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7 tháng 4 2015
Ta có \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(2\left(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2}{9}\)
\(2\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(2\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{2}{9}\div2\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\left(1\right)\)
Có \(\left(1\right)\Leftrightarrow\left(x+1\right).1=1.18\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=18-1\)
\(\Rightarrow x=17\)
DH
0
XO
24 tháng 6 2021
Ta có \(1\frac{1}{5}x+\frac{2}{3}x=-\frac{56}{125}\)
<=> \(\frac{6}{5}x+\frac{2}{3}x=-\frac{56}{125}\)
<=> \(\frac{28}{15}x=-\frac{56}{125}\)
<=> \(x=-\frac{2}{15}\)
b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
<=> \(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=0\)
<=> \(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
<=> x + 100 = 0
<=> x = -100
TP
1
VK
6 tháng 9 2017
Ta có: 2x + 2x+1 + 2x+2 = 56
=> 2x + 2x . 21 + 2x . 22 = 56
=> 2x . 1 + 2x . 2 + 2x . 4 = 56
=> 2x .(1 + 2 + 4) = 56
=> 2x . 7 = 56
=> 2x = 56 : 7
=> 2x = 8
Mà 8 = 23 = (-2)3
=> x = 2 hoặc x = -2
Vậy x = 2 hoặc x = -2.
20 tháng 1 2017
\(2\left|x+1\right|=10\)
Th1 :
\(\left|x+1\right|\ge0\Rightarrow x\ge-1\)
\(2\left(x+1\right)=10\)
\(\Rightarrow2x+2=10\)
\(\Rightarrow2x=8\)
\(\Rightarrow x=4\)
Th2 : \(\left|x+1\right|< 0\Rightarrow x< -1\)
\(2\left(-1-x\right)=10\)
\(\Rightarrow-2-2x=10\)
\(\Rightarrow-2x=8\)
\(\Rightarrow x=-4\)
7 tháng 7 2019
a) \(\left(x-1\right)^2=0\)
=> \(x-1=0\)
=> \(x=1\)
b) \(x\left(2x+1\right)=0\)
=> \(\orbr{\begin{cases}x=0\\2x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)
c) \(\left(-12\right)^2.x=56+10.13x\)
=> \(144x=56+130x\)
=> \(144x-130x=56\)
=> \(14x=56\)
=> \(x=56:14\)
=> \(x=4\)
d) \(x-\left(17-x\right)=x-7\)
=> \(x-17+x=x-7\)
=> \(2x-17=x-7\)
=> \(2x-x=-7+17\)
=> \(x=10\)
S
7 tháng 7 2019
a) (x-1)2 = 0
=> x - 1 = 0
=> x = 1
vậy_
b) x(2x+1)=0
=> x = 0 hoặc 2x + 1 = 0
=> x = 0 hoặc 2x = -1
=> x = 0 hoặc x = -1/2
vậy_
c) (-12)^2.x=56+10.13.x
=> 144x = 56 + 130x
=> 144x - 130x = 56
=> 14x = 56
=> x = 4
vậy_
d) x-(17-x)=x-7
=> x - 17 + x = x - 7
=> x + x - x = -7 + 17
=>x = 10
vậy_
e) (x+4) ⋮ (x+1)
=> x + 1 + 3 ⋮ x + 1
=> 3 ⋮ x + 1
=> x + 1 thuộc Ư(3)
=> x + 1 thuộc {-1; 1; -3; 3}
=> x thuộc {-2; 0; -4; 2}
vậy_
g) (4x+3)⋮(x-2)
=> 4x - 4 + 7 ⋮ x - 2
=> 2(x - 2) + 7 ⋮ x - 2
=> 7 ⋮ x - 2
=> x - 2 thuộc Ư(7)
=> x - 2 thuộc {-1; 1; -7; 7}
=> x thuộc {1; 3; -5; 9}
vậy_
TM
0
HA
4
6 tháng 4 2018
Ta có :
\(\frac{x-2}{12}+\frac{x-2}{20}+\frac{x-2}{30}+\frac{x-2}{42}+\frac{x-2}{56}+\frac{x-2}{72}=\frac{16}{9}\)
\(\Leftrightarrow\)\(\left(x-2\right)\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)
\(\Leftrightarrow\)\(\left(x-2\right)\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\frac{16}{9}\)
\(\Leftrightarrow\)\(\left(x-2\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\Leftrightarrow\)\(\left(x-2\right)\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\Leftrightarrow\)\(\left(x-2\right).\frac{2}{9}=\frac{16}{9}\)
\(\Leftrightarrow\)\(x-2=\frac{16}{9}:\frac{2}{9}\)
\(\Leftrightarrow\)\(x-2=\frac{16}{9}.\frac{9}{2}\)
\(\Leftrightarrow\)\(x-2=\frac{8}{1}.\frac{1}{1}\)
\(\Leftrightarrow\)\(x-2=8\)
\(\Leftrightarrow\)\(x=8+2\)
\(\Leftrightarrow\)\(x=10\)
Vậy \(x=10\)
Chúc bạn học tốt ~
TT
26 tháng 7 2020
1
\(\left(x-2\right):2.3=6\)
\(\Leftrightarrow\left(x-2\right):2=2\)
\(\Leftrightarrow\left(x-2\right)=4\)
\(\Leftrightarrow x=4+2=6\)
c) ta có
\(\left[\left(2x+1\right)+1\right]m:2=625\)
\(\Leftrightarrow\left[\left(2x+1\right)+1\right]\left\{\left[\left(2x+1\right)-1\right]:2+1\right\}=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-1:2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2+1=1250\)
\(\Leftrightarrow\left(2x+1\right)^2+1-2=1249\)
\(\Leftrightarrow\left(2x+1\right)^2+1=1251\)
\(\Leftrightarrow\left(2x+1\right)^2=1250\)
...
2
\(\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{7}{4}-\frac{1}{2}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right).\frac{5}{3}=\frac{5}{4}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}:\frac{5}{3}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)=\frac{5}{4}.\frac{3}{5}\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}\)
2\(^x\) + 2\(^{x+1}\) + 2\(^{x+2}\) = 56
2\(^x\).(1 + 2 + 22) = 56
2\(^x\).(1 + 2 + 4) = 56
2\(^x\).7 = 56
2\(^x\) = 56 : 7
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
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