|x|=5.6 tìm x
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(5,28+x):24=5,6
5,28+x=5,6×24
5,28+x=134,4
x=134,4-5,28
x=129,12
\(\frac{1}{3.4}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{3}-\frac{1}{x+1}\)
\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{x\left(x+1\right)}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+1}\)
\(=\frac{1}{3}-\frac{1}{x+1}\)
a) (3 x X - 7.5) + 2.3 = 5.6
3 x X - 7.5 = 5.6 - 2.3
3 x X - 7.5 = 3.3
3 x X = 3.3 + 7.5
3 x X = 10.8
X = 10.8 : 3
X = 3.6
b) 5 x X - 3 x X + 69 = 285
(5 - 3) x X = 285 - 69
2 x X = 216
X = 216 : 2
X = 108
Mình sửa đề luôn ^^
\(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{3}{10}\)
\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)
\(\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)
\(\frac{1}{x+1}=\frac{1}{30}\)
=> x + 1 = 30
=> x = 29
Vậy,..... @_@ ^^
(1/3-1/4+1/4-1/5+1/5-.......+1/x.(x+1)=3/10
1/3-1/x+1=3/10
tự làm...
1/3.4+1/4.5+1/5.6+1/6.7+....+1/x(x+1)=3/10
<=> \(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{\left(x+1\right)x}=\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)
<=> \(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}=\frac{1}{30}\)=> x+1=30=>x=29
\(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{x\left(x+1\right)}=\frac{3}{10}\)
\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)
\(\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)
\(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}\)
\(\frac{1}{x+1}=\frac{1}{30}\)
\(\Rightarrow x+1=30\)
\(x=30-1=29\)
x : 4,37 = 5,6 ( dư 1,53 )
x = 5,6 x 4,37 + 1,53
x = 24,472 + 1,53
x = 26,002
x : 4,37 = 5,6 ( dư 1,53 )
x = 5,6 . 4,37 + 1,53
x = 26,002
Ta có :
| x | = 5 . 6
=> | x | = 30
=> x ∈ { -30 ; 30 }
Vậy ............