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11 tháng 7 2018
\(a,A=\left(x+5\right)^3\)
\(b,B=\left(x-3\right)^3\)
\(c,C=\frac{x^3}{8}+\frac{x^2y}{4}+\frac{xy^2}{6}+\frac{y^3}{27}=\left(\frac{x}{2}+\frac{y}{3}\right)^3\)
Mk nghĩ đề bài phần c fải như trên ,cn đâu bn tự thay số vào nha.
9 tháng 7 2019
\(a,2\left(x-3\right)-5\left(2x-4\right)=0\)
=> \(2x-6-10x-20=0\)
=> \(\left(2x-10x\right)-\left(6+20\right)=0\)
=> \(-8x-26=0\)
=> \(-8x=26\)
=> \(x=26:-8=-\frac{13}{4}\)
Vậy \(x\in\left\{-\frac{13}{4}\right\}\)
\(b,3+\frac{1}{x-8}=0\)
=> \(\frac{1}{x-8}=0-3=-3\)
=> \(x-8=-\frac{1}{3}\)
=> \(x=-\frac{1}{3}+8=\frac{23}{3}\)
Vậy \(x\in\left\{\frac{23}{3}\right\}\)
\(c,\frac{8}{3}-\frac{2x+3}{5}=\frac{-7}{3}\)
=> \(15.\frac{8}{3}-15.\frac{2x+3}{5}=15.\frac{-7}{3}\)
Chiệt tiêu
=> \(5.8-3\left(2x+3\right)=5.\left(-7\right)\)
=> \(40-\left(6x+9\right)=-35\)
=> \(40-6x-9=-35\)
=>\(31=6x=-35\)
=> \(6x=41-\left(-35\right)=66\)
=> \(x=66:6=11\)
Vậy \(x\in\left\{11\right\}\)
\(d,\frac{1}{9}=\frac{5}{3x-5}=0\)
=> \(\frac{1}{9}=0\left(sai\right)\)
=> \(x\in\varnothing\)
15 tháng 3 2020
1, \(\frac{4y^2}{11x^4}.\left(-\frac{3x^2}{8y}\right)\)\(=\frac{4y.y}{11x^2.x^2}.\frac{-3x^2}{2.4y}\)\(=\frac{y}{11x^2}.\frac{-3}{2}=\frac{-3y}{22x^2}\)
2, \(\frac{4x^2}{5y^2}:\frac{6x}{5y}:\frac{2x}{3y}\)\(=\frac{4x^2}{5y^2}.\frac{5y}{6x}.\frac{3y}{2x}\)\(=\frac{2x.2x}{5y.y}.\frac{5y}{3.2x}.\frac{3y}{2x}\)\(=\frac{2x}{y}.\frac{1}{3}.\frac{3y}{2x}\)
\(\frac{2x}{3y}.\frac{3y}{2x}=1\)
3, \(\frac{x^2-4}{3x+12}.\frac{x+4}{2x-4}\)\(=\frac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}.\frac{x+4}{2\left(x-2\right)}\)\(=\frac{\left(x+2\right)}{3}.\frac{1}{2}=\frac{x+2}{6}\)
4, \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}\)\(=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\left(-\frac{2\left(x-2\right)}{x+2}\right)=\frac{5}{4}.\frac{-2}{1}=-\frac{5}{2}\)
5, \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}.\frac{3}{-\left(x-6\right)}=\frac{x+6}{2\left(x+5\right)}.\frac{-3}{1}=\frac{-3\left(x+6\right)}{2\left(x+5\right)}\)
6, \(\frac{x^2-9y^2}{x^2y^2}.\frac{3xy}{2x-6y}=\frac{\left(x-3y\right)\left(x+3y\right)}{\left(xy\right)^2}.\frac{3xy}{2\left(x-3y\right)}=\frac{x+3y}{xy}.\frac{3}{2}=\frac{3\left(x+3y\right)}{2xy}\)
7, \(\frac{3x^2-3y^2}{5xy}.\frac{15x^2y}{2y-2x}=\frac{3\left(x-y\right)\left(x+y\right)}{5xy}.\frac{5xy.3x}{-2\left(x-y\right)}=\frac{3\left(x+y\right)}{1}.\frac{3x}{-2}=\frac{-9x\left(x+y\right)}{2}\)
24 tháng 2 2020
\(1,\frac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\frac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}=\frac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\frac{x^3+y^3}{x\left(x^3-y^3\right)}\)
\(2,=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\frac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a+c-b\right)}=\frac{a+b-c}{a+c-b}\)
pt thành nhân tử là ra
H
30 tháng 6 2019
\(a,\frac{-x}{4}+6=8\)\(\Leftrightarrow\frac{x}{-4}=2\Leftrightarrow x=-8\)
b,\(\frac{-4}{x}-7=-5\Leftrightarrow\frac{-4}{x}=2\Leftrightarrow x=-2\)
c,\(12+\frac{-6}{5x}=17\Leftrightarrow-\frac{6}{5x}=5\Leftrightarrow x=-\frac{6}{25}\)
d,\(\frac{3-x}{7}=\frac{x+5}{4}\Leftrightarrow12-4x=7x+35\Leftrightarrow-11x=23\Leftrightarrow x=-\frac{23}{11}\)
e,\(7-2x=-\frac{3}{3x}=-\frac{1}{x}\Leftrightarrow7x-2x^2+1=0\)
\(\Leftrightarrow-2\left(x^2+\frac{7}{2}x+\frac{49}{16}\right)+\frac{57}{8}=0\Leftrightarrow\left(x+\frac{7}{4}\right)^2=\frac{57}{16}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{7}{4}=\frac{\sqrt{57}}{16}\\x+\frac{7}{4}=-\frac{\sqrt{57}}{16}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{57}-28}{16}\\x=\frac{-\sqrt{57}-28}{16}\end{matrix}\right.\)
28 tháng 6 2019
a, \(-\frac{x}{4}+6=8\)
=> \(-\frac{x}{4}=8-6=2\)
=> \(x=2.-4=-8\)
Vậy \(x\in\left\{-8\right\}\)
\(b,\frac{4}{-x}-7=-5\)
=> \(\frac{4}{-x}=-5+\left(-7\right)=-12\)
=> \(x=4:12=\frac{1}{3}\)
Vậy \(x\in\left\{\frac{1}{3}\right\}\)
\(c,12+\frac{-6}{5x}=17\)
=> \(-\frac{6}{5x}=17-12=5\)
=> \(5x=-6:5=-\frac{6}{5}\)
=> \(x=-\frac{6}{5}:5=\frac{6}{25}\)
Vậy \(x\in\left\{\frac{6}{25}\right\}\)
\(d,\frac{3-x}{7}=\frac{x+5}{4}\)
=>\(4\left(3-x\right)=7\left(x+5\right)\)
=> \(12-4x=7x+35\)
=> \(-4x-7x=35-12\)
=> \(-11x=23\)
=> \(x=23:\left(-11\right)=-\frac{23}{11}\)
Vậy \(x\in\left\{-\frac{23}{11}\right\}\)
e, \(7-2x=-\frac{3}{3x}\)
=> \(7-2x=-\frac{1}{x}\)
=> \(7=2x+\left(-\frac{1}{x}\right)\)
=> \(7=2x-\frac{1}{x}\)
=> \(7=\frac{2x^2}{x}-\frac{1}{x}\)
=> \(7=\frac{2x^2-1}{x}\)
=> :))
XU
8 tháng 2 2020
a, \(\frac{x-3}{5}\) = 6 - \(\frac{1-2x}{3}\)
⇔ 3(x - 3) = 90 - 5(1 - 2x)
⇔ 3x - 9 = 90 - 5 + 10x
⇔ 3x - 10x = 90 - 5 + 9
⇔ -7x = 94
⇔ x = \(\frac{-94}{7}\)
S = { \(\frac{-94}{7}\) }
b, \(\frac{3x-2}{6}\) - 5 = \(\frac{3-2\left(x+7\right)}{4}\)
⇔ 2(3x - 2) - 60 = 9 - 6(x + 7)
⇔ 6x - 4 - 60 = 9 - 6x - 42
⇔ 6x + 6x = 9 - 42 + 60 + 4
⇔ 12x = 31
⇔ x = \(\frac{31}{12}\)
S = { \(\frac{31}{12}\) }
c, \(\frac{x+8}{6}\) - \(\frac{2x-5}{5}\) = \(\frac{x+1}{3}\) - x + 7
⇔ 5(x+ 8) - 6(2x - 5) = 10(x+1) - 30x+210
⇔ 5x+ 40 - 12x+ 30 = 10x+ 10 - 30x+210
⇔ 5x - 12x - 10x+ 30x = 10+ 210 - 30- 40
⇔ 13x = 150
⇔ x = \(\frac{150}{13}\)
S = { \(\frac{150}{13}\) }
d, \(\frac{7x}{8}\) - 5(x - 9) = \(\frac{2x+1,5}{6}\)
⇔ 21x - 120(x - 9) = 4(2x + 1,5)
⇔ 21x - 120x + 1080 = 8x + 6
⇔ 21x - 120x - 8x = 6 - 1080
⇔ -107x = -1074
⇔ x = \(\frac{1074}{107}\)
S = { \(\frac{1074}{107}\) }
e, \(\frac{5\left(x-1\right)+2}{6}\) - \(\frac{7x-1}{4}\) = \(\frac{2\left(2x+1\right)}{7}\) - 5
⇔ 140(x-1)+56 - 42(7x-1) = 48(2x+1)-840
⇔ 140x -140+56 -294x+42= 96x+48 -840
⇔ 140x -294x -96x = 48 -840 -42 -56+140
⇔ -250x = -750
⇔ x = 3
S = { 3 }
f, \(\frac{x+1}{3}\) + \(\frac{3\left(2x+1\right)}{4}\) = \(\frac{2x+3\left(x+1\right)}{6}\) + \(\frac{7+12x}{12}\)
⇔ 4(x+1)+9(2x+1) = 4x+6(x+1)+7+12x
⇔ 4x+4+18x+9 = 4x+6x+6+7+12x
⇔ 4x+18x - 4x - 6x - 12x = 6+7- 9 - 4
⇔ 0x = 0
S = R
Chúc bạn học tốt !
22 tháng 4 2020
Bạn ơi giải giúp mình 2 bài này với ạ : https://hoc24.vn/hoi-dap/question/969683.html
Mình cảm ơn trước nhaa
\(-A\cdot\frac12xy^3=-\frac78x^3y^6\)
\(\Rightarrow A=-\frac78x^3y^6:\frac12xy^3\)
\(\Rightarrow A=-\frac74x^2y^3\)
A = 7/4 x^2 y^3