tĩm x
5 phần 9 - x tất cả mũ hai = 16/81
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\(\left(\dfrac{9}{4}\right)^{x-6}=\left(\dfrac{27}{8}\right)^{x-2}\)
\(\Leftrightarrow\left[\left(\dfrac{3}{2}\right)^2\right]^{x-6}=\left[\left(\dfrac{3}{2}\right)^3\right]^{x-2}\)
\(\Leftrightarrow\left(\dfrac{3}{2}\right)^{2x-12}=\left(\dfrac{3}{2}\right)^{3x-6}\)
\(\Leftrightarrow2x-12=3x-6\)
\(\Leftrightarrow-12+6=3x-2x\)
\(\Leftrightarrow x=-6\)
Vậy ............
P/S : Lần sau gỗ bằng công thức toán nhs bn :)
a.(3^2+4^2).x=10^2
(9+16).x =100
25.x =100
x =100:25
x =4
b.(x-5)^2 =81
x-5 =9
x =9+5
x =14
c.(2x+1)^3 = 343
2x+1 = 7
2x =7-1
2x =6
x =6:2
x = 3
Ta có :
\(\frac{1}{243^9}=\frac{1}{\left(81.3\right)^9}=\frac{1}{81^9.27^3}>\frac{1}{81^9.81^3}=\frac{1}{81^{11}}>\frac{1}{8^{12}}>\frac{1}{8^{13}}\)
\(\Rightarrow\frac{1}{243^9}>\frac{1}{83^{13}}\)
mình chắc chắn luôn
a, \(\left(x+\frac{4}{3}y^2\right)^2\)
\(=x^2+\frac{8}{3}xy^2+\frac{16}{9}y^4\)
b, \(\left(2x-3y\right)^2\)
\(=4x^2-12xy+9y^2\)
c, \(\left(x^2+2x\right)\left(2x-x^2\right)\)
\(=\left(2x+x^2\right)\left(2x-x^2\right)\)
\(=4x^2-x^4\)
d, \(\left(x+\frac{1}{2}\right)^3\)
\(=x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\)
e, \(\left(2x-\frac{6}{5}y\right)^3\)
\(=8x^3-\frac{72}{5}x^2y+\frac{216}{25}xy^2-\frac{216}{125}y^3\)
Bài 6 :
a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)
c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)
d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)
Bài 7 :
a) \(3^x+3^{x+2}=9^{17}+27^{12}\)
\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)
\(\Rightarrow10.3^x=3^{34}+3^{36}\)
\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)
\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)
b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)
\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)
\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)
c) Bài C bạn xem lại đề
d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)
\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)
\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)
\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)
\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)
1. \(2^x-26=6\)
\(\Rightarrow2^x=6+26\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
2. \(64\cdot4^x=16^8\)
\(\Rightarrow4^3\cdot4^x=4^{16}\)
\(\Rightarrow4^x=4^{16}:4^3\)
\(\Rightarrow4^x=4^{13}\)
\(\Rightarrow x=13\)
3. \(\left(2x-1\right)^4=16\)
\(\Rightarrow\left(2x-1\right)^4=2^4\)
\(\Rightarrow2x-1=2\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\)
4. \(\left(2x+1\right)^3=125\)
\(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
\(\left(\frac{5}{9}-x\right)^2=\frac{16}{81}\)
=> \(\left(\frac{5}{9}-x\right)^2=\left(\frac{4}{9}\right)^2\)
=> \(\orbr{\begin{cases}\frac{5}{9}-x=\frac{4}{9}\\\frac{5}{9}-x=-\frac{4}{9}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{9}-\frac{4}{9}=\frac{1}{9}\\x=\frac{5}{9}-\left(-\frac{4}{9}\right)=\frac{5}{9}+\frac{4}{9}=1\end{cases}}\)
Vậy x = 1/9 hoặc x = 1