3(2x-1)-5(x-3)=24-6(3x-4)
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1) 3x-4+24+6=x+27+3x
3x-x-3x=4-24-6+27
-x=1
x=-1
2) -9x-6+x=12-8x
-9x+x+8x=12+6
0x=18 (vô nghiệm )
3) 10x-15-20x+28=19-2x-22
10x-20x+2x=19-22+15-28
-8x=-16
x=2
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1) 3x-4+24+6=x+27+3x
\(\Leftrightarrow\)3x-3x-x=4-24-6+27
\(\Leftrightarrow\)-x=1
\(\Leftrightarrow\)x=-1
2) -9x-(6-x)=4(3-2x)
\(\Leftrightarrow\)-9x-6+x=12-8x
\(\Leftrightarrow\)-9x+x+8x=12+6
\(\Leftrightarrow\)0x=18
vậy pt vô nghiệm
3) 5(2x-3)-4(5x-7)=19-2(x+11)
\(\Leftrightarrow\)10x-15-20x+28=19-2x-22
\(\Leftrightarrow\)10x+2x-20x=19-22-28+15
\(\Leftrightarrow\)-8x=-16
\(\Leftrightarrow\)x=2
2:
a: =>x-1=0 hoặc 3x+1=0
=>x=1 hoặc x=-1/3
b: =>x-5=0 hoặc 7-x=0
=>x=5 hoặc x=7
c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)
d: =>x=0 hoặc x^2-1=0
=>\(x\in\left\{0;1;-1\right\}\)
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Mình làm được khổ nỗi lại chưa biết nghiệm là gì? @ thu hương có thể giải thích cho minh không
hiihhi
\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)
\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)
\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)
\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)
\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)
\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)
\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)
\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)
\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)
\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)
a,6x-3-5x+15+18x-24=24
19x-12=24
19x=36
x=36/19
c,10x-6x2+6x2-10x+21=3
0x=-18
không có x
d,3x2+3x-2x2-4x=-1-x
x2-x=-1-x
x2-x+x=-1
x2=-1
không có x thỏa mãn
\(3.\left(2x-1\right)-5.\left(x-3\right)=24-6.\left(3x-4\right)\)
=) \(\left(6x-3\right)-\left(5x-15\right)=24-\left(18x-24\right)\)
=) \(6x-3-5x+15=24-18x+24\)
=) \(6x-5x-3+15=48-18x\)
=) \(x-\left(3-15\right)=48-18x\)
=) \(x+12=48-18x\)
=) \(x+18x=48-12\)
=) \(19x=36\)=) \(x=\frac{36}{19}\)
Ta có 3(2x - 1) - 5(x - 3) = 24 - 6(3x - 4)
<=> 6x - 3 - 5x + 15 = 24 - 18x + 24
<=> x + 12 = 48 - 18x
=> x + 18x = 48 - 12
=> 19x = 36
=> x = \(\frac{36}{19}\)