A=|x-3/4|+1
B=|3x+1|-2
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a) \(4x-\sqrt[]{3\left(3x-1\right)}=3x-1\)
\(\Leftrightarrow\sqrt[]{3\left(3x-1\right)}=x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\3\left(3x-1\right)=\left(x+1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\9x-3=x^2+2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\left(a\right)\\x^2-7x+4=0\left(1\right)\end{matrix}\right.\)
Giải \(pt\left(1\right):\)
\(\Delta=49-16=33\Rightarrow\sqrt[]{\Delta}=\sqrt[]{33}\)
Phương trình (1) có 2 nghiệm phân biệt
\(\left[{}\begin{matrix}x=\dfrac{7+\sqrt[]{33}}{2}\\x=\dfrac{7-\sqrt[]{33}}{2}\end{matrix}\right.\) (thỏa \(\left(a\right)\))
a: \(\left(x+\dfrac{1}{4}\right)+\left(3x-4\right)+2\left(x-3\right)=1\)
=>\(x+\dfrac{1}{4}+3x-4+2x-6=1\)
=>\(6x-\dfrac{39}{4}=1\)
=>\(6x=1+\dfrac{39}{4}=\dfrac{43}{4}\)
=>\(x=\dfrac{43}{4}:6=\dfrac{43}{24}\)
b: \(2\left(x-3\right)=3\left(x+2\right)-x+1\)
=>\(2x-6=3x+6-x+1\)
=>2x-6=2x+7
=>-6=7(vô lý)
c: \(x\left(x+3\right)+x\left(x-2\right)=2x\left(x-1\right)\)
=>\(x^2+3x+x^2-2x=2x^2-2x\)
=>3x-2x=-2x
=>3x=0
=>x=0
d: \(\left(x-1\right)\cdot3x-2\left(x+2\right)-2x=x\left(x-1\right)\)
=>\(3x^2-3x-2x-4-2x=x^2-x\)
=>\(3x^2-7x-4-x^2+x=0\)
=>\(2x^2-6x-4=0\)
=>\(x^2-3x-2=0\)
=>\(x=\dfrac{3\pm\sqrt{17}}{2}\)
\(a)\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1.\left(x\ne2;4\right).\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1.\\ \Rightarrow x^2-4x-3x+12+x^2-4x+4+x^2-4x-2x+8=0.\\ \Leftrightarrow3x^2-17x+24=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}.\\x=3.\end{matrix}\right.\) (TM).
\(b)3x+12=0.\\ \Leftrightarrow3x=-12.\\ \Leftrightarrow x=-4.\)
\(c)5+2x=x-5.\\ \Leftrightarrow2x-x=-5-5.\\ \Leftrightarrow x=-10.\)
\(d)2x\left(x-2\right)+5\left(x-2\right)=0.\\ \Leftrightarrow\left(2x+5\right)\left(x-2\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{2}.\\x=2.\end{matrix}\right.\)
\(e)\dfrac{3x-4}{2}=\dfrac{4x+1}{3}.\\ \Rightarrow3\left(3x-4\right)-2\left(4x+1\right)=0.\\ \Leftrightarrow9x-12-8x-2=0.\\ \Leftrightarrow x=14.\)
\(f)\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1.\left(x\ne\pm1\right).\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x}{x^2-1}=1.\\ \Leftrightarrow x^2+3x-x^2+1=0.\\ \Leftrightarrow3x+1=0.\\ \Leftrightarrow x=\dfrac{-1}{3}.\)
\(g)\dfrac{2x}{x-1}+\dfrac{3-2x}{x+2}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}.\left(x\ne1;-2\right).\\ \Leftrightarrow\dfrac{2x^2+4x+\left(3-2x\right)\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}.\\ \Rightarrow2x^2+4x+3x-3-2x^2+2x-6=0.\\ \Leftrightarrow9x=9.\)
\(\Leftrightarrow x=1\left(koTM\right).\)
A.
\(\Leftrightarrow\) 9x - 2x - 6 = 3x + 1
\(\Leftrightarrow\) 4x = 7
\(\Leftrightarrow\) x = \(\dfrac{7}{4}\)
B.
\(\Leftrightarrow\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-13}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\) 5x + 15 - 4x +12 = x - 13
\(\Leftrightarrow\) 0x = -40 ( phương trình vô nghiệm)
C.
\(\Leftrightarrow\) 7x + 8 \(\ge\) 3x -3
\(\Leftrightarrow\) 4x \(\ge\) - 11
\(\Leftrightarrow\)\(x\ge\dfrac{-11}{4}\)
a)
\(\left|2-3x\right|=-1\) (vô lí vì \(\left|2-3x\right|\ge0\) )
b)
`3x-2,42+0,8=3,38-0,2x`
`<=>3x+0,2x=3,38+2,42-0,8`
`<=>3,2x=5`
`<=>x=25/16`
c)
\(\dfrac{3}{x-1}+\dfrac{2}{x^2+x+1}=\dfrac{3x^2}{x^3-1}\left(x\ne1\right)\)
\(< =>\dfrac{3}{x-1}+\dfrac{2}{x^2+x+1}=\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(< =>\dfrac{3\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}\)
suy ra
`3x^2 +3x+3+2x-2=3x^2`
`<=>3x^2 -3x^2 +3x+2x=-3+2`
`<=>5x=-1`
`<=>x=-1/5(tmđk)`
Với \(x\ge\dfrac{1}{6}\Leftrightarrow A=5x^2-6x+1-1=5x^2-6x\)
\(A=5\left(x^2-2\cdot\dfrac{3}{5}x+\dfrac{9}{25}\right)-\dfrac{9}{5}=5\left(x-\dfrac{3}{5}\right)^2-\dfrac{9}{5}\ge-\dfrac{9}{5}\\ A_{min}=-\dfrac{9}{5}\Leftrightarrow x=\dfrac{3}{5}\left(1\right)\)
Với \(x< \dfrac{1}{6}\Leftrightarrow A=5x^2+6x-1-1=5x^2+6x-2\)
\(A=5\left(x^2+2\cdot\dfrac{3}{5}x+\dfrac{9}{25}\right)-\dfrac{19}{5}=5\left(x+\dfrac{3}{5}\right)^2-\dfrac{19}{5}\ge-\dfrac{19}{5}\\ A_{min}=-\dfrac{19}{5}\Leftrightarrow x=-\dfrac{3}{5}\left(2\right)\\ \left(1\right)\left(2\right)\Leftrightarrow A_{min}=-\dfrac{19}{5}\Leftrightarrow x=-\dfrac{3}{5}\)
Với \(x\ge\dfrac{1}{3}\Leftrightarrow B=9x^2-6x-4\left(3x-1\right)+6=9x^2-18x+10\)
\(B=9\left(x^2-2x+1\right)+1=9\left(x-1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow x=1\left(1\right)\)
Với \(x< \dfrac{1}{3}\Leftrightarrow B=9x^2-6x+4\left(3x-1\right)+6=9x^2+6x+2\)
\(B=\left(9x^2+6x+1\right)+1=\left(3x+1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow x=-\dfrac{1}{3}\left(2\right)\)
\(\left(1\right)\left(2\right)\Leftrightarrow B_{min}=1\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
a: (x-2)(x+2)-(x+1)2=1
=>\(x^2-4-\left(x^2+2x+1\right)=1\)
=>\(x^2-4-x^2-2x-1=1\)
=>-2x-5=1
=>-2x=6
=>\(x=\dfrac{6}{-2}=-3\)
b: Sửa đề:\(x^3-8-\left(x-2\right)\left(x-4\right)=0\)
=>\(\left(x^3-8\right)-\left(x-2\right)\left(x-4\right)=0\)
=>\(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x-4\right)=0\)
=>\(\left(x-2\right)\left(x^2+2x+4-x+4\right)=0\)
=>\(\left(x-2\right)\left(x^2+x\right)=0\)
=>x(x+1)(x-2)=0
=>\(\left[{}\begin{matrix}x=0\\x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=2\end{matrix}\right.\)
c: 3x(x-1)+1-x=0
=>3x(x-1)-(x-1)=0
=>(x-1)(3x-1)=0
=>\(\left[{}\begin{matrix}x-1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
`a)1/3x-1/4x=1`
`(1/3-1/4)x=1`
`1/12x=1`
`x=1:1/12=12`
______________________________
`b)4/5+5/7:x=1/6`
`5/7:x=1/6-4/5`
`5/7:x=-19/30`
`x=5/7:(-19/30)`
`x=-150/133`
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`c)[2x]/3-x/4=5/6`
`x(2/3-1/4)=5/6`
`x. 5/12=5/6`
`x=5/6:5/12=2`