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2:
a: x=2,4-0,4=2
b: =>2x=-1,5+0,8=-0,7
=>x=-0,35
c: =>x-16=-15
=>x=1
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(3,5 + 5,7) . x + (3,5 + 5,7) = 0
[(3,5 + 5,7) . (x + 1)] = 0
9,2 . (x + 1) = 0
x + 1 = 0 : 9,2 = 0
=> x = 0 - 1 = -1
Vậy x = -1
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\(\left(\frac{1}{4}x-1\right)+\left(\frac{5}{6}x-2\right)-\left(\frac{3}{8}x+5\right)=3,5\)
\(\Rightarrow\frac{1}{4}x-1+\frac{5}{6}x-2-\frac{3}{8}x-5=\frac{7}{2}\)
\(\Rightarrow\left(\frac{1}{4}+\frac{5}{6}-\frac{3}{8}\right)x+\left[\left(-1\right)+\left(-2\right)+\left(-5\right)\right]=\frac{7}{2}\)
\(\Rightarrow\frac{17}{24}x+\left(-8\right)=\frac{7}{2}\)
\(\Rightarrow\frac{17}{24}x=\frac{7}{2}-\left(-8\right)=\frac{23}{3}\)
\(\Rightarrow x=\frac{23}{2}:\frac{17}{24}=\frac{276}{17}\)
~ Hok tốt ~
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Từ 0,2 đến 4,7 có số số hạng là: \(\left(4,7-0,2\right):0,5+1=10\left(số\right)\)
⇒ Tổng 10 số đó là: \(\dfrac{\left(4,7+0,2\right)\cdot10}{2}=24,5\)
\(\left(x+0,2\right)+\left(x+0,7\right)+...+\left(x+4,7\right)=65,5\\ \Leftrightarrow10x+24,5=65,5\\ \Leftrightarrow10x=41\\ \Leftrightarrow x=4,1\)
Vậy x = 4,1
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a) \(1\frac{2}{7} = 1 + \frac{2}{7} = \frac{9}{2}\)
\(\begin{array}{l}x:1\frac{2}{7} = - 3,5\\x:\frac{9}{7} = - \frac{7}{2}\\x = - \frac{7}{2}.\frac{9}{7}\\x = - \frac{9}{2}\end{array}\)
b) \(0,4.x - \frac{1}{5}.x = \frac{3}{4}\)
\(\begin{array}{l}\frac{2}{5}.x - \frac{1}{5}.x = \frac{3}{4}\\\left( {\frac{2}{5} - \frac{1}{5}} \right).x = \frac{3}{4}\\\frac{1}{5}.x = \frac{3}{4}\\x = \frac{3}{4}:\frac{1}{5}\\x = \frac{3}{4}.5\\x = \frac{{15}}{4}\end{array}\)
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a: =15+11,2=26,2
b: =-3
c: =-1,9-0,02=-1,92
d: =103-96-0,08=7-0,08=6,92
e: =15-0,99=14,01
f: -0,707+2,3=1,593
p: =8,26-25,14-8,26=-25,14
q: =-5,4(3,6+6,4)
=-5,4*10=-54
o: =12,7+(-7)
=5,7
m: =19,7-9=10,7
=> |3,5 - x| = -0.7 : [-2]
=> |3,5 - x| = 7/20
\(\Rightarrow\hept{\begin{cases}3,5-x=\frac{7}{20}\\3,5-x=-\frac{7}{20}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{63}{20}\\x=\frac{77}{20}\end{cases}}\)
Vậy x = 63/20 hoặc x = 77/20
Nếu \(3,5-x\ge0\Leftrightarrow x\le3,5\)thì phương trình trở thành
\(-2\left(3,5-x\right)=-0,7\Leftrightarrow3,5-x=0,35\Leftrightarrow x=3,15\)
Nếu \(x>3,5\)thì phương trình trở thành
\(-2\left(x-3,5\right)=0,7\Leftrightarrow x-3,5=0,35\Rightarrow x=3,85\)