Tìm X:
c, (3x+1)^2+(x+1)^2= 10.(x-1).(x+1)
Giúp mik nhé! <3333
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N
15 tháng 9 2021
\(c.\left(1-2x\right)^2-\left(3x-2\right)^2=0\)
\(\left(1-2x-3x+2\right)\left(1-2x+3x-2\right)=0\)
\(\left(-5x+3\right)\left(x-1\right)=0\)
\(\left[{}\begin{matrix}-5x+3=0\\-x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=1\end{matrix}\right.\)
\(d.\left(x-2\right)^2-\left(5-2x\right)^2=0\)
\(\left(x-2-5+2x\right)\left(x-2+5-2x\right)=0\)
\(\left(3x-7\right)\left(-x+3\right)=0\)
\(\left[{}\begin{matrix}3x-7=0\\-x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)
15 tháng 9 2021
\(c,\Leftrightarrow1-4x+4x^2=9x^2-12x+4\\ \Leftrightarrow5x^2-8x+3=0\\ \Leftrightarrow\left(x-1\right)\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{5}\end{matrix}\right.\\ d,\Leftrightarrow\left(x-2-5+2x\right)\left(x-2+5-2x\right)=0\\ \Leftrightarrow\left(3x-7\right)\left(3-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)
TM
10 tháng 3 2022
c. \(2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)-\dfrac{3}{2}=\dfrac{1}{4}\)
\(2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{1}{4}+\dfrac{3}{2}=\dfrac{7}{4}\)
\(\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{4}:2=\dfrac{7}{8}\)
\(\dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}=\dfrac{29}{24}\)
\(x=\dfrac{29}{24}:\dfrac{1}{2}=\dfrac{29}{12}\)
Vậy : ...
d. \(\dfrac{4}{5}-\dfrac{1}{2}x=\dfrac{1}{10}\)
\(-\dfrac{1}{2}x=\dfrac{1}{10}-\dfrac{4}{5}=-\dfrac{7}{10}\)
\(x=-\dfrac{7}{10}:\left(-\dfrac{1}{2}\right)=\dfrac{7}{5}\)
Vậy : ...
N
10 tháng 3 2022
c) \(2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)-\dfrac{3}{2}=\dfrac{1}{4}\)
\(\Rightarrow2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{1}{4}+\dfrac{3}{2}\)
\(\Rightarrow2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{1}{4}+\dfrac{6}{4}\)
\(\Rightarrow2\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{7}{4}\)
\(\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{7}{4}:2\)
\(\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{7}{4}.\dfrac{1}{2}\)
\(\Rightarrow\left(\dfrac{1}{2}x-\dfrac{1}{3}\right)=\dfrac{7}{8}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{21}{24}+\dfrac{8}{24}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{29}{24}\)
\(\Rightarrow x=\dfrac{29}{24}:\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{29}{24}.2\)
\(\Rightarrow x=\dfrac{29}{12}\)
d) \(\dfrac{4}{5}-\dfrac{1}{2}x=\dfrac{1}{10}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{4}{5}-\dfrac{1}{10}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{8}{10}-\dfrac{1}{10}\)
\(\Rightarrow\dfrac{1}{2}x=\dfrac{7}{10}\)
\(\Rightarrow x=\dfrac{7}{10}:\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{7}{10}.2\)
\(\Rightarrow x=\dfrac{7}{5}\)
TA
16 tháng 7 2017
a) \(\left(x+3\right)^3-x.\left(3x+1\right)^2+\left(2x+1\right).\left(4x^2-2x+1\right)-3x^2=54\)
\(\Leftrightarrow x^3+9x^2+27x+27-x.\left(9x^2+6x+1\right)+8x^3+1-3x^2=54\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=54\)
\(\Leftrightarrow26x+28=54\Leftrightarrow26x=54-28\Leftrightarrow26x=26\Leftrightarrow x=1\)
Vậy nghiệm của phương trình là x=1
b) \(\left(x-3\right)^3-\left(x-3\right).\left(x^2+3x+9\right)+6.\left(x+1\right)^2+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+6.\left(x^2+2x+1\right)+3x^2=-33\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\)
\(\Leftrightarrow27x+12x+6=-33\Leftrightarrow39x=-33-6\Leftrightarrow39x=-39\Leftrightarrow x=-1\)
Vậy nghiệm của phương trình là x = -1
16 tháng 7 2017
Trần Anh: Hí hí =)) ÀI LỚP DIU CHIU CHIU CHÍU :3 CẢM ƠN PẠN NHIỀU NHÁ ;) ;) ;)
11 tháng 10 2021
\(c,\Rightarrow\left(x-2\right)-\left(x-2\right)^2=0\\ \Rightarrow\left(x-2\right)\left(1-x+2\right)=0\\ \Rightarrow\left(x-2\right)\left(3-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\\ d,\Rightarrow\left(x^2+3\right)\left(x+1\right)+\left(x+1\right)=0\\ \Rightarrow\left(x^2+3+1\right)\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2+4=0\left(vô.nghiệm\right)\\x+1=0\end{matrix}\right.\Rightarrow x=-1\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 7
a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)
(\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)
- \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)
\(x\) = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))
\(x=\) - \(\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 7
b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)
3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)
3\(x\) - 3,7 = - \(\dfrac{19}{2}\)
3\(x\) = - \(\dfrac{19}{2}\) + 3,7
3\(x\) = - \(\dfrac{29}{5}\)
\(x\) = - \(\dfrac{29}{5}\) : 3
\(x\) = - \(\dfrac{29}{15}\)
Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\)
TA
8 tháng 7 2017
a) Cậu xem lại đề đi
b) \(3x.\left(x-2\right)-5x.\left(1-x\right)-8.\left(x^2-3\right)=4\)\(\Leftrightarrow3x^2-6x-5x+5x^2-8x^2+24-4=0\Leftrightarrow-11x+20=0\Leftrightarrow-11x=-20\Leftrightarrow x=\frac{20}{11}\)
c) \(2x^2+3.\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\Leftrightarrow2x^2+3\left(x^2-1\right)-5x\left(x+1\right)=0\)
\(\Leftrightarrow2x^2+3x^2-3-5x^2-5x=0\Leftrightarrow-5x=3\Leftrightarrow x=-\frac{3}{5}\)
8 tháng 7 2017
Trần Anh: Cảm ơn bạn nhiều nhé :)) Phần a đúng là có sai đề pạn ạ mik làm hoài mà cux ko ra hì hì !!~~ Dù sao mik cux cảm ơn pạn nhiều nhiều nhé :3
VQ
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
2 tháng 8 2023
(3\(x\) - 2)(\(x+4\)) - (1- \(x\))(2-\(x\)) =(\(x+1\))(\(x-2\))
3\(x^2\) + 12\(x\) - 2\(x\) - 8 - (\(x+1\))(\(x-2\)) - [-(\(x-2\))](1- \(x\)) = 0
3\(x^2\) + 10\(x\) - 8 - (\(x-2\))( \(x\) + 1 - 1 + \(x\)) = 0
3\(x^2\) + 10\(x\) - 8 - (\(x-2\)). 2\(x\) = 0
3\(x^2\) + 10\(x\) - 8 - 2\(x^2\) + 4\(x\) = 0
\(x^2\) + 14\(x\) - 8 = 0
\(x^2\) + 7\(x\) + 7\(x\) + 49 - 57 = 0
\(x\)( \(x\) + 7) + 7(\(x\) + 7) = 57
(\(x+7\))(\(x\) + 7) =57
(\(x+7\))2 = 57
\(\left[{}\begin{matrix}x+7=\sqrt{57}\\x+7=-\sqrt{57}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-7+\sqrt{57}\\x=-7-\sqrt{57}\end{matrix}\right.\)
Vậy \(x\) \(\in\) { -7 - \(\sqrt{57}\); - 7 + \(\sqrt{57}\)}
12 tháng 6 2023
2:
=>x^3-1-2x^3-4x^6+4x^6+4x=6
=>-x^3+4x-7=0
=>x=-2,59
4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50
=>-62x+12=-50
=>x=1
1 tháng 10 2023
(x-1)(x+2)-x(x-2)-3x
=x2+x-2-x2+2x-3x
=-2 <=>Bt =-2 mọi x
Vậy bt =-2 khi x=-100
( 3x + 1 )2 + ( x + 1 )2 = 10( x - 1 )( x + 1 )
<=> 9x2 + 6x + 1 + x2 + 2x + 1 = 10( x2 - 1 )
<=> 10x2 + 8x + 2 = 10x2 - 10
<=> 10x2 + 8x - 10x2 = -10 - 2
<=> 8x = -12
<=> x = -12/8 = -3/2
c. => 9x2 + 6x + 1 + x2 + 2x + 1 = 10 . ( x2 - 1 )
=> 10x2 + 8x + 2 = 10x2 - 10
=> 10x2 + 8x + 2 - 10x2 + 10 = 0
=> 8x + 12 = 0
=> 8x = 12
=> x = 3/2