(2/3-4/7)x2/5
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27 tháng 8 2021
`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`
27 tháng 8 2021
`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`
HK
2
NK
19 tháng 3 2023
\(\dfrac{4}{7}:\dfrac{4}{3}+5:\dfrac{4}{3}=\left(\dfrac{4}{7}+5\right):\dfrac{4}{3}=\dfrac{39}{7}:\dfrac{4}{3}=\dfrac{39}{7}\times\dfrac{3}{4}=\dfrac{117}{28}\)
\(\dfrac{4}{7}:\dfrac{4}{3}+5:\dfrac{4}{3}=\dfrac{3}{7}+\dfrac{15}{4}=\dfrac{12}{28}+\dfrac{105}{28}=\dfrac{117}{28}\)
\(\dfrac{3}{4}\times\dfrac{2}{5}-\dfrac{3}{4}\times\dfrac{2}{7}=\dfrac{3}{4}\times\left(\dfrac{2}{5}-\dfrac{2}{7}\right)=\dfrac{3}{4}\times\dfrac{4}{35}=\dfrac{12}{140}=\dfrac{3}{35}\)
\(\dfrac{3}{4}\times\dfrac{2}{5}-\dfrac{3}{4}\times\dfrac{2}{7}=\dfrac{3}{10}-\dfrac{3}{14}=\dfrac{42}{140}-\dfrac{30}{140}=\dfrac{12}{140}=\dfrac{3}{35}\)
\(\left(\dfrac{2}{3}-\dfrac{4}{7}\right)\times\dfrac{2}{5}=\dfrac{2}{21}\times\dfrac{2}{5}=\dfrac{4}{105}\)
\(\left(\dfrac{2}{3}-\dfrac{4}{7}\right)\times\dfrac{2}{5}=\dfrac{2}{3}\times\dfrac{2}{5}-\dfrac{4}{7}\times\dfrac{2}{5}=\dfrac{4}{15}-\dfrac{8}{35}=\dfrac{140}{525}-\dfrac{120}{525}=\dfrac{20}{525}=\dfrac{4}{105}\)
#YVA
19 tháng 3 2023
4/7:4/3+5:4/3
= 3/7 + 15/4
=4 5/28
4/7:4/3+5:4/3
= (4/7+5) x 3/4
=4 5/28
3/4x2/5-3/4x2/7
=3/10 - 3/14
=3/35
3/4x2/5-3/4x2/7
=3/4x(2/5-2/7)
=3/35
(2/3-4/7)x2/5
=2/21x2/5
=4/105
(2/3-4/7)x2/5
=2/3x2/5-4/7x2/5
=4/105
26 tháng 7 2021
1) \(\left(\dfrac{1}{2}x+3\right)\left(x^2-4x-6\right)\)
\(=\dfrac{1}{2}x^3-2x^2-3x+3x^2-12x-18\)
\(=\dfrac{1}{2}x^3+x^2-15x-18\)
2) \(\left(6x^2-9x+15\right)\left(\dfrac{2}{3}x+1\right)\)
\(=4x^3+6x^2-6x^2-9x+10x+15\)
\(=4x^3+x+15\)
3) Ta có: \(\left(3x^2-x+5\right)\left(x^3+5x-1\right)\)
\(=3x^5+15x^2-3x^2-x^4-5x^2+x+5x^3+25x-5\)
\(=3x^5-x^4+5x^3+10x^2+26x-5\)
4) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\)
\(=\left(x^2-1\right)\left(x-2\right)\)
\(=x^3-2x^2-x+2\)
27 tháng 4 2023
`@` `\text {dnammv}`
`a,`
`4x(x^2-x-1)-(x^2-2)(x+3)`
`= 4x^3-4x^2-4x- [x^2(x+3)-2(x+3)]`
`= 4x^3-4x^2-4x- (x^3+3x^2-2x-6)`
`= 4x^3-4x^2-4x-x^3-3x^2+2x+6`
`= 3x^3 - 7x^2-2x+6`
`b,`
`(x+5)(x+7)-7x(x+3)`
`= x(x+7)+5(x+7)-7x^2-21x`
`= x^2+7+5x+35-7x^2-21x`
`= -6x^2-16x+35`
`c,`
`x(x^2-x-2)-(x+5)(x-1)`
`= x^3-x^2-2x- [x(x-1)+5(x-1)]`
`= x^3-x^2-2x- (x^2-x+5x-5)`
`= x^3-x^2-2x - x^2 + x -5x+5`
`= x^3-2x^2- 4x+5`
`d,`
`(x+5)(x+7)-(x-4)(x+3)`
`= x(x+7)+5(x+7)- [x(x+3)-4(x+3)]`
`= x^2+7x+5x+35 - (x^2+3x-4x-12)`
`= x^2+12x+35 - x^2+x+12`
`= 13x+47`
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
15 tháng 8
A = - 3\(x\).(\(x-5\)) + 3(\(x^2\) - 4\(x\)) - 3\(x\) - 10
A = - 3\(x^2\) + 15\(x\) + 3\(x^2\) - 12\(x\) - 3\(x\) - 10
A = (- 3\(x^2\) + 3\(x^2\)) + (15\(x\) - 12\(x\) - 3\(x\)) - 10
A = 0 + (3\(x-3x\)) - 10
A = 0 - 10
A = - 10
13 tháng 11 2021
\(1,\Leftrightarrow x^2+10x+25=x^2-4x-21\\ \Leftrightarrow14x=-46\\ \Leftrightarrow x=-\dfrac{23}{7}\\ 2,\Leftrightarrow x^3+8=15+x^3+2x\\ \Leftrightarrow2x=-7\Leftrightarrow x=-\dfrac{7}{2}\\ 3,\Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x=-3\\ 4,\Leftrightarrow x^3-9x^2+27x-27=0\\ \Leftrightarrow\left(x-3\right)^3=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\\ 5,\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Leftrightarrow-12x=24\Leftrightarrow x=-2\\ 6,\Leftrightarrow x^2-3x+5x-15=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
VT
2
18 tháng 3 2023
b: \(=\left(\dfrac{-3}{4}\cdot\dfrac{-4}{3}\right)\cdot\dfrac{12}{7}=\dfrac{12}{7}\)
c: \(=-\dfrac{9}{5}\cdot\dfrac{5}{2}\cdot\dfrac{2}{5}=\dfrac{-9}{5}\)
=2/21.2/5
=4/105
HT
=4/105