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\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
1 ) 3x^2 - 11x + 6 = 3x^2 - 9x - 2x + 6 = 3x( x- 3 ) - 2( x - 3) = ( 3x - 2 )( x - 3 )
2) 8x^2 - 2x - 1 = 8x^2 - 4x + 2x - 1 = 4x( 2x - 1 ) + 2x - 1 = ( 4x + 1 )( 2x - 1 )
3; 8x^2 - 2x - 1 =8x^2 - 4x + 2x - 1 = 4x( 2x - 1 ) + 2x - 1 = ( 4x + 1 )( 2x - 1 )
4; x^4 - 3x^2 - 4 = x^4 - 4x^2 + x^2 - 4 = x^2 ( x ^2 - 4 ) + x^2 - 4 = ( x^2 + 1 )( x^2 - 4 ) = ( x^2 + 1 )( x - 2 )( x + 2)
5) = x^2 ( x + 2 ) - 3 ( x+ 2 ) = ( x^2 - 3 )( x + 2 )
Nhiều quá
Bài 1:
- \(\dfrac{11}{2}x\) + 1 = \(\dfrac{1}{3}x-\dfrac{1}{4}\)
- \(\dfrac{11}{2}\)\(x\) - \(\dfrac{1}{3}\)\(x\) = - \(\dfrac{1}{4}\) - 1
-(\(\dfrac{33}{6}\) + \(\dfrac{2}{6}\))\(x\) = - \(\dfrac{5}{4}\)
- \(\dfrac{35}{6}\)\(x\) = - \(\dfrac{5}{4}\)
\(x=-\dfrac{5}{4}\) : (- \(\dfrac{35}{6}\))
\(x\) = \(\dfrac{3}{14}\)
Vậy \(x=\dfrac{3}{14}\)
Bài 2: 2\(x\) - \(\dfrac{2}{3}\) - 7\(x\) = \(\dfrac{3}{2}\) - 1
2\(x\) - 7\(x\) = \(\dfrac{3}{2}\) - 1 + \(\dfrac{2}{3}\)
- 5\(x\) = \(\dfrac{9}{6}\) - \(\dfrac{6}{6}\) + \(\dfrac{4}{6}\)
- 5\(x\) = \(\dfrac{7}{6}\)
\(x\) = \(\dfrac{7}{6}\) : (- 5)
\(x\) = - \(\dfrac{7}{30}\)
Vậy \(x=-\dfrac{7}{30}\)
Bài 1 :
Gọi 3 số tự nhiên cần tìm là n - 1 ; n ; n + 1 ( với n \(\in\) N* )
Theo bài ra , ta có :
[ ( n -1 )n ] + [ ( n + 1 )n ] + [ (n - 1)( n + 1 ) ] = 242
=> ( \(n^2\) - n ) + ( \(n^2\) + n ) + ( \(n^2\) + n - n - 1 ) = 242
=> \(3n^2\) - 1 = 242
=> \(3n^2\) = 243
=> \(n^2\) = 81
=> n = 9 hoặc n = -9
Mà n là số tự nhiên \(\Rightarrow\) n = 9
Vậy 3 số cần tìm là 8 ; 9 ; 10
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
Ủa mấy cái này tưởng mấy em được học rồi nhỉ?
a, \(|3x-4|+|4y+1|=0\)
\(\Rightarrow\hept{\begin{cases}|3x-4|=0\\|4y+1|=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x-4=0\\4y+1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{4}{3}\\y=-\frac{1}{4}\end{cases}}}\)
b, Lập bảng xét dấu giá trị tuyệt đối
\(x\) \(-\frac{5}{2}\) \(\frac{1}{3}\)
\(2x+5\) \(-5-2x\) \(0\) \(2x+5\) \(||\) \(2x+5\)
\(3x-1\) \(1-3x\) \(||\)\(1-3x\) \(0\)\(3x-1\)
\(VT\) \(||\) \(||\)
TH1: \(x< -\frac{5}{2}\)\(\Rightarrow\hept{\begin{cases}|2x+5|=-5-2x\\|3x-1|=1-3x\end{cases}}\)
\(\Rightarrow-5-2x+1-3x=3\)\(\Leftrightarrow-4-5x=3\Leftrightarrow x=-\frac{7}{5}\left(L\right)\)
TH2: \(-\frac{5}{2}\le x\le\frac{1}{3}\)\(\Rightarrow\hept{\begin{cases}|2x+5|=2x+5\\|3x-1|=1-3x\end{cases}}\)
\(\Rightarrow2x+5+1-3x=3\)\(\Leftrightarrow6-x=3\Leftrightarrow x=3\left(L\right)\)
TH3: \(x>\frac{1}{3}\)\(\Rightarrow\hept{\begin{cases}2x+5|=2x+5\\|3x-1|=3x-1\end{cases}}\)
\(\Rightarrow2x+5+3x-1=3\)\(\Leftrightarrow5x+4=3\Leftrightarrow5x=-1\Leftrightarrow x=-\frac{1}{5}\left(L\right)\)
Vậy PT đã cho vô nghiệm.
P/S: Không hiểu ở đâu thì nhắn chị nhé.
c) \(\left(3x-1\right).\left(2x+7\right)-\left(x+1\right).\left(6x-5\right)=\left(x+2\right)-\left(x-5\right)\)
\(\Leftrightarrow6x^2+21x-2x-7-\left(6x^2-5x+6x-5\right)=x+2-x+5\)
\(\Leftrightarrow18x-2-7=0\)
\(\Rightarrow x=\dfrac{9}{18}=\dfrac{1}{2}\)
b) \(2.\left(3x-1\right).\left(2x+5\right)-6.\left(2x-1\right).\left(x+2\right)=1\)
\(\Leftrightarrow\left(6x-2\right).\left(2x+5\right)-\left(12x-6\right).\left(x+2\right)=1\)
\(\Leftrightarrow12x^2+30x-4x-10-\left(12x^2+24x-6x-12\right)=1\)
\(\Leftrightarrow12x^2+26x-10-12x^2-18x +12=1\)
\(\Leftrightarrow8x+2=1\)
\(\Rightarrow x=\dfrac{-1}{8}\)
c)3(2x-1)-5(x-3)+6(3x-4)=24
<=>6x-3-5x-15+18x-24=24
<=>19x-12=24
<=>19x=36
<=>x=\(\frac{36}{19}\)
d)2x(5-3x)+2x(3x-5)-3(x-7)=3
<=>10x-6x2+6x2-10x-3x-21=3
<=>-3(x-7)=3
<=>21-3x=3
<=>-3x=-18
<=>x=6
- \(\dfrac{5}{6}\) + 3\(x\) = \(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)\(x\)
3\(x\) + \(\dfrac{1}{2}\)\(x\) = \(\dfrac{2}{3}\) + \(\dfrac{5}{6}\)
\(\dfrac{7x}{2}\) = \(\dfrac{3}{2}\)
\(x\) = \(\dfrac{3}{2}\) : \(\dfrac{7}{2}\)
\(x\) = \(\dfrac{3}{7}\)
:)) anh Hải pro là ai vậy bn :))