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\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{x\left(x+1\right)}=\frac{99}{100}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)
\(1-\frac{1}{x+1}=\frac{99}{100}\)
=> \(\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)
=> x+1 = 100
=> x = 100 - 1
=> x = 99
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Câu 2:
a: x-158=32
=>x=158+32
=>x=190
b: \(x\cdot24=264\)
=>\(x=\dfrac{264}{24}\)
=>x=11
c: \(6x+9=3^7:3^4\)
=>\(6x+9=3^3\)
=>6x+9=27
=>6x=18
=>x=18/6=3
Câu 1:
a: \(86\cdot19+14\cdot19\)
\(=19\left(86+14\right)\)
\(=19\cdot100=1900\)
b: \(4\cdot\left(-5\right)^2-104\cdot\left(-5\right)^2\)
\(=4\cdot25-104\cdot25\)
\(=25\left(4-104\right)=-100\cdot25=-2500\)
c: \(7\cdot\left(-2\right)\cdot8\left(-5\right)\)
\(=7\cdot2\cdot8\cdot5\)
\(=56\cdot10=560\)
d: \(59-\left[59+\left(-76\right)\right]\)
\(=59-59+76\)
=76
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\(\frac{x-1}{65}+\frac{x-3}{63}=\frac{x-5}{61}+\frac{x-7}{59}\)
\(\frac{128x-258}{4095}=\frac{120x-722}{3599}\)
\(<=>\left(128x-258\right)3599=4095\left(120x-722\right)\)
\(=>x=66\)
2.x.3 + 5 = 59
2.x.3 = 59 - 5
2.x.3 = 54
x.3 = 54 : 2
x.3 = 27
x = 27 : 3
x = 9
\(2.x.3+5=59\Rightarrow6x+5=59\Rightarrow6x=59-5=54\)
\(\Rightarrow x=54:6=9\)
Vậy x=9