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a) Bổ sung cho đầy đủ đề
b) (3x - 1)/4 = (2x - 5)/3
3(3x - 1) = 4(2x - 5)
9x - 3 = 8x - 20
9x - 8x = -20 + 3
x = -17
c) Điều kiện: x ≠ -1/3
3/(-2) = (x - 3)/(3x + 1)
3.(3x + 1) = -2(x - 3)
9x + 3 = -2x + 6
9x + 2x = 6 - 3
11x = 3
x = 3/11 (nhận)
Vậy x = 3/11
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}-\dfrac{1}{x}=2010\)
\(\Rightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}-\dfrac{1}{x}=\dfrac{1}{2010}\)\(\Rightarrow\dfrac{-1}{x+3}=\dfrac{1}{2010}\)
\(\Rightarrow x+3=-2010\)
\(\Rightarrow x=-2013\)
Vậy x = -2013
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}-\dfrac{1}{x}=\dfrac{1}{2010}\)
=> \(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}-\dfrac{1}{x}=\dfrac{1}{2010}\)
=> \(-\dfrac{1}{x+3}=\dfrac{1}{2010}\)
=> \(-2010=x+3\)
=> \(x=-2013\)
\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}=\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-1\right)\left(x-4\right)}\) Đk: \(x\ne1;x\ne2;x\ne3;x\ne4\)
\(\Leftrightarrow\dfrac{2x-4}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\dfrac{2x-4}{\left(x-1\right)\left(x-3\right)\left(x-4\right)}\)
\(\Leftrightarrow\dfrac{2x-4}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}-\dfrac{2x-4}{\left(x-1\right)\left(x-3\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x-4\right)\left(2x-4\right)-\left(2x-4\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2-4x-8x+16-2x^2+4x+4x-8}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)}=0\)
\(\Leftrightarrow-4x+8=0\)
\(\Rightarrow x=2\) (KTM )
=> Pt vô nghiệm
`#3107.101107`
a)
`-2/3(x + 1) = 1/6 - x`
`=> -2/3x - 2/3 = 1/6 - x`
`=> -2/3x + x = 1/6 + 2/3`
`=> 1/3x = 5/6`
`=> x = 5/6 \div 1/3`
`=> x =5/2`
Vậy, `x = 5/2`
b)
`3(x + 1/3) - 1/2(x + 2) = 5/2x - 1`
`=> 3x + 1 - 1/2x - 1 = 5/2x - 1`
`=> 3x - 1/2x - 5/2x = -1`
`=> 0x = -1` (vô lý)
Vậy, `x` không có giá trị thỏa mãn.
a: \(\Leftrightarrow-\dfrac{2}{3}x-\dfrac{2}{3}=\dfrac{1}{6}-x\)
=>\(\dfrac{1}{3}x=\dfrac{1}{6}+\dfrac{2}{3}=\dfrac{5}{6}\)
=>\(x=\dfrac{5}{6}\cdot3=\dfrac{5}{2}\)
b: \(\Leftrightarrow3x+1-\dfrac{1}{2}x-1=\dfrac{5}{2}x-1\)
=>\(\dfrac{5}{2}x=\dfrac{5}{2}x-1\)
=>0=-1(vô lý)
a: \(x=\left(-\dfrac{2}{3}\right)^5:\left(-\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)
b: =>x-1/2=1/3
=>x=5/6
c: =>2/3x-1=0 hoặc 3/4x+1/2=0
=>x=3/2 hoặc x=-1/2:3/4=-1/2*4/3=-4/6=-2/3
d =>4/9:x=10/3:9/4=10/3*4/9=40/27
=>x=4/9:40/27=4/9*27/40=108/360=3/10
Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)
$\Rightarrow -11x\geq 0$
$\Rightarrow x\leq 0$
Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$
PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$
$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$
$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$
$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$
$\frac{1}{2}(1-\frac{1}{21})=-x$
$\frac{10}{21}=-x$
$\Rightarrow x=\frac{-10}{21}$
Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)
$\Rightarrow -11x\geq 0$
$\Rightarrow x\leq 0$
Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$
PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$
$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$
$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$
$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$
$\frac{1}{2}(1-\frac{1}{21})=-x$
$\frac{10}{21}=-x$
$\Rightarrow x=\frac{-10}{21}$
\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}-\dfrac{1}{x}=\dfrac{1}{2010}\)
\(\Rightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}-\dfrac{1}{x}=\dfrac{1}{2010}\Rightarrow\dfrac{-1}{x+3}=\dfrac{1}{2010}\)
\(\Rightarrow x+3=\left(-2010\right)\)
\(\Rightarrow x=\left(-2013\right)\)
Vậy .......................
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