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a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
\(\Rightarrow\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=-4x+1\end{cases}}\Rightarrow\orbr{\begin{cases}4x-\frac{3}{2}x-1=\frac{1}{2}\\-4x-\frac{3}{2}x+1=\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{2}x=\frac{3}{2}\\-\frac{11}{2}x=-\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
phần b ở đề bài mình ghi sai, là bằng 0 chứ ko phải bằng 10
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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 = \(\frac{8}{17}\)
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1.a) có: \(|x-\frac{3}{2}|,|x+1|,\left|x-2\right|\ge0\Rightarrow4x\ge0\Rightarrow x\ge0\)
\(x\ge0\Rightarrow x-\frac{3}{2}\ge\frac{-3}{2}\Rightarrow\left|x-\frac{3}{2}\right|\ge\left|\frac{-3}{2}\right|=\frac{3}{2}\Rightarrow\left|x-\frac{3}{2}\right|=x-\frac{3}{2}\)
cmtt: \(|x-2|=x-2\)
\(\Rightarrow3x-\frac{3}{2}+1-2=4x\)
\(\Rightarrow3x-\frac{5}{2}=4x\)
\(\Rightarrow x=\frac{-5}{2}\left(ko,t/m\right)\)
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\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{7}{12}-\left(\frac{5}{2}-\frac{13}{6}\right)\)
\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{7}{12}-\frac{1}{3}\)
\(\frac{1}{3}-\left(\frac{2}{3}-x+\frac{5}{4}\right)=\frac{1}{4}\)
\(\frac{2}{3}-x+\frac{5}{4}=\frac{1}{3}-\frac{1}{4}\)
\(\frac{2}{3}-x+\frac{5}{4}=\frac{1}{12}\)
\(\frac{2}{3}-x=\frac{1}{12}-\frac{5}{4}\)
\(\frac{2}{3}-x=-\frac{7}{6}\)
\(x=\frac{2}{3}-\left(-\frac{7}{6}\right)\)
\(x=\frac{2}{3}+\frac{7}{6}\)
\(x=\frac{11}{6}\)
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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
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Ta có: \(\hept{\begin{cases}\left|x+1\right|\ge0\\\left|x+3\right|\ge0\\\left|x+5\right|\ge0\end{cases}}\Rightarrow VT\ge0\)
\(\Leftrightarrow3x-4\ge\Leftrightarrow x\ge\frac{4}{3}\)
\(\Rightarrow pt\Leftrightarrow3x+9=3x-4\Leftrightarrow9=-4\)(vô lí)
Vậy pt vô nghiệm
\(\left||2x-3|-x+3\right|=4x-1\)(1)
*Nếu \(x\le3\)thì \(\left(1\right)\Leftrightarrow\left|2x-3\right|+3-x=4x-1\)
\(\Leftrightarrow\left|2x-3\right|=5x-4\)(2)
+) TH1: \(x\ge\frac{3}{2}\)thì \(\left(2\right)\Leftrightarrow2x-3=5x-4\)
\(\Leftrightarrow-3x=-1\Leftrightarrow x=\frac{1}{3}\left(L\right)\)
+) TH2: \(x< \frac{3}{2}\)thì \(\left(2\right)\Leftrightarrow3-2x=5x-4\)
\(\Leftrightarrow-7x=-7\Leftrightarrow x=1\left(TM\right)\)
*Nếu \(x>3\)thì \(\left(1\right)\Leftrightarrow\left|2x-3\right|-3+x=4x-1\)
\(\Leftrightarrow\left|2x-3\right|=3x+2\)(3)
+) TH1: \(x\ge\frac{3}{2}\)thì \(\left(3\right)\Leftrightarrow2x-3=3x+2\Leftrightarrow-x=5\Leftrightarrow x=-5\left(L\right)\)
+) TH2: \(x< \frac{3}{2}\)thì \(\left(3\right)\Leftrightarrow3-2x=3x+2\Leftrightarrow-5x=-1\Leftrightarrow x=\frac{1}{5}\left(L\right)\)
Vậy x = 1
Ta có : \(3.\left|1-\frac{1}{2}x\right|=\frac{4}{5}\left(4x-1\right)\)
=> \(\left|1-\frac{1}{2}x\right|=\frac{4}{15}\left(4x-1\right)\)
=> \(\left|1-\frac{1}{2}x\right|=\frac{16}{15}x-\frac{4}{15}\) (1)
ĐK: \(\frac{16}{15}x-\frac{4}{15}\ge0\Rightarrow x\ge\frac{1}{4}\)
Khi đó (1) <=> \(\orbr{\begin{cases}1-\frac{1}{2}x=\frac{16}{15}x-\frac{4}{15}\\1-\frac{1}{2}x=-\frac{16}{15}x+\frac{4}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{47}{30}x=\frac{19}{15}\\-\frac{17}{30}x=\frac{11}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{38}{47}\left(tm\right)\\x=-\frac{22}{17}\left(\text{loại}\right)\end{cases}}\)
Vậy x = 38/47