8x+x=3^2.12
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26 tháng 12 2018
a) 47 - [ (45.24 - 52.12) : 14 ]
= 47 - ( 420 : 14 )
= 47 - 30
= 17
22 tháng 12 2019
\(x^2-1=2^2.12\)
\(x^2-1=4.12\)
\(x^2-1=48\)
\(x^2\) \(=48+1\)
\(x^2\) \(=49\)
\(x^2\) \(=7^2\)
=>\(x=7\)
19 tháng 7 2016
1. (7.x-15):3=2 2. 12.(x+37)=504 3. 85-3.(7+x)=64
7.x-15=2.3 x+37=504:12 3.(7+x)=85-64
7.x-15=6 x+37=42 3.(7+x)=21
7.x=6+15 x=42-37 7+x=21:3
7.x=21 x=5 7+x=7
x=21:7 x=7-7
x=3 x=0
19 tháng 7 2016
1. (7.x-15):3 = 2
(7.x-15)=2.3
(7.x-15)=6
7x=15+6
7x=21
x=21:7
x=3
2.12.(x+37=504
x+37=504:12
x+37=42
x = 42-37
x=5
3. 85-3.(7+x)=64
3.(7+x)=85-64
3.(7+x)=21
7+x=21:3
7+x=7
x=7-7
x=0
VM
22 tháng 12 2019
\(x^2-1=2^2.12\)
\(\Rightarrow x^2-1=4.12\)
\(\Rightarrow x^2-1=48\)
\(\Rightarrow x^2=48+1\)
\(\Rightarrow x^2=49\)
\(\Rightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)
Vậy \(x\in\left\{7;-7\right\}.\)
Chúc bạn học tốt!
HL
23 tháng 12 2019
(x^2)-1=(2^2).12
(x^2)-1=4.12
(x^2)-1=48
x^2=48+1
x^2=49
suy ra;x^2=7^2
suy ra;x=7
NK
0
NH
1
5 tháng 3 2020
\(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+\frac{1}{3\cdot103}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+\frac{100}{3\cdot103}+...+\frac{100}{100\cdot110}\right)x=10\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+....+\frac{1}{10}-\frac{1}{110}\right)x=10\)\(\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+....+\frac{1}{110}\right)\right]x=10\)\(\left[\left(1+\frac{1}{2}+....+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+....+\frac{1}{100}\right)-\left(\frac{1}{11}+\frac{1}{12}+....+\frac{1}{110}\right)\right]\)
\(\Rightarrow\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{110}\right)\right]x=10\)
\(\left[\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right]\)
\(\Rightarrow x=10\)
NA
1
5 tháng 3 2019
Ta có:
$(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{10.110}).x=\frac{1}{1.11}+\frac{1}{2.12}+...+\frac{1}{100.110}$
$\Leftrightarrow \frac{1}{100}\left ( \frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )x=\frac{1}{10}\left ( \frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110} \right )$
$\Leftrightarrow \left ( \frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )x=10\left ( \frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110} \right )$
Đặt $A=\frac{1}{1}-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}$
$\Rightarrow A=\left ( 1+\frac{1}{2}+...+\frac{1}{10} \right )+\left ( \frac{1}{11}+\frac{1}{12}+...+\frac{1}{100} \right )-\left ( \frac{1}{11}+\frac{1}{12}+...+\frac{1}{100} \right )-\left (\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110} \right )$
$\Rightarrow A=\left ( 1+\frac{1}{2}+...+\frac{1}{10} \right )-\left (\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110} \right )$
$\Rightarrow A=\frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}$
Thay vào phương trình, ta có:
$\left ( \frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )x=10\left ( \frac{1}{1}-\frac{1}{100}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110} \right )$
$\Leftrightarrow x=10$
5 tháng 4 2015
\(\Rightarrow\left(\frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110}\right).x=10.\left(\frac{10}{1.10}+\frac{10}{2.12}+...+\frac{10}{100.110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{10}-\frac{1}{110}\right).x=10.\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right).x=10.\left(\left(1+\frac{1}{2}+..+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)-\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}\right)\right)\)
\(\Rightarrow\left(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right).x=10.\left(\left(1+\frac{1}{2}+..+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right)\)=> x = 10
5 tháng 4 2015
Nhân 100 vào 2 vế ta được :
(100/1.101 + 100/2.102 + 100/3.103 +....+100/10.110) . x = (10/1.11 + 10/2.12 + 10/100.110 )10
=>(1+1/2+1/3+...+1/10-1/101-...-1/110)x=(1+1/2+1/3+...+1/10+1/11+...+1/100-1/11-...-1/100-1/101-...-1/110)10
=>(1+1/2+1/3+...+1/10-1/101-...-1/110)x=(1+1/2+1/3+...+1/10-1/101-...-1/110)10
=>x=10
Hay thì like nha ! hj hj
TT
1
11 tháng 4 2017
Đặt :
\(A=\dfrac{1}{1.101}+\dfrac{1}{2.102}+\dfrac{1}{3.103}+..................+\dfrac{1}{10.110}\)
\(A=\dfrac{1}{100}\left(\dfrac{100}{1.101}+\dfrac{100}{2.102}+..................+\dfrac{100}{10.101}\right)\)
\(A=\dfrac{1}{100}\left(1-\dfrac{1}{101}+\dfrac{1}{2}-\dfrac{1}{102}+..............+\dfrac{1}{10}-\dfrac{1}{101}\right)\)
\(A=\dfrac{1}{100}\left[\left(1+\dfrac{1}{2}+...........+\dfrac{1}{10}\right)-\left(\dfrac{1}{101}+\dfrac{1}{102}+..........+\dfrac{1}{101}\right)\right]\)
Đặt :
\(B=\dfrac{1}{1.11}+\dfrac{1}{2.12}+...............+\dfrac{1}{100.101}\)
\(B=\dfrac{1}{10}\left(\dfrac{10}{1.11}+\dfrac{10}{2.12}+.............+\dfrac{10}{100.101}\right)\)
\(B=\dfrac{1}{10}\left(1-\dfrac{1}{11}+\dfrac{1}{2}-\dfrac{1}{12}+..............+\dfrac{1}{100}-\dfrac{1}{101}\right)\)
\(B=\dfrac{1}{10}\left[\left(1+\dfrac{1}{2}+...........+\dfrac{1}{100}\right)-\left(\dfrac{1}{11}+\dfrac{1}{12}+...............+\dfrac{1}{101}\right)\right]\)
\(=\dfrac{1}{10}\left[\left(1+\dfrac{1}{2}+.........+\dfrac{1}{10}\right)-\left(\dfrac{1}{101}+\dfrac{1}{102}+...........+\dfrac{1}{101}\right)\right]\)
\(\Rightarrow B=10A\)
\(\Rightarrow A.x=10A\)
\(x=10A:A\)
\(x=10\) (thỏa mãn)
Vậy \(x=10\) là giá trị cần tìm
~ Chúc bn học tốt ~
\(8x+x=3^2\cdot12\)
\(9x=9\cdot12\)
\(x=12\)
Ta có : 8x + x = 3\(^2\). 12
\(\Rightarrow\)9x = 9 . 12
\(\Rightarrow\)9x = 108
\(\Rightarrow\)x = 12