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\(|\dfrac{4}{3}x-\dfrac{3}{4}|=\left|-\dfrac{1}{3}\right|.\left|x\right|\Leftrightarrow|\dfrac{4}{3}x-\dfrac{3}{4}|=\dfrac{1}{3}.\left|x\right|\left(1\right)\)
Tìm nghiệm \(\dfrac{4}{3}x-\dfrac{3}{4}=0\Leftrightarrow\dfrac{4}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.\dfrac{3}{4}\Leftrightarrow x=\dfrac{9}{16}\)
\(x=0\)
Lập bảng xét dấu :
\(x\) \(0\) \(\dfrac{9}{16}\)
\(\left|\dfrac{4}{3}x-\dfrac{3}{4}\right|\) \(-\) \(0\) \(-\) \(0\) \(+\)
\(\left|x\right|\) \(-\) \(0\) \(+\) \(0\) \(+\)
TH1 : \(x< 0\)
\(\left(1\right)\Leftrightarrow-\dfrac{4}{3}x+\dfrac{3}{4}=\dfrac{1}{3}.\left(-x\right)\)
\(\Leftrightarrow-\dfrac{4}{3}x+\dfrac{3}{4}=-\dfrac{1}{3}.x\)
\(\Leftrightarrow\dfrac{4}{3}x-\dfrac{1}{3}x=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{3}{4}\) (loại vì không thỏa \(x< 0\))
TH2 : \(0\le x\le\dfrac{9}{16}\)
\(\left(1\right)\Leftrightarrow-\dfrac{4}{3}x+\dfrac{3}{4}=\dfrac{1}{3}x\)
\(\Leftrightarrow\dfrac{4}{3}x+\dfrac{1}{3}x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.\dfrac{3}{5}\Leftrightarrow x=\dfrac{9}{20}\) (thỏa điều kiện \(0\le x\le\dfrac{9}{16}\))
TH3 : \(x>\dfrac{9}{16}\)
\(\left(1\right)\Leftrightarrow\dfrac{4}{3}x-\dfrac{3}{4}=\dfrac{1}{3}x\)
\(\Leftrightarrow\dfrac{4}{3}x-\dfrac{1}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}\) (thỏa điều kiện \(x>\dfrac{9}{16}\))
Vậy \(x\in\left\{\dfrac{9}{20};\dfrac{3}{4}\right\}\)
a) x - 7 = -5
=> x = -5 + 7
=> x = 2
b) 128 - 3 . (x + 4) = 23
=> 3.( x + 4) = 128 - 23
=> 3.( x + 4) = 105
=> x + 4 = 105 : 3
=> x + 4 = 35
=> x = 35 - 4
=> x = 31
c) [ ( 6x - 39 ) : 7 ].4 = 12
( 6x -39) : 7 = 12 :4
=> ( 6x - 39 ): 7 = 3
=> 6x - 39 = 3 x 7
=> 6x - 39 = 21
=> 6x = 21 + 39
=> 6x = 60
=> x = 60: 6
=> x = 10
d) ( x: 3 - 4) , 5 = 15
=> x: 3 - 4 = 15 : 5
=> x : 3 - 4 = 3
=> x: 3 = 3+4
=> x: 3 = 7
=> x = 7x 3
=> x =21
Theo đề, ta có: \(\frac{x}{2}=\frac{y}{-5}\)và \(x-y=-7\)
Theo TC dãy tỉ số bằng nhau, ta có:
\(\frac{x}{2}=\frac{y}{-5}=\frac{x-y}{2-\left(-5\right)}=\frac{-7}{7}=-1\)
\(\Leftrightarrow\hept{\begin{cases}x=-1.2=-2\\y=-1.-5=5\end{cases}}\)
Cái dấu chéo / là gì vậy bạn ( có phải la dấu GTTĐ | ko)
a, ( 8x - 3 ) ( 3x + 2 ) - ( 4x + 7 ) ( x + 4 ) = ( 2x + 1 ) ( 5x - 1 )
( 24x2 + 16x - 9x - 6 ) - ( 4x2 - 16x - 7x + 28 ) = 10x2 - 2x + 5x -1
24x2 + 16x - 9x - 6 -4x2 - 16x - 7x - 10x2 + 2x - 5x = 6 + 28 - 1
10x2 -19x = 33
10x2 - 19x -33 = 0 \(\Leftrightarrow\)10x( x+ 3 ) + 11 ( x- 3 ) = 0
=> ( x- 3 ) ( 10x + 11 ) = 0\(\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-11}{10}\end{cases}}\)
b, 4( x - 1 ) ( x + 5 ) - ( x + 2 ) ( x + 5 ) = 3( x - 1 ) ( x + 2 )
4( x2 - 5x - x + 5 ) - ( x2 + 5x + 2x + 10 ) = 3( x2 + 2x - x - 2 )
4x2 - 20x - 4x + 20 - x2 - 5x - 2x - 10 = 3x2 + 6x - 3x - 6
( 4x2 - x2 ) + ( -20x - 4x - 5x - 2x ) + 20 - 10 = 3x2 + ( 6x - 3x ) - 6
3x2 - 31x - 3x2 - 3x = -6-10
-34x = -16
x = \(\frac{8}{17}\)
\(128-3\left(x+7\right)=4^3\\ 128-3\left(x+7\right)=64\\ 3\left(x+7\right)=128-64\\ 3\left(x+7\right)=64\\ x+7=\dfrac{64}{3}\\ x=\dfrac{64}{3}-7\\ x=\dfrac{43}{3}\)
\(128-3\left(x+7\right)=4^3\)
\(\Leftrightarrow128-3x-21=64\)
\(\Leftrightarrow107-3x-21=64\)
\(\Leftrightarrow-3x+43=0\)
\(\Leftrightarrow x=\dfrac{43}{3}\)