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\(\left[\left(6x-72\right):2-84\right].28=5628\)\(\Rightarrow\left(6x-72\right):2-84=5628:28=201\)
\(\Rightarrow\left(6x-72\right):2=201+84=285\)
\(\Rightarrow6x-72=285.2=570\)
\(\Rightarrow6x=570+72=642\)
\(\Rightarrow x=\frac{642}{6}=107\)

\(720:\left[41-\left(2x-5\right)\right]=2^3.5\)
\(\Leftrightarrow720:\left[41-\left(2x-5\right)\right]=40\)
\(\Leftrightarrow41-\left(2x-5\right)=720:40\)
\(\Leftrightarrow41-\left(2x-5\right)=18\)
\(\Leftrightarrow2x-5=41-18\)
\(\Leftrightarrow2x-5=23\)
\(\Leftrightarrow2x=28\)
\(\Leftrightarrow x=14\)
Ta có :\(720:\left[41-\left(2X-5\right)\right]=2^3\cdot5\)
\(\Leftrightarrow41-\left(2X-5\right)=18\)
\(\Leftrightarrow2X-5=23\)
\(\Leftrightarrow2X=28\)
\(\Leftrightarrow X=14\)(TMĐK)
Vậy X= 14 là giá trụ cần tìm.

= \(\frac{3^2.5^4.7^9}{3^3.5^2.7^5.3^3.5^2.7^5}\)
=\(\frac{3^2.5^4.7^9}{3^6.5^4.7^{10}}\)
= \(\frac{1.1.1}{3^4.1.7}\)
= \(\frac{1}{567}\)
\(\frac{\left(3^2.5.7^9\right).\left(3^5.5^3\right)}{\left(3^3.5^2.7^5\right)^2}=\frac{\left(3^2.3^5\right).\left(5.5^3\right).7^9}{3^6.5^4.7^{10}}=\frac{3^7.5^4.7^9}{3^6.5^4.7^{10}}=\frac{3}{7}\)

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\\ =\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =1-\dfrac{1}{101}\\ =\dfrac{100}{101}\)
\(\dfrac{5}{1\cdot3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{99\cdot101}\\ =\dfrac{5}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\right)\\ =\dfrac{5}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{5}{2}\cdot\left(1-\dfrac{1}{101}\right)\\ =\dfrac{5}{2}\cdot\dfrac{100}{101}\\ =\dfrac{250}{101}\)
\(a,\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...\dfrac{1}{99}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{100}{101}\)

\(B=\dfrac{2^2}{1\cdot3}+\dfrac{3^2}{2\cdot4}+\dfrac{4^2}{3\cdot5}+...+\dfrac{99^2}{98\cdot100}\\ =\dfrac{1\cdot3+1}{1\cdot3}+\dfrac{2\cdot4+1}{2\cdot4}+\dfrac{3\cdot5+1}{3\cdot5}+...+\dfrac{98\cdot100+1}{98\cdot100}\\ =\dfrac{1\cdot3}{1\cdot3}+\dfrac{1}{1\cdot3}+\dfrac{2\cdot4}{2\cdot4}+\dfrac{1}{2\cdot4}+\dfrac{3\cdot5}{3\cdot5}+\dfrac{1}{3\cdot5}+...+\dfrac{98\cdot100}{98\cdot100}+\dfrac{1}{98\cdot100}\\ =1+\dfrac{1}{1\cdot3}+1+\dfrac{1}{2\cdot4}+1+\dfrac{1}{3\cdot5}+...+1+\dfrac{1}{98\cdot100}\\ =\left(1+1+1+...+1\right)+\left(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\right)\\ =98+\left(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\right)\\ \)Gọi \(\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\) là A
\(A=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{98\cdot100}\\ =\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{2\cdot4}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{98\cdot100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{1}+\dfrac{1}{2}-\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{3}{2}-\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{295}{198}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\cdot\dfrac{14651}{9900}=\dfrac{14651}{19800}\)
\(B=98+A=98+\dfrac{14651}{19800}=98\dfrac{14651}{19800}\)
Dễ thấy phần nguyên của B là 98
Vậy phần nguyên của B là 98

mk ko ghi đb nhé
\(=\frac{1\cdot3+1}{1\cdot3}+\frac{2\cdot4+1}{2\cdot4}+...+\frac{99\cdot101+1}{99\cdot101}.\)
\(=1+\frac{1}{1\cdot3}+1+\frac{1}{2\cdot4}+...+1+\frac{1}{99\cdot101}\)
\(=99+\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{2\cdot4}+...+\frac{2}{99\cdot101}\right)\)
\(=99+\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{100}+\frac{1}{99}-\frac{1}{101}\right)\)
\(=99+\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
phần còn lại bn tự tính nha

mình đang xem đây, nhưng chỉ có điều kiện, không có đáp án nên thấy khó hiểu.
Gọi d là ƯCLN(18n+3, 21n+7) (d ∈ N)
Để phân số \(\frac{18n+3}{21n+7}\) có thể rút gọn được, d phải khác 1.
Ta có:
\(6\left(21n+7\right)-7\left(18n+3\right)⋮d\)
\(\Rightarrow21⋮d\) ⇒ d ∈{1,3,7,21}
Mà d phải khác 1 và 21n+7 không chia hết cho 3 và 21 suy ra d=7
Vậy mọi số tự nhiên n thỏa mãn ƯCLN(18n+3, 21n+7) là 7 thì phân số có thể rút gọn đc.
Mk ko chắc lắm :v

( 4 x - 13 ) 4 + 4 3 = 145
( 4 x - 13 ) 4 + 64 = 145
( 4 x - 13 ) 4 = 81
( 4 x - 13 ) 4 = 3 4
=> 4 x - 13 = 3
4 x = 16
x = 4
3 x + 2 - 3 x = 72
3 x ( 3 2 - 1 ) = 72
3 x . 8 = 72
3 x = 9
3 x = 3 2
=> x = 2
2 x + 2 - 2 x - 1 = 224
2 x ( 2 2 - 2 -1 ) = 224
2 x . 3 , 5 = 224
2 x = 64
2 x = 2 6
=> x = 6
( 4 x - 13 ) 4 + 4 3 = 145
( 4 x - 13 ) 4 + 64 = 145
( 4 x - 13 ) 4 = 81
( 4 x - 13 ) 4 = 3 4
=> 4 x - 13 = 3
4 x = 16
x = 4
3 x + 2 - 3 x = 72
3 x ( 3 2 - 1 ) = 72
3 x . 8 = 72
3 x = 9
3 x = 3 2
=> x = 2
2 x + 2 - 2 x - 1 = 224
2 x ( 2 2 - 2 -1 ) = 224
2 x . 3 , 5 = 224
2 x = 64
2 x = 2 6
=> x = 6

a) \(\left(4x-13\right)^4+4^3=145\)
\(\Rightarrow\left(4x-13\right)^4+64=145\)
\(\Rightarrow\left(4x-13\right)^4=81\)
\(\Rightarrow4x-13=\pm3\)
+) \(4x-13=3\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
+) \(4x-13=-3\)
\(\Rightarrow4x=10\)
\(\Rightarrow x=\frac{5}{2}\)
Vậy \(x=4\) hoặc \(x=\frac{5}{2}\)
b) \(3^{x+2}-3^x=72\)
\(\Rightarrow3^x.3^2-3^x=72\)
\(\Rightarrow3^x.\left(3^2-1\right)=72\)
\(\Rightarrow3^x.8=72\)
\(\Rightarrow3^x=9\)
\(\Rightarrow3^x=3^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
c) \(2^{x+2}-2^{x-1}=224\)
\(\Rightarrow2^{x-1+3}-2^{x-1}=224\)
\(\Rightarrow2^{x-1}.2^3-2^{x-1}=224\)
\(\Rightarrow2^{x-1}.\left(2^3-1\right)=224\)
\(\Rightarrow2^{x-1}.7=224\)
\(\Rightarrow2^{x-1}=32\)
\(\Rightarrow2^{x-1}=2^5\)
\(\Rightarrow x-1=5\)
\(\Rightarrow x=6\)
Vậy x = 6

Trọng mỗi số 4;8;16;11;20 thì 4;8;11;20 là ước của A
Vì a=23.52.11\(\Rightarrow\)a=2200
2200\(⋮\)4;8;11;20
\(\Rightarrow\)trong mỗi số 4;8;16;11:20 thì 4;8;11;20 là ước của a
Bằng2024
bạn giải ra một bài toán được ko mik ko biết