Tìm x, biết \(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}...\dfrac{30}{62}.\dfrac{31}{64}=2^x\)
Tìm số nguyên n, biết rằng:
\(\dfrac{1}{4} . \dfrac{2}{6} . \dfrac{3}{8} .\dfrac{4}{10} . \dfrac{5}{12} .... \dfrac{30}{62} . \dfrac{31}{64} = 2^{n}\)\(\)
\(\dfrac{1}{2.2}.\dfrac{2}{2.3}.....\dfrac{31}{64}=2^x\\ =>\dfrac{1}{2.2.2.....2.64}=2^x\\ \dfrac{1}{2^{30}.26}=2^x\\ =>\dfrac{1}{2^{36}}=2^x\\ =>2^{-36}=2^x\\ =>x=-36\)
Ta có: \(2^n=\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}....\dfrac{30}{62}.\dfrac{31}{64}\)
⇔ \(2^n=\dfrac{1.2.3.4....31}{2.\left(2.3.4.....1\right).64}=\)
⇔ \(2^n=\dfrac{1}{2}.\dfrac{1}{64}=\dfrac{1}{128}\) \(\Leftrightarrow\) \(2^n=\dfrac{1}{2^6}\)
⇔ \(2^{x+6}=1\)
⇔ \(x+6=0\)
⇒ \(\left\{{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
Tìm x :
\(\dfrac{1}{4}\) . \(\dfrac{2}{6}\) . \(\dfrac{3}{8}\) . \(\dfrac{4}{10}\) . .... . \(\dfrac{30}{62}\) . \(\dfrac{31}{64}\)
`1/4. 2/6. 3/8. 4/10........ 30/62. 31/64=2^x`
`=>\underbrace{1/2. 1/2. 1/2. 1/2..............1/2}_{\text{30 số 2}}. 1/64=2^x`
`=>(1/2)^{30}.(1/2)^{6}=2^x`
`=>(1/2)^{36}=2^x`
`=>2^{-36}=2^x`
`=>x=-36`
Vậy `x=-36`
\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{12}..........\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2\)x
Tìm x, biết
\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{12}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}\)=\(4^x\)
Lời giải:
Ta có:
\(\text{VT}=\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.....\frac{30}{62}.\frac{31}{64}=\frac{1.2.3....31}{2.4.6.8...64}\)
Xét mẫu số:
\(2.4.6.8.....62.64=(2.1)(2.2)(2.3)(2.4)....(2.31)(2.32)\)
\(=2^{32}(1.2.3....31.32)\)
Suy ra:
\(\text{VT}=\frac{1.2.3....31}{2^{32}.(1.2.3...31.32)}=\frac{1}{2^{32}.32}=\frac{1}{2^{37}}\)
Do đó \(4^x=\frac{1}{2^{37}}\Leftrightarrow 2^{2x}=\frac{1}{2^{37}}\Leftrightarrow 2^{2x+37}=1\)
\(\Leftrightarrow 2x+37=0\Leftrightarrow x=-\frac{37}{2}\)
Vậy \(x=\frac{-37}{2}\)
Tìm x biết :
a)\(\dfrac{x+1}{2x+1}=\dfrac{\dfrac{1}{2}x+2}{x+3}\)
b) \(x^2+5x=0\)
c)\(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}...\dfrac{30}{62}.\dfrac{31}{64}=2^x\)
a: \(\Leftrightarrow\dfrac{x+1}{2x+1}=\dfrac{x+4}{2x+6}\)
=>(x+1)(2x+6)=(2x+1)(x+4)
\(\Leftrightarrow2x^2+6x+2x+6=2x^2+8x+x+4\)
=>9x+4=8x+6
=>x=2
b: \(x^2+5x=0\)
=>x(x+5)=0
=>x=0 hoặc x=-5
Tìm x,biết:
a) -5(x+\(\dfrac{1}{5}\))-\(\dfrac{1}{2}\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}x-\dfrac{5}{6}\)
b) \(\dfrac{1}{4}\).\(\dfrac{2}{6}\).\(\dfrac{3}{8}\).\(\dfrac{4}{10}\)...\(\dfrac{30}{62}\).\(\dfrac{31}{64}\)=2x
a) \(-5\cdot\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\cdot\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}\cdot x-\dfrac{5}{6}\\ -5\cdot x+1-\dfrac{1}{2}\cdot x-\dfrac{1}{3}=\dfrac{3}{2}\cdot x-\dfrac{5}{6}\\ x\cdot\left(-5-\dfrac{1}{2}\right)+\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{3}{2}\cdot x\\ x\cdot\dfrac{-11}{2}+\dfrac{7}{6}=\dfrac{3}{2}\cdot x\\ \dfrac{3}{2}\cdot x-\dfrac{-11}{2}\cdot x=\dfrac{7}{6}\\ x\cdot\left(\dfrac{3}{2}-\dfrac{-11}{2}\right)=\dfrac{7}{6}\\ x\cdot7=\dfrac{7}{6}\\ x=\dfrac{7}{6}:7\\ x=\dfrac{1}{6}\)
Vậy x = \(\dfrac{1}{6}\)
b, \(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2^x\\ \dfrac{1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot30\cdot31}{2^{30}\cdot\left(1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot30\cdot31\right)\cdot64}=2^x\\ \dfrac{1}{2^{30}\cdot2^6}=2^x\\ \dfrac{1}{2^{36}}=2^x\\ 2^{-36}=2^x\\ \Rightarrow x=-36\)
1. Tìm x:
a. |x| = 5,6 b. |x-3,5| = 5 c.|x+\(\dfrac{3}{4}\)| - \(\dfrac{1}{2}\)= 0 d. |4x| -|-13,5| = |\(\dfrac{9}{4}\)|
e. (x-1)3 = 27 f. (2x-3)2=36 g. 5x + 2= 625 h. (2x -1) 3 = -8
i.* \(\dfrac{1}{4}\).\(\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}...\dfrac{30}{62}.\dfrac{31}{64}=2^x\)
a.
| x | = 5,6
=>\(\left[{}\begin{matrix}x=5,6\\x=-5,6\end{matrix}\right.\)
Vậy \(x\in\left\{-5,6;5,6\right\}\)
b, \(\left|x-3,5\right|=5\)
=>\(\left[{}\begin{matrix}x-3,5=5\\x-3,5=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8,5\\x=-1,5\end{matrix}\right.\)
Vậy \(x\in\left\{-1,5;8,5\right\}\)
c,\(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=> \(\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{4};\dfrac{5}{4}\right\}\)
d,\(\left|4x\right|-\left(\left|-13,5\right|\right)=\left|\dfrac{1}{4}\right|\)
=> \(\left|4x\right|-13,5=\dfrac{1}{4}\)
=> \(\left|4x\right|=13,75\)
=>\(\left[{}\begin{matrix}4x=13,75\\4x=-13,75\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3,4375\\x=-3,4375\end{matrix}\right.\)
Vậy \(x\in\left\{-3,4375;3,4375\right\}\)
e, ( x - 1 ) 3 = 27
=> x - 1 = 3
=> x = 4
Vậy x = 4
f, ( 2x - 3)2 = 36
=> \(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=4,5\\x=-1,5\end{matrix}\right.\)
Vậy x\(\in\left\{-1,5;4,5\right\}\)
g, \(5^{x+2}=625\)
=> \(5^{x+2}=5^4\)
=> x + 2 = 4
=> x = 2
Vậy x = 2
h, ( 2x - 1)3 = -8
=> 2x - 1 = -2
=> x = \(\dfrac{-1}{2}\)
Vậy x = \(\dfrac{-1}{2}\)
i, \(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}...\dfrac{30}{62}.\dfrac{31}{64}=2^x\)
=> \(\dfrac{1.2.3.4.5...30.31}{4.6.8.10.12...62.64}=2^x\)
=>\(\dfrac{1.2.3.4.5...30.31}{\left(2.3.4.5...30.31.32\right)\left(2.2.2.2...2.2_{ }\right)}=2^x\)(có 31 số 2)
=> \(\dfrac{1}{32.2^{31}}=2^x\)
=> \(\dfrac{1}{2^{36}}=2^x\)
=> x = -36
Vậy x = -36
Tính \(H=\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}...............................\dfrac{30}{62}.\dfrac{31}{64}\)
H = \(\dfrac{1}{4}\).\(\dfrac{2}{6}\).\(\dfrac{3}{8}\)...........\(\dfrac{30}{62}\).\(\dfrac{31}{64}\)
H = \(\dfrac{1}{2.2}\).\(\dfrac{2}{2.3}\).........\(\dfrac{31}{2.32}\)
H = \(\dfrac{1.2.3......31}{2.2.2.3.........2.32}\)
H = \(\dfrac{1}{256}\)
1,Tìm x, biết:
a/\(^{\left(x-1\right)^{x+2}}=^{\left(x-1\right)^{x+4}}\)
b/\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\)
c/\(4^x+3^x=2^x+6^x\left(STN\right)\)
GIÚP MÌNH NHANH NHA MÌNH TICK CHO
a)x=1;2;-2(bạn nên tự giải)
b)=>\(\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}\)=2x
=>\(\dfrac{2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31}{60\left(2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31\right)\cdot64}=2x\)
=>\(\dfrac{1}{60\cdot64}=2x\)=> 1/3840 =2x
=>x = 1/7680
c)=>4x - 2x = 6x - 3x
=>2x (2x-1)= 3x(2x-1)
=> 2x = 3x
=>x = 0