\(\sqrt{25x-25}-\dfrac{15}{2}\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}\left(x\ge1\right)\)

\(< =>5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{\sqrt{x-1}}{3}=6+\sqrt{x-1}\)

\(< =>30\sqrt{x-1}-15\sqrt{x-1}=36+6\sqrt{x-1}\)

\(< =>9\sqrt{x-1}=36\\ < =>\sqrt{x-1}=4\\ < =>x-1=16\\ < =>x=17\left(tm\right)\)

`(3x-12)xx6=72`

`3x-12=72:6`

`3x-12=12`

`3x=12+12`

`3x=24`

`x=8`

cơ `-0,01=-1/100`

có `-1/100=-5/500`

có `211<500`

`=>5/221>5/500`

`=>-5/221<-5/500`

`=>-5/221<-0,01`

`x:3=1 1/2`

`x:3=(1xx2+1)/2`

`x:3=3/2`

`x=3/2xx3`

`x=9/2`

`1/(3xx4)+1/(4xx5)+1/(5xx6)+...+1/(38xx39)`

`=1/3-1/4+1/4-1/5+1/5-1/6+...+1/38-1/39`

`=1/3-1/39`

`=4/13`

\(-3+6=-3\) =))

`(2x-1/3)^2=4`

\(< =>\left[{}\begin{matrix}2x-\dfrac{1}{3}=2\\2x-\dfrac{1}{3}=-2\end{matrix}\right.\\ < =>\left[{}\begin{matrix}2x=2+\dfrac{1}{3}\\2x=-2+\dfrac{1}{3}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}2x=\dfrac{7}{3}\\2x=-\dfrac{5}{3}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{7}{3}:2\\x=-\dfrac{5}{3}:2\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{7}{3}\cdot\dfrac{1}{2}\\x=-\dfrac{5}{3}\cdot\dfrac{1}{2}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\)

`7(x-1/2)^2=9`

`(x-1/2)^2=9/7`

\(=>\left[{}\begin{matrix}x-\dfrac{1}{2}=\sqrt{\dfrac{9}{7}}\\x-\dfrac{1}{2}=-\sqrt{\dfrac{9}{7}}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{3}{\sqrt{7}}+\dfrac{1}{2}\\x=-\dfrac{3}{\sqrt{7}}+\dfrac{1}{2}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{6+\sqrt{7}}{2\sqrt{7}}\\x=\dfrac{-6+\sqrt{7}}{2\sqrt{7}}\end{matrix}\right.\)

a)

dãy số trên có số số hạng là

`(61-5):4+1=15` (số hạng)

b)

gọi số thứ 45 là: x

`=>5;9;13;...;x` có 45 số hạng

số thứ 45 là

`(x-5):4+1=45`

`=>(x-5):4=45-1`

`=>(x-5):4=44`

`=>x-5=44xx4`

`=>x-5=176`

`=>x=176+5`

`=>x=181`

Vậy số thứ 45 là 181

`2^x :4=8`

`2^x =8xx4`

`2^x =32`

`2^x =2^5`

`=>x=5`