(36-3x)mũ 2=81
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\(\left[-\frac{1}{3}\right]^3\cdot x=\frac{1}{81}\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{3}\right]^3\)
\(\Leftrightarrow x=\frac{1}{81}:\left[-\frac{1}{27}\right]\)
\(\Leftrightarrow x=\frac{1}{81}\cdot(-27)=-\frac{1}{3}\)
\(\left[x-\frac{1}{2}\right]^3=\frac{1}{27}\)
\(\Leftrightarrow\left[x-\frac{1}{2}\right]^3=\left[\frac{1}{3}\right]^3\)
=> Làm nốt
Mấy bài kia cũng làm tương tự
(- \(\dfrac{1}{3}\))3.\(x\) = \(\dfrac{1}{81}\)
\(x=\dfrac{1}{81}\) : (- \(\dfrac{1}{3}\))3
\(x\) = - (\(\dfrac{1}{3}\))4 :(\(\dfrac{1}{3}\))3
\(x=-\dfrac{1}{3}\)
Vậy \(x=-\dfrac{1}{3}\)
*) So sánh 354 và 281
Ta có:
354 = (32)27 = 927
281 = (23)27 = 827
Do 9 > 8 \(\Rightarrow\) 927 > 827
Vậy 354 > 281
*) So sánh 536 và 1124
Ta có:
536 = (53)12 = 12512
1124 = (112)12 = 12112
Do 125 > 121 nên 12512 > 12112
Vậy 536 > 1124
*) So sánh 7.213 và 216
Ta có:
216 = 23.213 = 8.213
Do 7 < 8 nên 7.213 < 8.213
Vậy 7.213 < 216
Tìm x hả bạn ?
a ) \(\left(3x+\frac{1}{4}\right)^3=-27\)
\(\left(3x+\frac{1}{4}\right)^3=\left(-3\right)^3\)
\(\Rightarrow3x+\frac{1}{4}=-3\)
\(\Rightarrow3x=-3-\frac{1}{4}=-\frac{13}{4}\)
\(\Rightarrow x=-\frac{13}{4}:3=-\frac{13}{12}\)
Vậy x = \(-\frac{13}{12}\)
cho em hoi câu này xin các anh chị:
10mux x+4y = 2013
27 mũ 11 và 81 mũ 8
625 mũ 5 và 125 mũ 7
5 mũ 36 và 11 mũ 24
5 mũ 23 và 6,5 mũ 22
7.2 mũ 13 và 2 mũ 16
a) \(81-\left(3x+2\right)^2=9^2-\left(3x+2\right)^2=\left(9-3x-2\right)\left(9+3x+2\right)=\left(7-3x\right)\left(11+3x\right)\)
b) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)
\(=\left(5x-5\right)\left(9x-3\right)=15\left(x-1\right)\left(3x-1\right)\)
c) \(9\left(x-5y\right)^2-16\left(x+y\right)^2=\left[3\left(x-5y\right)-4\left(x+y\right)\right]\left[3\left(x-5y\right)+4\left(x+y\right)\right]\)
\(=\left(-x-19y\right)\left(7x-11y\right)\)
1/ a) \(2.3.12.12.3=2.3.2^2.3.2^2.3.3=2^5.3^4\)
b) \(3.5.27.125=3.5.3^3.5^3=3^4.5^4=\left(3.5\right)^4\)
2/ a) \(\left(27^3\right)^4=27^{3.4}=27^{12}\)
Vậy \(\left(27^3\right)^4=27^{12}\)
b) \(5^{36}=\left(5^6\right)^6\) và \(11^{24}=\left(11^4\right)^6\)
Do đó \(5^6=15625\) và \(11^4=14641\)
Vì 15625>14641 nên\(\left(5^6\right)^6>\left(11^4\right)^6hay5^{36}>11^{24}.\)
3/ a) \(x^3=125=>x=5\)
b) \(\left(3x-14\right)^3=2^5.5^2+200\)
\(\left(3x-14\right)^3=1000\)
\(3x-14=10^3\)
\(3x=10^3+14\)
\(3x=1014\)
\(x=\frac{1014}{3}=338\)
c) \(\left(2x-1\right)^4=81\)
\(\left(2x-1\right)^4=3^4\)
\(2x-1=3\)
\(2x=3+1\)
\(x=\frac{4}{2}=2\)
d) \(5x+3^4=2^2.7^2\)
\(5x+3^4=\left(2.7\right)^2=14^2\)
\(5x+81=196\)
\(5x=196-81\)
\(5x=115\)
\(x=\frac{115}{5}=23\)
e) \(4^x=1024=>x=5\).
Bài làm :
\(a\text{)}\Leftrightarrow3^{3x+1}=3^4\Leftrightarrow3x+1=4\Leftrightarrow3x=3\Leftrightarrow x=1\)
\(b\text{)}\Leftrightarrow7x+1=15\Leftrightarrow7x=14\Leftrightarrow x=2\)
\(3^{3x+1}=81\)
\(3^{3x+1}=3^4\)
\(\Rightarrow3x+1=4\)
\(\Rightarrow3x=3\)
\(\Rightarrow x=1\)
\(\left(7x+1\right)^2+5=230\)
\(\left(7x+1\right)^2=230-5\)
\(\left(7x+1\right)^2=225\)
\(\left(7x+1\right)^2=15^2\)
\(\Rightarrow\orbr{\begin{cases}7x+1=15\\7x+1=-15\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{-16}{7}\end{cases}}\)
\(x^3-4x^2-9x+36=0\)
=> \(x^2\left(x-4\right)-9\left(x-4\right)=0\)
=> \(\left(x-4\right)\left(x^2-9\right)=0\)
=> \(\orbr{\begin{cases}x-4=0\\x^2-9=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=\pm3\end{cases}}\)
\(\left(x^2-9\right)^2-\left(x-3\right)^2=0\)
=> \(\left(x^2-9+x-3\right)\left[x^2-9-\left(x-3\right)\right]=0\)
=> \(\left(x^2+x-12\right)\left(x^2-9-x+3\right)=0\)
=> \(\left(x^2+x-12\right)\left(x^2-x-6\right)=0\)
=> \(\left(x^2-3x+4x-12\right)\left(x^2+2x-3x-6\right)=0\)
=> \(\left[x\left(x-3\right)+4\left(x-3\right)\right]\left[x\left(x+2\right)-3\left(x+2\right)\right]=0\)
=> \(\left(x-3\right)\left(x+4\right)\left(x-3\right)\left(x+2\right)=0\)
=> \(\left(x-3\right)^2\left(x+4\right)\left(x+2\right)=0\)
=> \(\hept{\begin{cases}\left(x-3\right)^2=0\\x+4=0\\x+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=3\\x=-4\\x=-2\end{cases}}\)
\(x^3-3x+2=0\)
=> \(x^3-x-2x+2=0\)
=> \(x^2\left(x-1\right)-2\left(x-1\right)=0\)
=> \(\left(x-1\right)\left(x^2-2\right)=0\)
=> x = 1
TL
x=9,
x=15
HT
\(\sqrt{(36-3x)2}\) = \(\sqrt{81}\) <=> |36-3x| = 9 (*)
-TH1: 36-3x > 0 <=> x<12
(*) <=> 36-3x = 9 <=> x=9 (nhận)
-TH2: 36-3x< 0 <=> x>12
(*) <=> 3x-36 = 9 <=> x=15 (nhận)
S={9;15}