(102–150)–50
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Ta có : \(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)
\(\Leftrightarrow\dfrac{5x-150}{50}-1+\dfrac{5x-102}{49}-2+\dfrac{5x-56}{48}-3+\dfrac{5x-12}{47}-4+\dfrac{5x-660}{46}+10=0\)
\(\Leftrightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)
\(\Leftrightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)
\(\Leftrightarrow5x-200=0\)
\(\Leftrightarrow x=40\)
Vậy ...
Ta có: \(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)
\(\Leftrightarrow\dfrac{5x-150}{50}-1+\dfrac{5x-102}{49}-2+\dfrac{5x-56}{48}-3+\dfrac{5x-12}{47}-4+\dfrac{5x-660}{46}+10=0\)
\(\Leftrightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)
\(\Leftrightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)
mà \(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}>0\)
nên 5x-200=0
\(\Leftrightarrow5x=200\)
hay x=40
Vậy: S={40}
a: \(20-\left[30-\left(5-1\right)^2\right]\)
\(=20-\left[30-4^2\right]\)
\(=20-14=6\)
b: \(71+\dfrac{50}{5+3\left(57-6\cdot7\right)}\)
\(=71+\dfrac{50}{5+3\cdot\left(57-42\right)}\)
\(=71+\dfrac{50}{5+3\cdot15}=71+\dfrac{50}{50}=72\)
c: \(4\cdot\left\{270:\left[50-\left(2^5+45:5\right)\right]\right\}\)
\(=4\cdot\left\{270:\left[50-32-9\right]\right\}\)
\(=4\cdot\left\{\dfrac{270}{50-41}\right\}=4\cdot\dfrac{270}{9}=4\cdot30=120\)
d: \(411-\left[\dfrac{\left(107+3\right)}{5}-2^2\right]\)
\(=411-\left[\dfrac{110}{5}-4\right]\)
=410-22+4
=410-18
=392
e: \(450-5\left[3^2\left(7^5:7^3-41\right)-12\right]+18\)
\(=450-5\left[9\cdot\left(7^2-41\right)-12\right]+18\)
\(=450-5\cdot\left[9\cdot8-12\right]+18\)
=468-5*60
=468-300
=168
f:
\(102-150:\left[18-2\cdot\left(10-8\right)^2\right]+1018^0\)
\(=102-150:\left[18-2\cdot4\right]+1\)
\(=103-\dfrac{150}{18-8}=103-15=88\)
\(pt\Leftrightarrow\frac{5x-150}{50}+\frac{5x-102}{49}+\frac{5x-56}{48}+\frac{5x-12}{47}+\frac{5x-16}{46}-14=0\)
\(\Leftrightarrow\frac{5x-150}{50}-1+\frac{5x-102}{49}-2+\frac{5x-56}{48}-3+\frac{5x-12}{47}-4+\frac{5x-16}{46}-4=0\)
\(\Leftrightarrow\frac{5x-200}{50}+\frac{5x-200}{49}+\frac{5x-200}{48}+\frac{5x-200}{47}+\frac{5x-200}{46}=0\)
\(\Leftrightarrow\left(5x-200\right)\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}\right)=0\)
Do \(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}\ne0\) nên \(5x-200=0\Rightarrow x=\frac{200}{5}=40\)
Vậy x= 40
\(\frac{5x-150}{50}+\frac{5x-102}{49}+\frac{5x-56}{48}+\frac{5x-12}{47}+\frac{5x-660}{46}=0\)
\(\Leftrightarrow\)\(\left(\frac{5x-150}{50}-1\right)+\left(\frac{5x-102}{49}-2\right)+\left(\frac{5x-56}{48}-3\right)+\left(\frac{5x-12}{47}-4\right)+\left(\frac{5x-660}{46}+10\right)=0\)
\(\Leftrightarrow\)\(\frac{5x-200}{50}+\frac{5x-200}{49}+\frac{5x-200}{48}+\frac{5x-200}{47}+\frac{5x-200}{46}=0\)
\(\Leftrightarrow\)\(\left(5x-200\right)\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}\right)=0\)
\(\Leftrightarrow\)\(5x-200=0\)
\(\Leftrightarrow\)\(5x=200\)
\(\Leftrightarrow\)\(x=40\)
Vậy x = 40
Câu b hướng làm đó là tách con 1/3 và 1/2 ra thành 50 phân số giống nhau. E tách 1/3=50/150 rồi so sánh 1/101, 1/102,...,1/149 với 1/150. Còn vế sau 1/2=50/100 tách tương tự rồi so sánh thôi
2a.
$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}$
$< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}$
$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{50-49}{49.50}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}$
$=1-\frac{1}{50}< 1$ (đpcm)
150-{102-(14-11)2.20070}
=150-(102-32.20070)
=150-(100-9-1)
=150-90
=60
150− [102 − (14 − 11)2.20070]
= 150 − [100 − 32.1]
= 150 − 91
= 69
8102-2102=2102.4102-2102=2102.(4102-1)
Do 4 mủ chẵn có tận cùng là 6 =>4102 có tận cùng là 6 =>4102-1 tận cùng là 5
=>4102-1 chia hết cho 5
2102 chia hết cho 2
=>2102.(4102-1) chia hết cho 10
Vậy 8102-2102 chia hết cho 10
Ta có 8102 = (84)25 . 82 = (....6)25 . 82 = ...6 . 64 = ....4
2102 = (22)51 = 451 = .....4
Ta thấy .....4 - .....4 = ......0 chia hết cho 10
Vậy 8102 - 2102 chia hết cho 10
-98 nhé
/HT\
98 nha