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![](https://rs.olm.vn/images/avt/0.png?1311)
ĐK : 6x \(\ge0\Rightarrow x\ge0\)
Khi đó |x + 1| = x + 1
|x + 2| = x +2
|x + 3| = x +3
|x + 4| = x + 4
|x + 5| = x +5
Khi đó |x + 1| + |x + 2| + |x + 3| + |x + 4| + |x + 5| = 6x
<=> x + 1 + x + 2 + x + 3 + x + 4 + x + 5 = 6x
<=> 5x + 15 = 6x
<=> x = 15 (tm)
Vậy x = 15
b) 3x + 2 - 3x + 1 - 3x = 15.340
=> 3x(32 - 3 - 1) = 15.340
<=> 3x . 5 = 15.340
<=> 3x = 341
<=> x = 41
Vậy x = 41
a,vì /x+1/,/x+2/,/x+3/,/x+4/,/x+5/\(\ge\)0 mà /x+1/+/x+2/+/x+3/+/x+4/+/x+5/=6x suy ra x>0
nên /x+1/+/x+2/+/x+3/+/x+4/+/x+5/=x+1+x+2+x+3+x+4+x+5=6x ( giải thích: /x/=x khi x \(\ge0\))
suy ra 5x+21=6x suy ra x=21
b, \(3^{x+2}-3^{x+1}-3^x=15.3^{40}\)
suy ra \(3^x\left(9-3-1\right)=5.3^{41}\)
suy ra \(3^x.5=5.3^{41}\Rightarrow x=41\)
![](https://rs.olm.vn/images/avt/0.png?1311)
60-[15*X+4]=15/2:1/2
60-[15*X+4]=15
15*X+4=60-15
15*X+4=45
15*X=45-4
15*X=41
X=41:15
X=41/15
ko ghi đề
\(60-\left(15.x+4\right)=\frac{15}{2}.\frac{2}{1}\)
\(60-\left(15.x+4\right)=15\)
\(15.x+4=60-15\)
\(15.x+4=45\)
\(15.x=45-4\)
\(15.x=41\)
\(x=41:15\)
\(x=\text{2.7333}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
-x + 20 = - (-15) - (8) + 13
-x + 20 = 15 - 8 + 13
-x + 20 = 7 + 13
- x + 20 = 20
x = 20 - 20
x = 0
-(-10) + x = -13 + (-9) + (-6)
10 + x = -13 - 9 - 6
10 + x = -28
x = -28 - 10
x = -38
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
a) \(2\dfrac{17}{20}-1\dfrac{15}{11}+6\dfrac{9}{20}:3\)
\(=\dfrac{57}{20}-\dfrac{26}{11}+\dfrac{129}{20}:3\)
\(=\dfrac{107}{220}+\dfrac{43}{20}\)
\(=\dfrac{29}{11}\)
b) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\)
\(=\dfrac{31}{7}:\left(\dfrac{7}{5}.\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}\)
\(=\dfrac{5}{7}\)
c) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\)
\(=\left(\dfrac{29}{9}.\dfrac{15}{23}.\dfrac{36}{29}\right):\dfrac{5}{23}\)
\(=\dfrac{60}{23}:\dfrac{5}{23}\)
\(=12\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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![](https://rs.olm.vn/images/avt/0.png?1311)
a) -12.(x - 5) + 7(3 - x) = 5
=> -12x + 60 + 21 - 7x = 5
=> -19x + 81 = 5
=> -19x = 5 - 81
=> -19x = -76
=> x = -76 : (-19)
=> x = 4
b) (x + 1) + (x + 2) + (x + 3) + ... + (x + 20) = 250
=> (x + x + x + ... + x) + (1 + 2 + 3 + ... + 20) = 250
=> 20x + 210 = 250
=> 20x = 250 - 210
=> 20x = 40
= > x = 40 : 20
=> x = 2
\(-12\left(x-5\right)+7\left(3-x\right)=5\)
\(\Leftrightarrow-12x+60+21-7x=5\)
\(\Leftrightarrow-19x+81=5\)
\(\Leftrightarrow81-5=19x\)
\(\Leftrightarrow19x=76\)
\(\Leftrightarrow x=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{-2}{3}\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)
\(\Leftrightarrow\dfrac{-2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{-2}{3}x-\dfrac{2}{3}x=\dfrac{-1}{3}-\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{-4}{3}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{3}{8}\)
Vậy \(x=\dfrac{3}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-3\right)\left(x-12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=12\end{cases}}\)
\(\Rightarrow x\in\left\{3;12\right\}\)
\(\left(x^2-81\right)\left(x^2+9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-81=0\\x^2+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x\in\varnothing\end{cases}}\Leftrightarrow x=9\)
\(\Rightarrow x=9\)
\(\left(x-4\right)\left(x+2\right)< 0\)
\(\Rightarrow\hept{\begin{cases}x-4\\x+2\end{cases}}\)trái dấu
\(TH1:\hept{\begin{cases}x-4>0\\x+2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\x< -2\end{cases}}\Leftrightarrow x\in\varnothing\)
\(TH2:\hept{\begin{cases}x-4< 0\\x+2>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 4\\x>-2\end{cases}}\Leftrightarrow x\in\left\{-1;0;1;2;3\right\}\)
Vậy \(x\in\left\{-1;0;1;2;3\right\}\)
\(\left(x-1\right)+\left(x-2\right)+...+\left(x-20\right)=15\)
\(\left(x+x+...+x\right)-\left(1+2+...+20\right)=15\)
\(20x-\dfrac{20.21}{2}=15\)
\(20x-210=15\)
\(20x=210+15\)
\(20x=225\)
\(x=225:20\)
\(x=11,25\)
Đề thiếu 1 vế rồi em, phải = cái gì đó nữa chứ