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h) \(5^x+5^{x+2}=650\)
\(\Leftrightarrow5^x+5^x.5^2=650\)
\(\Leftrightarrow5^x\left(1+25\right)=650\)
\(\Leftrightarrow5^x.26=650\)
\(\Leftrightarrow5^x=25\)
\(\Leftrightarrow x=2\)
haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=
bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây
a) ( x + 5 )3 = -64
x + 5 = - 4
x = - 4 - 5
x = -9
b) (2x - 3)2=9
2x - 3 = 3
2x = 3+3
2x = 6
x = 6 : 2
x = 3
e) \(\dfrac{8}{2x}=4\)
=> 4 . 2x = 8
8x =8
x = 8 : 8
x = 1
g) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\)
\(\left(\dfrac{1}{2}\right)^{2x}:\left(\dfrac{1}{2}\right)^1=\dfrac{1}{8}\)
\(\left(\dfrac{1}{2}\right)^{2x}:\dfrac{1}{2}=\dfrac{1}{8}\)
\(\left(\dfrac{1}{2}\right)^{2x}=\dfrac{1}{8}.\dfrac{1}{2}\)
\(\left(\dfrac{1}{2}\right)^{2x}=\dfrac{1}{16}\)
\(\left(\dfrac{1}{2}\right)^{2x}=\left(\dfrac{1}{2}\right)^{2.2}\)
=> x = 2
h) \(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\)
\(\dfrac{1}{4}.x=\dfrac{1}{32}\)
x = \(\dfrac{1}{32}:\dfrac{1}{4}\)
x = \(\dfrac{1}{8}\)
i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\)
\(x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)\)
\(x=\dfrac{-1}{27}\)
a) (x + 5)3 = -64
=> (x + 5)3 = (-4)3
x + 5 = -4
x = -4 - 5
x = -9
b) (2x - 3)2 = 9
=> (2x - 3)2 = (\(\pm\)3)2
=> 2x - 3 = 3 hoặc 2x - 3 = -3
*2x - 3 = 3
2x = 3 + 3
2x = 9
x = \(\dfrac{9}{2}\)
*2x - 3 = -3
2x = -3 + 3
2x = 0
x = 0 : 2
x = 0
Vậy x \(\in\left\{\dfrac{9}{2};0\right\}\)
c) \(\dfrac{x}{\dfrac{4}{2}}=\dfrac{4}{\dfrac{x}{2}}\)
=> \(x.\dfrac{x}{2}=4.\dfrac{4}{2}\)
\(\dfrac{x}{2}=8\)
x = 8 : 2
x = 4
d) \(\dfrac{-32}{\left(-2\right)^n}=4\)
\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)
=> (-2)n . (-2)2= (-2)5
(-2)n = (-2)5 : (-2)2
(-2)n = (-2)3
Vậy n = 3
e) \(\dfrac{8}{2x}=4\)
=> 2x . 4 = 8
2x = 8 : 4
2x = 2
x = 1
g) \(\left(\dfrac{1}{2}\right)^{2x-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^3\)
2x - 1 = 3
2x = 3 + 1
2x = 4
x = 4 : 2
x = 2
h) \(\left(\dfrac{1}{2}\right)^2.x=\left(\dfrac{1}{2}\right)^5\)
\(x=\left(\dfrac{1}{2}\right)^5:\left(\dfrac{1}{2}\right)^2\)
\(x=\left(\dfrac{1}{2}\right)^3\)
\(x=\dfrac{1}{8}\)
i) \(\left(\dfrac{-1}{3}\right)x=\dfrac{1}{81}\)
\(x=\dfrac{1}{81}:\left(\dfrac{-1}{3}\right)\)
\(x=\left(\dfrac{-1}{3}\right)^4:\left(\dfrac{-1}{3}\right)\)
\(x=\left(\dfrac{-1}{3}\right)^3\)
\(x=\dfrac{-1}{27}\).
\(a,x^2=16\)
\(x^2=4^2=\left(-4\right)^2\)
\(x=2\) hoặc \(x=-2\)
\(b,x^3=-8\)
\(x^3=\left(-2\right)^3\)
\(x=-2\)
\(c,\left(x+2\right)^2=4\)
\(\left(x+2\right)^2=2^2=\left(-2\right)^2\)
\(x+2=2\Rightarrow x=0\) hoặc \(x+2=-2\Rightarrow x=-4\)
\(d,\left(1-x\right)^3=1\)
\(1-x=1\)
\(x=0\)
e,phần này mk chưa nghĩ ra,sorry bn nha!
a: \(\Leftrightarrow2x-3=x\)
=>x=3
b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+\dfrac{5}{4}\cdot2^x=\dfrac{7}{32}\)
=>2^x=1/8
=>x=-3
c: =>2x+7=-4
=>2x=-11
=>x=-11/2
d: =>(4x-3)^2*(4x-4)(4x-2)=0
hay \(x\in\left\{\dfrac{3}{4};1;\dfrac{1}{2}\right\}\)
a/ \(4\dfrac{1}{3}:\dfrac{x}{4}=6:0,3\)
\(\Leftrightarrow\dfrac{13}{3}:\dfrac{x}{4}=20\)
\(\Leftrightarrow\dfrac{52}{3x}=20\)
\(\Leftrightarrow x=\dfrac{13}{15}\)
Vậy..
b/ \(\left(x-1\right)^5=-32\)
\(\Leftrightarrow\left(x-1\right)^5=\left(-2\right)^5\)
\(\Leftrightarrow x-1=-2\)
\(\Leftrightarrow x=-1\)
Vậy..
c/ \(\left(2^3:4\right).2^{x+1}=64\)
\(\Leftrightarrow2.2^{x+1}=64\)
\(\Leftrightarrow2^{x+2}=2^6\)
\(\Leftrightarrow x+2=6\)
\(\Leftrightarrow x=4\)
Vậy..
d/ \(\left|3-2x\right|-3=-3\)
\(\Leftrightarrow\left|3-2x\right|=0\)
\(\Leftrightarrow3-2x=0\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy..
e/ \(\left|x+\dfrac{4}{5}\right|-\dfrac{1}{7}=0\)
\(\Leftrightarrow\left|x+\dfrac{4}{5}\right|=\dfrac{1}{7}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{1}{7}\\x+\dfrac{4}{5}=-\dfrac{1}{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{23}{35}\\x=-\dfrac{33}{35}\end{matrix}\right.\)
Vậy..
b: =>(3x-1)(3x+1)(2x+3)=0
hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};-\dfrac{3}{2}\right\}\)
c: \(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{19}{12}\)
=>2x-1/3=19/12 hoặc 2x-1/3=-19/12
=>2x=23/12 hoặc 2x=-15/12=-5/4
=>x=23/24 hoặc x=-5/8
d: \(\Leftrightarrow-\dfrac{5}{6}\cdot x+\dfrac{3}{4}=-\dfrac{3}{4}\)
=>-5/6x=-3/2
=>x=3/2:5/6=3/2*6/5=18/10=9/5
e: =>2/5x-1/2=3/4 hoặc 2/5x-1/2=-3/4
=>2/5x=5/4 hoặc 2/5x=-1/4
=>x=5/4:2/5=25/8 hoặc x=-1/4:2/5=-1/4*5/2=-5/8
f: =>14x-21=9x+6
=>5x=27
=>x=27/5
h: =>(2/3)^2x+1=(2/3)^27
=>2x+1=27
=>x=13
i: =>5^3x*(2+5^2)=3375
=>5^3x=125
=>3x=3
=>x=1
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
c. \(^{ }\left(2x+3\right)^2=\dfrac{9}{121}\)
=> \(\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)
=> 2x +3 = \(\dfrac{3}{11}\) hoặc 2x+3 = \(\dfrac{-3}{11}\)
=> x= \(\dfrac{-15}{11}\) hoặc x = \(\dfrac{-18}{11}\)
d. \(\left(2x-1\right)^3=\dfrac{-8}{27}\)
=> \(\left(2x-1\right)^3=\left(\dfrac{-2}{3}\right)^3\)
=> 2x-1 = \(\dfrac{-2}{3}\)
=> x= \(\dfrac{1}{6}\)
a)\(16^x=32^8\)
\(\Rightarrow\left(2^4\right)^x=\left(2^5\right)^8\)
\(\Rightarrow2^{4x}=2^{40}\)
\(\Rightarrow4x=40\)
\(\Rightarrow x=10\)
b)\(4^x=32^{40}\)
\(\Rightarrow\left(2^2\right)^x=\left(2^5\right)^{40}\)
\(\Rightarrow2^{2x}=2^{200}\)
\(\Rightarrow2x=200\)
\(\Rightarrow x=100\)
c)\(\left(\dfrac{2}{3}\right)^x=\left(\dfrac{4}{9}\right)^4\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left[\left(\dfrac{2}{3}\right)^2\right]^4\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^8\)
\(\Rightarrow x=8\)
d)\(2^{3x+1}=32^2\)
\(\Rightarrow2^{3x+1}=\left(2^5\right)^2=2^{10}\)
\(\Rightarrow3x+1=10\)
\(\Rightarrow3x=9\)
\(\Rightarrow x=3\)
e)\(\left(2x-1\right)^3:7=49\)
\(\Rightarrow\left(2x-1\right)^3=343\)
\(\Rightarrow\left(2x-1\right)^3=7^3\)
\(\Rightarrow2x-1=7\)
\(\Rightarrow2x=8\)
\(\Rightarrow x=4\)
a) Ta có: \(16^x=32^8\)
=> \(\left(2^4\right)^x=\left(2^5\right)^8\)
=> \(2^{4.x}=2^{5.8}\)
=> 4x = 40
=> x = 10
Vậy x =10
b) Ta có : \(4^x=32^{40}\)
=> \(\left(2^2\right)^x=\left(2^5\right)^{40}\)
=> \(2^{2x}=2^{5.40}\)
=> 2x = 200
=> x =100
Vậy x = 100
c) Ta có : \(\left(\dfrac{2}{3}\right)^x=\left(\dfrac{4}{9}\right)^4\)
=> \(\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^{2.4}\)
=> x = 8
Vậy x =8
d) Ta có : \(2^{3x+1}=32^2\)
=> \(2^{3x+1}=\left(2^5\right)^2\)
=> 3x+1 =5.2
=> 3x+1 = 10
=> 3x = 10-1=9
=> x= \(\dfrac{9}{3}\)=3
Vậy x = 3
e) (2x-1)\(^3\) : 7 = 49
(2x-1)\(^3\) = 49.7
(2x-1)\(^3\) = 343
(2x-1)\(^3\) = \(7^3\)
=> 2x-1 = 7
2x = 8
x = 8:2
x = 4
Vậy x = 4