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\(\frac{x+5}{3}=\frac{5}{9}\)
=> ( x + 5 ) x 9 = 3 x 5
=> x + 5 = \(\frac{3\cdot5}{9}\)
=> x + 5 = \(\frac{5}{3}\)
=> x = \(\frac{-10}{3}\)
Bài 5:Giải:
Ta có: \(\left\{{}\begin{matrix}a+3c=2016\left(1\right)\\a+2b=2017\left(2\right)\end{matrix}\right.\)
Từ \(\left(1\right)\Leftrightarrow a=2016-3c\)
Lấy \(\left(2\right)-\left(1\right)\) ta được:
\(2b-3c=1\Leftrightarrow b=\dfrac{1+3c}{2}\)
Khi đó:
\(P=a+b+c=\left(2016-3c\right)+\dfrac{1+3c}{2}\) \(+\) \(c\)
\(=\left(2016+\dfrac{1}{2}\right)+\dfrac{-6c+3c+2c}{2}\)
\(=2016\dfrac{1}{2}-\dfrac{c}{2}\) Vì \(a,b,c\ge0\) nên:
\(P=2016\dfrac{1}{2}-\dfrac{c}{2}\le2016\dfrac{1}{2}\)
Vậy \(P_{max}=2016\dfrac{1}{2}\Leftrightarrow c=0\)
|x-1|+5(x-2)=5x-4.|-2|
|x-1|+5x-10=5x-4.2=5x-8
|x-1|=5x-5x+10-8
|x-1|=2
x-1=2;-2
x-1=3;-1
k đúng cho mình nha
a) \(-\dfrac{2}{3}\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)
\(\Rightarrow-\dfrac{2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)
\(\Rightarrow\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{2}{3}x+\dfrac{2}{3}x\)
\(\Rightarrow\dfrac{1}{2}=\dfrac{4}{3}x\)
\(\Rightarrow x=\dfrac{1}{2}:\dfrac{4}{3}=\dfrac{3}{8}\)
Vậy \(x=\dfrac{3}{8}\).
\(\left(x+1\right)\left(x+7\right)< 0\)
thì \(x+1;x+7\)khác dấu
th1\(\hept{\begin{cases}x+1< 0\\x+7>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -1\\x>-7\end{cases}\Rightarrow}-7< x< -1\left(tm\right)}\)
th2\(\hept{\begin{cases}x+1>0\\x+7< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-1\\x< -7\end{cases}\Rightarrow}-1< x< -7\left(vl\right)}\)
vậy với\(-7< x< -1\)thì \(\left(x+1\right)\left(x+7\right)< 0\)
a) (2x - 3) = 5
<=> 2x - 3 = 5
<=> 2x = 5 + 3
<=> 2x = 8
<=> x = 4
=> x = 4
b) (5x - 3) = 1/2
<=> 5x - 3 = 1/2
<=> 5x = 1/2 + 3
<=> 5x = 7/2
<=> x = 7/10
=> x = 7/10
c) (x + 1)(x + 7) < 0
<=> x = -1; -7
<=> x < -7 <=> x = -8 <=> (-8 + 1)(-8 + 7) < 0 <=> 7 < 0 (loại)
<=> -7 < x < -1 <=> x = -6 <=> (-6 + 1)(-6 + 7) < 0 <=> -5 < 0 (nhận)
<=> x > -1 <=> x = 0 <=> (x + 1)(x + 7) < 0 <=> 7 < 0 (loại)
Vậy: -7 < x < -1
Ta có : \(\left(5x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\2x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x=3\\2x=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{5}{2}\end{cases}}\)
a) \(\frac{-x}{4}=\frac{-9}{x}\)
\(\Rightarrow x^2\left(-1\right)=-36\)
\(\Rightarrow x^2=6^2\)
\(\Rightarrow x=\pm6\)
Vậy \(x=\left[\begin{matrix}6\\-6\end{matrix}\right.\)
b) \(\frac{12}{x+1}=\frac{5}{30}\)
\(\Rightarrow\left(x+1\right)5=12.30\)
\(\Rightarrow5x+5=360\)
\(\Rightarrow5x=355\)
\(\Rightarrow x=355:5\)
\(\Rightarrow x=71\)
Vậy \(x=71.\)
c) \(\frac{x+42}{5}=\frac{5x+2}{20}\)
\(\Rightarrow\left(x+42\right)20=5\left(5x+2\right)\)
\(\Rightarrow20x+840=25x+10\)
\(\Rightarrow20x-25x=-840+10\)
\(\Rightarrow-5x=-830\)
\(\Rightarrow x=166\)
Vậy \(x=166.\)
a, \(\frac{-x}{4}=\frac{-9}{x}\Rightarrow-x.x=-9.4\)
\(\Rightarrow-x.x=-36=-6.6\Rightarrow x=6\)
Vậy x = 6
b, \(\frac{12}{x+1}=\frac{5}{30}=\frac{1}{6}\Rightarrow x+1=12.6\)
\(\Rightarrow x+1=72\Rightarrow x=72-1=71\)
Vậy x = 71
c, \(\frac{x+42}{5}=\frac{5x+2}{20}\Rightarrow20\left(x+42\right)=5\left(5x+2\right)\)
\(\Rightarrow20x+840=25x+10\Rightarrow20x-25x=-840+10\)
\(\Rightarrow\left(20-25\right)x=-830\Rightarrow-5x=-830\)
\(\Rightarrow x=-830\div\left(-5\right)\Rightarrow x=166\)
Vậy x = 166
\(|x-1|+5\left(x-2\right)=5x-8\)
\(\Rightarrow|x-1|+5x-10=5x-8\)
\(\Rightarrow|x-1|=5x-8-5x+10=2\)
\(\Rightarrow\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}}\)