a) Tính giá trị biểu thức sau:
\(\dfrac{9^4.8^6}{6^{10}.16^3}\)
b) Tìm x:
1/ \(\left(\dfrac{4}{5}x-1\right).\left(6x+\dfrac{1}{2}\right)=0\)
2/
\(\left(2x+1\right)^2=25\)
3/
x:8,5=0,69:(-1,15)
4/
3(3x+5)-2(6-2x)=10
5/
\(\sqrt{x}=2,5\)
6/
\(3^{X-1}+5.3^{x-1}=162\)
7/
\(\dfrac{X-5}{1}+\dfrac{X-5}{2}+\dfrac{X-5}{3}+\dfrac{X-5}{4}=0\)
x-5/1+x-5/2+x-5/3+x-5/4=0
=>(x-5)(1+1/2+1/3+1/4)
vì 1+1/2+1/3+1/4 khác 0
=>x-5=0=>x=5
\(\dfrac{9^4.8^6}{6^{10}.16^3}=\dfrac{\left(3^2\right)^4.\left(2^3\right)^6}{3^{10}.2^{10}.\left(2^4\right)^3}=\dfrac{3^8.2^{18}}{3^{10}.2^{22}}=\dfrac{1}{9.4}=\dfrac{1}{36}\)
\(\left(\dfrac{4}{5}x-1\right).\left(6x+\dfrac{1}{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{4}{5}x-1=0\\6x+\dfrac{1}{2}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{4}{5}x=1\\6x=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{-1}{12}\end{matrix}\right.\)
vậy \(x\in\left\{\dfrac{-1}{12};\dfrac{5}{4}\right\}\)
\(\left(2x+1\right)^2=25\)
\(\left(2x+1\right)^2=\left(\pm5\right)^2\)
\(2x+1=\pm5\)
+)\(2x+1=5\Rightarrow2x=4\Rightarrow x=2\)
+)\(2x+1=-5\Rightarrow2x=-6\Rightarrow x=-3\)
vậy \(x\in\left\{-3;2\right\}\)
\(x:8,5=0,69:\left(-1,15\right)\)
\(\Rightarrow x:8,5=-0,6\)
\(\Rightarrow x=-0,6.8,5=-5,1=\dfrac{-51}{10}\)
vậy \(x=\dfrac{-51}{10}\)
Câu 6 : \(3^{x-1}+5.3^{x-1}=162\)
⇒\(1.3^{x-1}+5.3^{x-1}=162\)
⇒\(3^{x-1}.\left(5+1\right)=162\)
⇒\(3^{x-1}.6=162\)
⇒\(3^{x-1}=162:6\)
⇒\(3^{x-1}=27\)
⇒\(3^{x-1}=3^3\)
⇒x-1=3
⇒x=3+1
⇒x=4
Vậy x=4
Câu 5
\(\sqrt{x}=2,5\)
⇒\(\sqrt{x}=\sqrt{6,25}\)
⇒x=6,25
Vậy x=6,25
Câu 4
3(3x+5)-2(6-2x)=10
⇒9x+15-12+4x = 10
⇒(9x+4x)+(15-12) = 10
⇒13x+3=10
⇒13x=10-3
⇒13x=7
⇒x=\(\dfrac{7}{13}\)
Vậy x=\(\dfrac{7}{13}\)