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sao chỉ có mỗi dấu cộng với dấu trừ thế bạn ? đề còn đâu
\(\Rightarrow\frac{1}{a}=\frac{5}{6}-\frac{b}{3}=\frac{5-2b}{6}\Rightarrow6=a\left(5-2b\right)\)
Đến đây bạn lm tiếp nha
a) x + 15 = 36 - 2x
x + 15 = 36 - (x + x )
15 =36 - ( x + x) - x
15 = 36 - x - x - x
15 = 36 - 3x
3x = 36 - 15
3x = 21
x = 21 : 3
=> x = 7
b) (x - 7) - (2x +5) = -14
x - 7 -( 2x + 5) = -14
x - (2x + 5) = -14 + 7 = -7
x - 2x - 5 = -7
x - 2x = -7 + 5 = -2
x - x + x = 2
x = 2 (-x + x cũng bằng chính nó)
=> x = 2
c) (x - 12) - 15 = (-7 + 20) - (18+x)
(x - 12) - 15 = 13 - (18 + x)
(x - 12) - 15 = 13 - 18 - x
(x - 12) - 15 = -5 - x
15 = (x - 12 ) - (-5 - x)
15 = x - 12 + 5 + x
15 = x + (-12) + 5 + x
15 = 2x + [(-12) + 5]
15 = 2x + -7
2x = -7 + 15
2x = 8
x = 8 : 2
=> x = 4
..................
a,18 chia hết cho n
=>n\(\in\)Ư(18)={-18,-9,-6,-3,-2,-1,1,2,3,6,9,18}
|x-10|+|x-11|+|x-12|+|x-13|=4
=>|x-10|+|x-13|+|x-11|+|x-12|=4
=>|x-10|+|13-x|+|x-11|+|12-x|=4
Ta có: |x-10|+|x-13|+|x-11|+|x-12|>=3+1=4(Bất đẳng thức giá trị tuyệt đối)
DBXRK 11<=x<=12=>x=11 hoặc x=12
Vậy x=11 hoặc x=12
\(1,\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}=\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}=6\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)=6\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).Tacocongthuc:\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\Rightarrow\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}=6\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-.....-\frac{1}{11}\right)=6\left(\frac{1}{3}-\frac{1}{11}\right)=\frac{48}{33}=\frac{16}{11}\)
\(2,\left(x+1\right)+\left(x+2\right)+.....+\left(x+211\right)=211x+\left(1+2+....+211\right)=211x+\frac{212.211}{2}=211x+22366=23632\Leftrightarrow211x=23632-22366=1266\Leftrightarrow x=6\)
a, \(14:\left(4\frac{2}{3}:1\frac{5}{9}\right)+14:\left(\frac{2}{3}+\frac{8}{9}\right)\)
=> \(14:\frac{28}{9}+14:\frac{14}{9}=>14.\frac{9}{28}+14.\frac{9}{14}\)
=> 14. ( \(\frac{9}{28}+\frac{9}{14}\) )
=> \(14.\frac{27}{28}=\frac{419}{28}\)
b, \(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)
=> \(\frac{4}{5}+\frac{12}{35}+\frac{4}{21}+\frac{4}{33}\)
=> \(\frac{8}{7}+\frac{24}{77}=\frac{16}{11}\)
bài 2 :
( x + 1 ) + ( x + 2 ) + ... + ( x + 211 ) = 23632
=> ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 211 ) = 23632
=> 211x + 22366 = 23632
=> 211x = 23632 - 22366
=> 211x = 1266
=> x = 1266 : 211
x = 6
\(a,2x+15=27\)
\(\Rightarrow2x=27-15\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=12\div2\)
\(\Rightarrow x=6\)
a) \(|3a-1|=|-14|\)
\(\Leftrightarrow|3a-1|=14\)
\(\Leftrightarrow\orbr{\begin{cases}3a-1=14\\3a-1=-14\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}a=5\\a=\frac{-13}{3}\end{cases}}\)
Vậy ..,.
\(b,-12\left|1-2a\right|=-36\)
\(\Rightarrow\left|1-2x\right|=-\frac{36}{-12}\)
\(\Rightarrow\left|1-2x\right|=3\)
\(\Rightarrow\orbr{\begin{cases}1-2x=3\\1-2x=-3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=-2\\2x=4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)