\(2^{x+4}-2^{x+3}-2^{x+2}-2^{x+1}-2=62\) ( tìm x)
Tìm x :
a, 1/4 x 2/6 x 3/8 x 4/10x....x30/62 x 31/64 = \(2^x\)
b, (x-1)3 = -27
c, (2x+1)2= 625
d, x2 + x = 0
Ta có: (1/4)*(2/6)*(3/8)*(4/10)*(5/12)*...*(30/62)*(31/64)=2^x. Tìm x
Ta có: \(2^x=\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{12}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}\)
\(\Leftrightarrow2^x=\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot31}{2\cdot\left(2\cdot3\cdot4\cdot...\cdot31\right)\cdot64}\)
\(\Leftrightarrow2^x=\dfrac{1}{2}\cdot\dfrac{1}{64}=\dfrac{1}{128}\)
\(\Leftrightarrow2^x=\dfrac{1}{2^6}\)
\(\Leftrightarrow2^{x+6}=1\)
\(\Leftrightarrow x+6=0\)
hay x=-6
Vậy: x=-6
`1/4 . 2/6 . 3/8 ... . 30/62 .31/64 =2^x`
`-> (1.2.3....30.31)/(4.6.8....62.64)=2^x`
`-> (1.(2.3...31))/(2.(2.3.4...31).32)=2^x`
`-> 1/(2.32)=2^x`
`-> 1/64=2^x`
`-> 1/(2^6)=2^x`
`-> x=-6`.
Tìm x, biết
1) (x - 1)x + 2 = (x - 1)x + 4
2) 1/4 . 2/6 . 3/8 . 4/10 ... 30/62 . 31/64 = 2x
3) 32 . 3-5 . 3x = 311
4) 2-1 . 2x + 2 . 2x = 5 . 25
Tìm x biết
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}......\frac{30}{62}.\frac{31}{62}=2^x\)
\(\frac{1}{4}.\frac{2}{6}............\frac{31}{64}=2^x\)
\(\Rightarrow\frac{1.2........31}{2.2.2.3...........2.31.64}=2^x\)
\(\Rightarrow\frac{1}{2^{30}.2^4}=2^x\)
\(\Rightarrow\frac{1}{2^{34}}=2^x\)
\(\Rightarrow x=-34\)
tìm x
a, x2 + x = 0
b, (x-1)x+2 = (x-1)x+4
c, 1/4. 2/6. 3/8. 4/10. 5/12..... 30/62. 31/64=2x
d, x-1/ x+5= 6/7
e, x2/ 6= 24/25
g, x-2/ x-1= x+4/ x+7
a) x2 + x = 0
=> x( x+ 1 ) = 0
=> x = 0
hoặc x = -1
b) b, (x-1)x+2 = (x-1)x+4
=> x + 2 = x + 4
=> 0x = 2 ( ktm)
Vậy ko có giá trị x nào thoả mãn đk
d) Ta có: x-1/x+5 = 6/7
=>(x-1).7 = (x+5).6
=>7x-7 = 6x+ 30
=> 7x-6x = 7+30
=> x = 37
Vậy x = 37
e, x2/ 6= 24/25
=> x2 . 25 = 6 . 24
⇒
Vậy
g, x-2/ x-1= x+4/ x+7
tìm x biết :
(1/4).(2/6).(3/8).(4/10).(5/12)....(30/62).(31/64) = 2^x
Có: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}...\frac{30}{62}.\frac{31}{64}=\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}...\frac{30}{2.31}.\frac{31}{2.32}=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}...\frac{1}{2}.\frac{1}{2}.\frac{1}{32}\)
\(=\frac{1}{2^{31}.2^5}=\frac{1}{2^{36}}=2^x\)\(\Rightarrow1=2^x.2^{36}=2^{36+x}\)\(\Rightarrow2^{36+x}=2^0\Rightarrow36+x=0\Rightarrow x=-36\)
tìm x :
1) 9.x = 62
2) (3+x) : 4 = 15
Bài 1: Tìm x là STN biết:
1/ 5x - 2 - 32 = 24 - ( 68 : 66 - 62)
2/ 3x + 42 = 196 : (193 x 192) - 3.12014
3/ 2x + 2x + 4 = 272
4/ 3 + 2x - 1 = 24 - \([4^2-(2^2-1)]\)
1: =>\(5^{x-2}-9=2^4-\left(6^2-6^2\right)\)
=>\(5^{x-2}=16+9=25\)
=>x-2=2
=>x=4
2: \(\Leftrightarrow3^x+16=19^6:19^5-3=19-3=16\)
=>3^x=0
=>x=0
3: \(\Leftrightarrow2^x+2^x\cdot16=272\)
=>2^x*17=272
=>2^x=16
=>x=4
4: \(\Leftrightarrow2^{x-1}+3=24-\left(4^2-2^2+1\right)=24-\left(16-4+1\right)\)
=>\(2^{x-1}+3=24-16+4-1=8+4-1=12-1=11\)
=>2^x-1=8
=>x-1=3
=>x=4
1/4 x 2/6 x3/8 x 4/10 x 5/12 x....x 30/62 x 31/64=2n. Tìm n
<=> \(\frac{1.2.3....31}{4.6.8....64}=2^n\Rightarrow\frac{1.2.3....30.31}{2\left(2.3.4.5...31\right).32}=2^n\Leftrightarrow\frac{1}{2.32}=2^n\Leftrightarrow\frac{1}{2^6}=2^n\)
=> 2^6.2^n = 1
=> 2^ (n + 6 ) = 2^0
=> n+ 6 = 0
=> n = - 6
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}....\frac{31}{64}=\frac{1.2.3....31}{4.6.8....64}=\frac{1.2.3....31}{2.3.2.4....2.32}=\frac{1.2.3....31}{2^{30}.\left(3.4....32\right)}=\frac{2}{2^{30}.32}=\frac{1}{2^{34}}=2^{-34}=2^n=>n=-34\)
2^n × 2³¹ = \(\dfrac{ }{ }\)2/4×4/6×6/8×...×62/64
2^n×2³¹=1/32=2^-5
2^n=2^-5 ÷ 2³¹=2^-36
=>n=-36