A) (x+2/3)3=1/8
B) 52x-1-125=0
C) 81/32x+1=3
Em cần gấp chiều nộp rồi ạ
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a). 5x = 125
=>5x = 53
=> x = 3
b) 32x = 81
=> 32x = 34
=> 2x = 4
=> x = 2
c). 52x-3 – 2.52 = 52 .3
=>52x: 53 = 52 .3 + 2.52
=>52x: 53 = 52 .5
=>52x = 52 .5.53
=>52x = 56
=> 2x = 6
=> x=3
a) 5x = 125
x = 125 : 5
x = 25
b) 32x = 81
x = 81 : 32
x = 81/ 32
c) 52x - 3 - 2.52 = 52.3
52x - 3 - 104 = 156
x - 3 - 104 = 156 : 52
x - 3 - 104 = 3
x - 3 = 3 + 104
x - 3 = 107
x = 107 + 3
x = 110
Ko chắc lắm , sai thì Sorry nhé .
b) 32x = 81
32x = 34
=> 2x = 4
=> x = 2
a). 52x-3 – 2.52 = 52 .3
52x: 53 = 52 .3 + 2.52
52x: 53 = 52 .5
52x = 52 .5.53
52x = 56
=> 2x = 6
=> x=3
\(5.5^{2x+2}=125\)
\(=>5^{2x+2}=125:5\)
\(=>5^{2x+2}=25=5^2\)
\(=>2x+2=2\)
\(=>2x=2-2\)
\(=>2x=0\)
\(=>x=0:2\)
\(=>x=0\)
Bài 2:
a: =>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
`a)|2x+1|=5`
`<=>` \(\left[ \begin{array}{l}2x+1=5\\2x+1=-5\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-6\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\)
`b)|2x+1|=0`
`<=>2x+1=0`
`<=>2x=-1`
`<=>x=-1/2`
`c)|2x+1|=7`
`<=>` \(\left[ \begin{array}{l}2x+1=7\\2x+1=-7\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=6\\2x=-8\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=4\\x=-4\end{array} \right.\)
`d)|2x+5|=|3x-7|`
`<=>` \(\left[ \begin{array}{l}2x+5=3x-7\\2x+5=7-3x\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=12\\5x=2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=12\\x=\dfrac25\end{array} \right.\)
`e)|2x+7|=1`
`<=>` \(\left[ \begin{array}{l}2x+7=1\\2x+7=-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=-6\\2x=-8\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=3\\x=-4\end{array} \right.\)
`g)|x-2|+|2x-3|=2`
Nếu `x>=2=>|x-2|=x-2,|2x-3|=2x-3`
`pt<=>x-2+2x-3=2`
`<=>3x-5=2`
`<=>3x=7`
`<=>x=7/3(tm)`
Nếu `x<=3/2=>|x-2|=2-x,|2x-3|=3-2x`
`pt<=>2-x+3-2x=2`
`<=>5-3x=2`
`<=>3x=3`
`<=>x=1(tm)`
Nếu `3/2<=x<=2=>|x-2|=2-x,|2x-3|=2x-3`
`pt<=>2-x+2x-3=2`
`<=>x-1=2`
`<=>x=3(l)`
`h)|x+2|+|1-x|=3x+2`
Vì `VT>=0=>3x+2>=0=>x>=-2/3`
`=>|x+2|=x+2`
`pt<=>x+2+|1-x|=3x+2`
`<=>|1-x|=2x(x>=0)`
`<=>` \(\left[ \begin{array}{l}2x=1-x\\2x=x-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}3x=1\\x=-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac13(TM)\\x=-1(KTM)\end{array} \right.\)
a.
$|2x+1|=5$
\(\Leftrightarrow \left[\begin{matrix}
2x+1=5\\
2x+1=-5\end{matrix}\right.\Leftrightarrow \left[\begin{matrix}
x=2\\
x=-3\end{matrix}\right.\)
b.
$|2x+1|=0$
$\Leftrightarrow 2x+1=0$
$\Leftrightarrow x=-\frac{1}{2}$
c.
$|2x+1|=7$
\(\Leftrightarrow \left[\begin{matrix} 2x+1=7\\ 2x+1=-7\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=3\\ x=-4\end{matrix}\right.\)
a) \(8x+56:14=60\)
\(\Rightarrow8x+4=60\)
\(\Rightarrow8x=56\)
\(\Rightarrow x=\dfrac{56}{8}\)
\(\Rightarrow x=7\)
b) Mình làm rồi nhé !
c) \(41-2^{x+1}=9\)
\(\Rightarrow2^{x+1}=41-9\)
\(\Rightarrow2^{x+1}=32\)
\(\Rightarrow2^{x+1}=2^5\)
\(\Rightarrow x+1=5\)
\(\Rightarrow x=4\)
d) \(3^{2x-4}-x^0=8\)
\(\Rightarrow3^{2x-4}-1=8\)
\(\Rightarrow3^{2x-4}=9\)
\(\Rightarrow3^{2x-4}=3^2\)
\(\Rightarrow2x-4=2\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
g) \(65-4^{x+2}=2014^0\)
\(\Rightarrow65-4^{x+2}=1\)
\(\Rightarrow4^{x+2}=64\)
\(\Rightarrow4^{x+2}=4^3\)
\(\Rightarrow x+2=3\)
\(\Rightarrow x=1\)
i) \(120+2\left(4x-17\right)=214\)
\(\Rightarrow2\left(4x-17\right)=214-120\)
\(\Rightarrow2\left(4x-17\right)=94\)
\(\Rightarrow4x-17=47\)
\(\Rightarrow4x=47+17\)
\(\Rightarrow4x=64\)
\(\Rightarrow x=16\)
a: \(8x+56:14=60\)
=>8x+4=60
=>8x=60-4=56
=>x=56/8=7
b: \(5^{2x-3}-2\cdot5^2=5^2\cdot3\)
=>\(5^{2x-3}=5^2\cdot3+2\cdot5^2=5^3\)
=>2x-3=3
=>2x=6
=>x=3
c: \(41-2^{x+1}=9\)
=>\(2^{x+1}=41-9=32\)
=>x+1=5
=>x=4
d: \(3^{2x-4}-x^0=8\)
=>\(3^{2x-4}-1=8\)
=>\(3^{2x-4}=8+1=9\)
=>2x-4=2
=>2x=6
=>x=3
g: \(65-4^{x+2}=2014^0\)
=>\(65-4^{x+2}=1\)
=>\(4^{x+2}=65-1=64\)
=>x+2=3
=>x=1
i: 120+2(4x-17)=214
=>2(4x-17)=214-120=94
=>4x-17=94/2=47
=>4x=64
=>\(x=\dfrac{64}{4}=16\)
ap dung bdt am gm
\(\sqrt{1+8a^3}=\sqrt{\left(1+2a\right)\left(4a^2-4a+1\right)}\)\(\le\frac{1+2a+4a^2-2a+1}{2}=\frac{4a^2+2}{2}=2a^2+1\)
\(\Rightarrow\frac{1}{\sqrt{1+8a^3}}\ge\frac{1}{2a^2+1}\)
tuongtu ta cung co \(\frac{1}{\sqrt{1+8b^3}}\ge\frac{1}{2b^2+1};\frac{1}{\sqrt{1+8c^3}}\ge\frac{1}{2c^2+1}\)
\(\Rightarrow\)VT\(\ge\frac{1}{2a^2+1}+\frac{1}{2b^2+1}+\frac{1}{2c^2+1}\)
tiep tuc ap dung bat cauchy-schwarz dang engel ta co
\(VT\ge\frac{1}{2a^2+1}+\frac{1}{2b^2+1}+\frac{1}{2c^2+1}\ge\frac{\left(1+1+1\right)^2}{2\left(a^2+b^2+c^2\right)+3}=\frac{3^2}{6+3}=1\)(dpcm)
dau = xay ra \(\Leftrightarrow a=b=c=1\)
a) \(\left(x+\frac{2}{3}\right)^3=\frac{1}{8}\)
\(\Rightarrow x+\frac{2}{3}=\frac{1}{2}\)
\(x=\frac{1}{2}-\frac{2}{3}\)
\(x=\frac{-1}{6}\)
b) 52x-1-125 = 0
52x-1 = 0+125
52x-1 = 125
<=> 52x-1 = 53
=> 2x-1=3
=> x = 2
c) \(\frac{8^1}{3^{2x+1}}=3\)
\(\Rightarrow8=3.3^{2x+1}=3^{2x+1+1}=3^{2x+2}\)
\(\Rightarrow8\ne3^{2x+2}\)
=> x vô nghiệm
a, \(\left(\frac{x+2}{3}\right)^3=\frac{1}{8}\)\(\Rightarrow\left(\frac{x+2}{3}\right)^3=\left(\frac{1}{2}\right)^3\)
\(\Rightarrow\frac{x+2}{3}=\frac{1}{2}\)\(\Rightarrow\left(x+2\right).2=3.1\)\(\Rightarrow x+2=\frac{3}{2}\)\(\Rightarrow x=-\frac{1}{2}\)