145 x 2 + 4 x 154 + 3 + 154 + 154 =
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a) 125.3+125.125.6
=125.3+125.750
=125.(3+750)
=125.753
=94125
b) (245.99+245)-(145.101-145)
=(245.(99+1))-(145.(101-1))
=245.100-145.100
=(245-145).100
=100.100
=10000
c) 154.(28+52)-45.(26+54)
=154.80-45.80
=(154-45).80
=109.80
=8720
a) 125x3+125+125x6
=125x(3+1+6)
=125x10
=1250
b) (245x99+245)-(145x101-145)
=(245x99+245)-(145x100)
=(245x100)-14500
=24500-14500
10000
c) 154x80-45x80
=80x(145-45)
=80x100
=8000
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x + 2587 = 7531
=> x = 7531 - 2587 = 4944
x + 154 = 458
=> x = 458 - 154 = 304
x - 145 = 182
=> x = 182 + 145 = 327
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Ta có: x + (x+1)+(x+2)+...+(x+10)=154
=> (x+x+..+x) + (1+2+...+10)=154
=> 10x + 55 = 154
=> 10x = 154 - 55
=> 10x =99
=> x = 99:10
=> x=9,9
Vậy x = 9,9
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(x+1)+(x+2)+(x+3)+....+(x+15)+(x+16)=154
=>(x+x+...+x)+(1+2+...+16)=154
=>16x+136=154
=>16x=18
=>x=1,125 (hình như đề sai)
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Ta có 2/40 + 2/88 + 2/154 + ... + 2/x( x + 3) = 202
=> 2/5 x 8 + 2/8 x 11 + ... + 2/x( x + 3 ) = 202
=> 1/5 x 8 + 1/8 x 11 + ... + 1/x( x + 3 ) = 202 : 2
=> 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/x - 1/x + 3 = 101
=> 1/5 - 1/x + 3 = 101
=> 1/x + 3 = 1/5 - 101
=> 1/X + 3 = 504/5
=> 504(x + 3 ) = 5
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\(\frac{2}{40}+\frac{2}{88}+\frac{2}{154}+...+\frac{2}{x\left(x+3\right)}=\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+...+\frac{2}{x\left(x+3\right)}\)
\(=\frac{2}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{2}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(=\frac{2}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)\)
Từ đó ta có:
\(\frac{2}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{202}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
\(x+3=308\)
x = 305
145x2+4x154+3+154+154
=154x[2+4+3+1]
=154x10
+1540
=1540 nha cậu