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f) (x – 5)2 = 9 
a) 3x – 15 = 25 – 5x
=> 3x + 5x = 25 + 15
=> 8x = 40
=> x = 5
b) 3x - 17 = 2x – 7
=> 3x - 2x = -7 + 17
=> x = 10
c) 2x – 17 = – (3x – 18)
=> 2x - 17 = -3x + 18
=> 2x + 3x = 18 + 17
=> 5x = 35
=> x = 7
d) 3x – 14 = 2(x – 9) + 1
=> 3x - 14 = 2x - 18 + 1
=> 3x - 2x = -18 + 1 + 14
=> x = -3
f) (x – 5)2 = 9
\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
a) Ta có: \(3x-15=25-5x\)
\(\Leftrightarrow3x-15-25+5x=0\)
\(\Leftrightarrow8x-40=0\)
\(\Leftrightarrow8x=40\)
hay x=5
Vậy: x=5
b) Ta có: \(3x-17=2x-7\)
\(\Leftrightarrow3x-17-2x+7=0\)
\(\Leftrightarrow x-10=0\)
hay x=10
Vậy: x=10
c) Ta có: \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=-3x+18\)
\(\Leftrightarrow2x-17+3x-18=0\)
\(\Leftrightarrow5x-35=0\)
\(\Leftrightarrow5x=35\)
hay x=7
Vậy: x=7
d) Ta có: \(3x-14=2\left(x-9\right)+1\)
\(\Leftrightarrow3x-14=2x-18+1\)
\(\Leftrightarrow3x-14-2x+18-1=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy: x=-3
f) Ta có: \(\left(x-5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;8\right\}\)
a)
<=> 3x - 3 + x - 2 = 2x - 2 - x + 1
<=> 3x + x - 2x + x = -2 + 1 + 3 + 2
<=> 3x = 4
<=> x = 4/3
Các câu sau làm tương tự
\(\left(3x-3\right)+\left(x-2\right)=\left(2x-2\right)-\left(x-1\right)\)
<=> \(3x-3+x-2=2x-2-x+1\)
<=> \(4x-5=x-1\)
<=> \(3x=4\)
<=> \(x=\frac{4}{3}\)
Vậy....
c) x.(1+2+3+4+...+100)=0
x.5050=0
x=0:5050=0
Vậy x=0
d) x.(1+2+3+4+5+...+100)=5050
x.5050=5050
x=1
Vậy x=1
e) x+1+x+2+x+3+x+4+...+x+100=5050
(x+x+x+x+...+x)+(1+2+3+4+...+100)=5050
100 số hạng x
x.100+5050=5050
x.100=0
x=0
Vậy x=0
a ) \(-5\times\left(-x+7\right)-3\times\left(-x-5\right)=-4\times\left(12-x\right)+48\)
\(\Leftrightarrow5x-35+3x+15=-48+4x+48\)
\(\Leftrightarrow5x-3x+4x=35-15-48+48\)
\(\Leftrightarrow2x=20\)
\(\Leftrightarrow x=10\)
b ) \(-2\times\left(15-3x\right)-4\times\left(-7x+8\right)=-5-9\times\left(-2x+1\right)\)
\(\Leftrightarrow-30+6x-28x-32=-5+18x-9\)
\(\Leftrightarrow6x-28x-18x=30+32-5-9\)
\(\Leftrightarrow-40x=48\)
\(\Leftrightarrow x=-1.2\)
a) \(\left|4+2x\right|=-4x\)
th1: \(4+2x\ge0\Leftrightarrow2x\ge-4\Leftrightarrow x\ge\dfrac{-4}{2}=-2\)
\(\Rightarrow\left|4+2x\right|=-4x\Leftrightarrow4+2x=-4x\Leftrightarrow4=-4x-2x=-6x\)
\(\Leftrightarrow x=\dfrac{4}{-6}=\dfrac{-2}{3}\left(tmđk\right)\)
th2: \(4+2x< 0\Leftrightarrow2x< -4\Leftrightarrow x< \dfrac{-4}{2}=-2\)
\(\Rightarrow\left|4+2x\right|=-4x\Leftrightarrow-4-2x=-4x\Leftrightarrow-4=-4x+2x=-2x\)
\(\Leftrightarrow x=\dfrac{-4}{-2}=2\left(loại\right)\)
vậy \(x=\dfrac{-2}{3}\)
b) \(\left|3x-1\right|+2=x\)
th1: \(3x-1\ge0\Leftrightarrow3x\ge1\Leftrightarrow x\ge\dfrac{1}{3}\)
\(\Rightarrow\left|3x-1\right|+2=x\Leftrightarrow3x-1+2=x\Leftrightarrow3x-x=1-2\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\left(loại\right)\)
th2: \(3x-1< 0\Leftrightarrow3x< 1\Leftrightarrow x< \dfrac{1}{3}\)
\(\Rightarrow\left|3x-1\right|+2=x\Leftrightarrow1-3x+2=x\Leftrightarrow x+3x=1+2\)
\(\Leftrightarrow4x=3\Leftrightarrow x=\dfrac{3}{4}\left(loại\right)\)
vậy phương trình vô nghiệm
c) \(\left|x+15\right|+1=3x\)
th1: \(x+15\ge0\Leftrightarrow x\ge-15\)
\(\Rightarrow\left|x+15\right|+1=3x\Leftrightarrow x+15+1=3x\Leftrightarrow3x-x=15+1\)
\(\Leftrightarrow2x=16\Leftrightarrow x=\dfrac{16}{2}=8\left(tmđk\right)\)
th2: \(x+15< 0\Leftrightarrow x< -15\)
\(\Rightarrow\left|x+15\right|+1=3x\Leftrightarrow-x-15+1=3x\Leftrightarrow3x+x=-15+1\)
\(\Leftrightarrow4x=-14\Leftrightarrow x=\dfrac{-14}{4}=\dfrac{-7}{2}\left(loại\right)\)
vậy \(x=8\)
d) \(\left|2x-5\right|+x=2\)
th1: \(2x-5\ge0\Leftrightarrow2x\ge5\Leftrightarrow x\ge\dfrac{5}{2}\)
\(\Rightarrow\left|2x-5\right|+x=2\Leftrightarrow2x-5+x=2\Leftrightarrow2x+x=2+5\)
\(\Leftrightarrow3x=7\Leftrightarrow x=\dfrac{7}{3}\left(loại\right)\)
th2: \(2x-5< 0\Leftrightarrow2x< 5\Leftrightarrow x< \dfrac{5}{2}\)
\(\Rightarrow\left|2x-5\right|+x=2\Leftrightarrow5-2x+x=2\Leftrightarrow-2x+x=2-5\)
\(\Leftrightarrow-x=-3\Leftrightarrow x=3\left(loại\right)\)
vậy phương trình vô nghiệm