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(x-45).27 = 0
=> x - 45 = 0
=> x = 45 + 0
=> x = 45
23.(42 - x) = 23
=> 42 - x = 23 : 23
=> 42 - x = 1
=> x = 42 - 1
=> x = 41
l--i-k-e cho mk
a) (x - 45) .27 = 0
=> x - 45 = 0
=> x = 45
b) 23 . (42 - x) = 23
42 - x = 1
x = 42 - 1
x = 41
a) ( x - 45 ) . 27 = 0
( x - 45 ) = 0 : 27
( x - 45 ) = 0
x = 0 + 45
x = 45
b) 23 . ( 42 - x ) = 23
( 42 - x ) = 23 : 23
( 42 - x ) = 1
x = 42 - 1
x = 41
\(\text{Ta có: (x - 45).27 = 0
}\)
\(\text{ x - 45 = 0 : 27
}\)
\(\text{ x - 45 = 0
}\)
\(\text{ x = 0 + 45
}\)
\(\text{ x = 45}\)
a) (x – 45).27 = 0
=> x - 45 = 0
=> x = 45
b) 23.(42- x) = 23
=> 42- x = 1
=> x = 41
c. 3x – 5=7
=> 3x = 12
=> x = 4
e. 15 – 5x=10
=> 5x = 5
=> x = 1
Ta có:
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}\right):2.x=\frac{23}{45}\)
\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right):2.x=\frac{23}{45}\)
\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right):2.x=\frac{23}{45}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{90}\right):2.x=\frac{23}{45}\)
\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}\div\frac{11}{45}=\frac{23}{11}\)
Vậy \(x=\frac{23}{11}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).\frac{1}{2}.x=\frac{23}{45}\)
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{8.9}+\frac{1}{9.10}\right).\frac{1}{2}.x=\frac{23}{45}\)
\(\left(\frac{1}{2}+\frac{1}{6}+....+\frac{1}{72}+\frac{1}{90}\right).\frac{1}{2}.x=\frac{23}{45}\)
\(\left(\frac{1}{2}-\frac{1}{90}\right).\frac{1}{2}.x=\frac{23}{45}\)
\(\frac{22}{45}.\frac{1}{2}x=\frac{23}{45}\)
\(\frac{11}{45}.x=\frac{23}{45}\)
\(x=\frac{23}{45}\div\frac{11}{45}\)
\(x=\frac{23}{11}\)
=> \(x=\frac{23}{11}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\Rightarrow2.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{46}{45}\)
\(=\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x\)
\(=\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\right).x\)
\(=\left(\frac{1}{2}-\frac{1}{90}\right).x=\left(\frac{45}{90}-\frac{1}{90}\right)x=\frac{44}{90}.x=\frac{22x}{45}=\frac{46}{45}\)
=> 22x=46
=> x=\(46:22=\frac{23}{11}\)
a) (x - 45) . 27 = 0
x - 45 = 0 : 27
x - 45 = 0
x = 0 + 45
x = 45
b) 23 . (42 - x) = 23
42 - x = 23 : 23
42 - x = 1
- x = 1 + (-42)
-x = -41
x = 41
a) ( x - 45 ) . 27 = 0
Suy ra: x - 45 = 0
x = 0 + 45
x = 45
b) 23 . ( 42 - x) = 23
42 - x = 23 : 23
42 - x = 1
x = 42 - 1
x = 41