x-3\right)}=\dfrac{x-3}{x+2}\)
8 tháng 3 2023
a: =1/2x^3*x^2-1/2x^3*6x-1/2x^3*10
=1/2x^5-3x^4-5x^3
b: =-3x^2*5x^3+3x^2*4x^2-3x^2*3x+3x^2*3x
=-15x^5+12x^4-9x^3+9x^2
c: \(=3x\cdot5x^2-3x\cdot2x-3x=15x^3-6x^2-3x\)
d: \(=\dfrac{1}{2}x^2y\cdot2x^3-\dfrac{1}{2}x^2y\cdot\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)
7 tháng 12 2021
\(a,\left(2x-5\right)\left(5-x\right)=5\left(2x-5\right)-x\left(2x-5\right)=10x-25-2x^2+5x=15x-2x^2-25\\ b,\dfrac{1}{3x-2}-\dfrac{1}{3x+2}=\dfrac{3x+2-3x+2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{4}{\left(3x-2\right)\left(3x+2\right)}\)
\(c,\dfrac{3}{x-3}-\dfrac{6x}{x^2-9}+\dfrac{x}{x+3}=\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x+9-6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2-6x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{x+3}\)
24 tháng 12 2021
b: \(=\dfrac{7x-42-x^2+36}{x\left(x-6\right)}=\dfrac{-x^2+7x-6}{x\left(x-6\right)}=\dfrac{-x+1}{x}\)
24 tháng 12 2021
\(\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x-3\right)}=\dfrac{x\left(x+3\right)-3\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+3x-3x-9}{x\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{1}{x}\)
a)
\(\dfrac{1}{2}{x^2}.\dfrac{6}{5}{x^3} = \dfrac{1}{2}.\dfrac{6}{5}.{x^2}.{x^3} = \dfrac{3}{5}{x^5}\);
b)
\(\begin{array}{l}{y^2}(\dfrac{5}{7}{y^3} - 2{y^2} + 0,25) = {y^2}.\dfrac{5}{7}{y^3} - {y^2}.2{y^2} + {y^2}.0,25)\\ = \dfrac{5}{7}{y^5} - 2{y^4} + 0,25{y^2}\end{array}\);
c)
\(\begin{array}{l}(2{x^2} + x + 4)({x^2} - x - 1) \\= 2{x^2}({x^2} - x - 1) + x({x^2} - x - 1) + 4({x^2} - x - 1)\\ = 2{x^4} - 2{x^3} - 2{x^2} + {x^3} - {x^2} - x + 4{x^2} - 4x - 4 \\= 2{x^4} - {x^3} + {x^2} - 5x - 4\end{array}\);
d)
\(\begin{array}{l}(3x - 4)(2x + 1) - (x - 2)(6x + 3) \\= 3x(2x + 1) - 4(2x + 1) - x(6x + 3) + 2(6x + 3)\\ = 6{x^2} + 3x - 8x - 4 - 6{x^2} - 3x + 12x + 6\\ = 4x + 2\end{array}\).