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a: \(\Leftrightarrow4^x\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)=4^8\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)\)
=>4^x=4^8
=>x=8
b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)
=>2^x=2^11
=>x=11
c: =>1/6*6^x+6^x*36=6^15(1+6^3)
=>6^x=6*6^15
=>x=16
d: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)
=>x=9
Bài 2
\(a,\left(x-3\right)^2=9\Leftrightarrow\left(x-3\right)^2=3^2\Leftrightarrow x-3=3\Leftrightarrow x=6\)
\(b,\left(\frac{1}{2}+x\right)^2=16\Leftrightarrow\left(\frac{1}{2}+x\right)^2=4^2\Leftrightarrow\frac{1}{2}+x=4\Leftrightarrow x=\frac{7}{2}\)
a, ( x-1)3= -27
=> x - 1 = -3
=> x = -2
b, ( 2x - 1)2=25
=> 2x - 1 = 5 hoặc 2x - 1 = -5
=> 2x = 6 hoặc 2x = -4
=> x = 3 hoặc x = -2
c, ( x - 3/4)2= ( 1/2)6
=> (x - 3/4)^2 = 1/64
=> x - 3/4 = 1/8 hoặc x - 3/4 = -1/8
=> x = 7/8 hoặc x = 5/8
d, 2 x + 2 x +2 = 80
=> 2^x + 2^x.4 = 80
=> 2^x(1 + 4) = 80
=> 2^x.5 = 80
=> 2^x = 16
=> x = 4
e, 4x + 4 x + 3 = 4160
=> 4^x(1 + 64) = 4160
=> 4^x.65 = 4160
=> 4^x = 64
=> x = 3
\(a,\frac{(-10)^5}{3\cdot(-6)^4}=\frac{(-2\cdot5)^5}{3\cdot(-2\cdot3)^4}=\frac{(-2)^5\cdot5^5}{3\cdot(-2)^4\cdot3^4}=\frac{(-2)^5\cdot5^5}{(-2)^4\cdot3^5}=-2\cdot\frac{5^5}{3^5}=\frac{-6250}{243}\)
\(b,\frac{2^{15}\cdot9^4}{6^6\cdot8^3}=\frac{\left[2^3\right]^5\cdot\left[3^2\right]^4}{\left[3\cdot2\right]^6\cdot\left[2^3\right]^3}=\frac{2^{15}\cdot3^8}{3^6\cdot2^6\cdot2^9}=\frac{2^{15}\cdot3^8}{3^6\cdot2^{15}}=\frac{3^8}{3^6}=3^2=9\)
\(c,\left[1+\frac{2}{3}-\frac{1}{4}\right]\cdot\left[\frac{4}{5}-\frac{3}{4}\right]^2\)
\(=\left[\frac{12}{12}+\frac{8}{12}-\frac{3}{12}\right]\cdot\left[\frac{16}{20}-\frac{15}{20}\right]^2\)
\(=\frac{17}{12}\cdot\left[\frac{1}{20}\right]^2=\frac{17}{12}\cdot\frac{1^2}{20^2}=\frac{17}{12}\cdot\frac{1}{400}=\frac{17}{4800}\)
\(d,2^3+3\cdot\left[\frac{1}{2}\right]^0+\left[(-2)^2:\frac{1}{2}\right]\)
\(=8+3\cdot\frac{1^0}{2^0}+\left[4:\frac{1}{2}\right]\)
\(=8+3\cdot1+8=8+3+8=19\)
a) 814=(23)14=23*14=242
1610=(8*2)10=810*210=(23)10*210=230*210=240
Vì 242 > 240 nên 814 > 1610
b) 233=(23)11=811
322=(32)11=911
Vì 811 < 911 nên 233 < 322
Ta có : 5(x - 2)(x + 3) = 1
=> (5x - 10)(x + 3) = 1
=> 5x2 - 10x + 15x - 30 = 1
=> 5x2 - 5x - 30 = 1
=> 5x(x - 1) = 31
=> x(x - 1) = 31/5 (chịu)
1) Tính
a) 253 : 52 = (52)3 : 52 = 56 : 52 = 54 = 625
\(b)\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^6=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^6=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^9\) d) 9 . 32 . \(\dfrac{1}{81}\) . 32 = 32 . 32 . \(\dfrac{1}{3^4}\) . 32 = 9
2) Tìm x thuộc Q, biết:
a) 3x + 2 = 27
=> 3x + 2 = 33
x + 2 = 3
x = 3 - 2
x = 1
b) \(\left(\dfrac{1}{2}x-3\right)^4=81\)
\(\Rightarrow\left(\dfrac{1}{2}x-3\right)^4=3^4\)
\(\dfrac{1}{2}x-3=3^{ }\)
\(\dfrac{1}{2}x=3+3\)
\(\dfrac{1}{2}x=9\)
\(x=9:\dfrac{1}{2}\)
\(x=18\)
c) \(\left(x-\dfrac{1}{2}\right)^3=-27\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(-3\right)^3\)
\(x-\dfrac{1}{2}=-3\)
\(x=-3+\dfrac{1}{2}\)
\(x=\dfrac{-5}{2}\)
d) 5 . 5x + 1 = 125
5x + 1 = 125 : 5
5x + 1 = 25
5x + 1 = 52
x + 1 = 2
x = 2 - 1
x = 1.
a: \(\dfrac{\left(-1\right)^2}{2^2}=\dfrac{1}{4};\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)
Do đó: \(\dfrac{\left(-1\right)^2}{2^2}=\left(-\dfrac{1}{2}\right)^2\)
b: \(\dfrac{3^3}{5^3}=\left(\dfrac{3}{5}\right)^3< \dfrac{3}{5}\)(do \(0< \dfrac{3}{5}< 1\))
d: \(\left(\dfrac{3}{4}\right)^7:\left(\dfrac{3}{4}\right)^3=\left(\dfrac{3}{4}\right)^4\)
Vì \(0< \dfrac{3}{4}< 1\)
nên \(\left(\dfrac{3}{4}\right)^4< \left(\dfrac{3}{4}\right)^2\)
=>\(\left(\dfrac{3}{4}\right)^7:\left(\dfrac{3}{4}\right)^3< \left(\dfrac{3}{4}\right)^2\)
e: \(\left(0,5\right)^6:\left(0,5\right)^2=\left(0,5\right)^{6-2}=\left(0,5\right)^4=\left(0,5\right)^{2\cdot2}=\left[\left(0,5\right)^2\right]^2\)
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