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(x+2)+(x+4)+(x+6)....+(x+50)=750
=(x+x+x+x+.....+x)+2+4+6+....+50
=>25x+650=750
=>25x=100
=>x=4
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Câu hỏi 1 : 29 Câu hỏi 2 : {-1;3} Câu hỏi 3 : -60 Câu hỏi 4: {-19;-15} Câu hỏi 5: đề sai
Câu hỏi 6: 99 Câu hỏi 7: 981 Câu hỏi 8: 36 Câu hỏi 9 sai đề Câu hỏi 10 sai đề
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a, (sửa đề )
\(1+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{x.\left(x+1\right)}=\frac{1999}{2000}\)
=\(1+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x.\left(x+1\right)}\right)=\frac{1999}{2000}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x+\left(x+1\right)}=1-\frac{1999}{2000}=\frac{1}{2000}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{2000}\)
=\(\frac{1}{1}-\frac{1}{x+1}=\frac{1}{2000}\)
=\(\frac{1}{x+1}=\frac{1}{1}-\frac{1}{2000}=\frac{1999}{2000}\)
=> \(x+1=1:\frac{1999}{2000}=\frac{2000}{1999}\)
=>\(x=\frac{2000}{1999}-1=\frac{1}{1999}\)
Vậy x ∈{ \(\frac{1}{1999}\)}
b, \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+.....+\frac{2}{x+\left(x+1\right)}=\frac{2}{9}\)
=> \(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+.....+\frac{2}{x+\left(x+1\right)}=\frac{2}{9}\)
=>\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+.....+\frac{2}{x+\left(x+1\right)}=\frac{2}{9}\)
=>2.(\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+....+\frac{1}{x.\left(x+1\right)}\))=\(\frac{2}{9}\)
=>\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+....+\frac{1}{x+\left(x+1\right)}=\frac{2}{9}:2=\frac{1}{9}\)
=>\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)
=>\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
=>\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}\)
=>\(x+1=18\)
=>\(x=18-1=17\)
=>x∈{17}
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4x+4x+2=272
4x+4x.42=272
4x.1+4x.16=272
4x.(16+1)=272
4x.17=272
4x=272:17
4x=16=42
=>x=2
\(x^{50}=x\)
\(=>x\in\left\{1;0\right\}\)
\(\left(2^x+1\right)^3=125\)
\(\left(2^x+1\right)^3=5^3\)
\(2^x+1=5\)
\(2^x=4\)
\(2^x=2^2\)
\(=>x=2\)
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a ) 6 |x - 7| = 18 : (-3)
\(\Rightarrow6\left|x-7\right|=-6\)
\(\Rightarrow\left|x-7\right|=-1\) (1)
Mà \(\hept{\begin{cases}\left|x-7\right|\ge0\forall x\\-1< 0\end{cases}}\)
\(\Rightarrow\) | x - 7| = - 1 ( vô lí ) (2)
Từ (1) và (2) \(\Rightarrow\) \(x\in\varnothing\)
Vậy \(x\in\varnothing\)
Câu c tương tự nhé
b) -7 | x + 4| = 21 : (-3)
\(\Rightarrow-7\left|x+4\right|=-7\)
\(\Rightarrow\left|x+4\right|=1\)
\(\Rightarrow\orbr{\begin{cases}x+4=1\\x+4=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=-5\end{cases}}\)
Vậy \(x\in\left\{-3;-5\right\}\)
@@ Hc tốt @@
## Chiyuki Fujito
6 | x - 7 | = 18 : ( -3 )
6 | x - 7 | = ( -6 )
| x - 7 | = ( -6 ) : 6
| x - 7 | = ( -1 )
\(\Rightarrow\)x - 7 = ( -1 ) hoặc x - 7 = 1
x = ( -1 ) + 7 x = 1 + 7
x = 6 x = 8
\(\Rightarrow\) x = 6 ; x = 8
-7 | x + 4 | = 21 : ( -3 )
-7 | x + 4 | = ( -7 )
| x + 4 | = ( -7 ) : ( -7 )
| x + 4 | = 1
\(\Rightarrow\)x + 4 = 1 hoặc x + 4 = ( -1 )
x = 1 - 4 x = ( -1 ) - 4
x = -3 x = -5
\(\Rightarrow\) x = ( -3 ) ; x = ( -5 )
3 | x + 5 | = ( -9 )
| x + 5 | = ( -9 ) : 3
| x + 5 | = ( -3 )
\(\Rightarrow\)x + 5 = ( -3 ) hoặc x + 5 = 3
x = ( -3 ) - 5 x = 3 - 5
x = ( -8 ) x = ( -2 )
\(\Rightarrow\)x = ( -8 ) ; x = ( -2 )
@ Học tốt @
Nhớ k cho mình nha !!!!! Thank
\(x^6=2\)
\(\Rightarrow x=\sqrt[6]{2}\)
Đổi: x^6 = 2 thành \(x^6=2\)
Suy ra :
x= \(\sqrt[6]{2}\)
Đs: