\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)

\(ĐK:x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)

\(\Leftrightarrow4x^2-9=4x+12\)

\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)

\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)

\(ĐK:x\ge5\)

\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)

ĐK:x>=1

\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)

\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)

\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)

\(ĐK:x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)

\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}=0\)    (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)

1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)

=>4x=18

hay x=9/2

2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)

=>4x=108

hay x=27

3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)

\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)

=>4x=12

hay x=3

1:

a: =7/5(40+1/4-25-1/4)-1/2021

=21-1/2021=42440/2021

b: =5/9*9-1*16/25=5-16/25=109/25

a, \(\left(\dfrac{5}{6}:x-1\dfrac{1}{4}\right)^4=\dfrac{81}{16}\)

\(\Leftrightarrow\left(\dfrac{5}{6}:x-1\dfrac{1}{4}\right)^4=\left(\dfrac{3}{2}\right)^4\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{6}:x-\dfrac{5}{4}=\dfrac{3}{2}\\\dfrac{5}{6}:x-\dfrac{5}{4}=\dfrac{-3}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{6}:x=\dfrac{11}{4}\\\dfrac{5}{6}:x=\dfrac{7}{12}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{35}{66}\\x=\dfrac{10}{7}\end{matrix}\right.\)

Vậy ...

b, \(\left|\dfrac{2}{5}x-0,2\right|.1\dfrac{1}{4}-2=1,5\)

\(\Leftrightarrow\left|\dfrac{2}{5}x-0,2\right|.\dfrac{5}{4}=\dfrac{3}{5}\)

\(\Leftrightarrow\left|\dfrac{2}{5}x-0,2\right|=\dfrac{12}{25}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}x-0,2=\dfrac{12}{25}\\\dfrac{2}{5}x-0,2=\dfrac{-12}{25}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}x=\dfrac{17}{25}\\\dfrac{2}{5}x=\dfrac{-7}{25}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{10}\\x=\dfrac{-7}{10}\end{matrix}\right.\)

Vậy ...

\(\left(\dfrac{5}{6}:x-1\dfrac{1}{4}\right)^4=\dfrac{81}{16}\)

\(\Rightarrow\left(\dfrac{5}{6}:x-1\dfrac{1}{4}\right)^4=\dfrac{3}{2}^4\)

\(\Rightarrow\dfrac{5}{6}:x-\dfrac{5}{4}=\dfrac{3}{2}\)

Ở đây tự làm không muốn viết mỏi tay.

a. x=5/6

\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)

\(a,\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}-\dfrac{2}{3}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{1}{3}\right)^3\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\)

\(x=\dfrac{1}{2}+\dfrac{1}{3}\)

\(x=\dfrac{1}{5}\)

\(a)\left(\dfrac{-2}{3}\right)^2.x=\left(\dfrac{-2}{3}\right)^5\)

\(\Rightarrow x=\left(\dfrac{-2}{3}\right)^5:\left(\dfrac{-2}{3}\right)^2\)

\(\Rightarrow x=\left(\dfrac{-2}{3}\right)^3\)

Vậy ...............

\(b)\left(-\dfrac{1}{3}\right)^3.x=\dfrac{1}{81}\)

\(\Rightarrow\dfrac{-1}{27}.x=\dfrac{1}{81}\)

\(\Rightarrow x=\dfrac{1}{81}:\dfrac{-1}{27}\)

\(\Rightarrow x=\dfrac{-1}{3}\)

Vậy ..................

Chúc bạn học tốt!

a: \(\Leftrightarrow\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{-3}{2}x+\dfrac{3}{4}\)

=>13/6x=3/4+5/6

=>13/6x=9/12+10/12=19/12

hay x=19/26

b: \(\left(5x-3\right)\left(2x+5\right)=0\)

=>5x-3=0 hoặc 2x+5=0

=>x=3/5 hoặc x=-5/2

c: \(\left(\dfrac{5}{6}:x-\dfrac{5}{4}\right)^4=\dfrac{81}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{6}:x-\dfrac{5}{4}=\dfrac{3}{2}\\\dfrac{5}{6}:x-\dfrac{5}{4}=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{6}:x=\dfrac{11}{4}\\\dfrac{5}{6}:x=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{33}\\x=-\dfrac{10}{3}\end{matrix}\right.\)

d: \(\left|\dfrac{2}{5}x-\dfrac{1}{5}\right|\cdot\dfrac{5}{4}-2=\dfrac{3}{2}\)

\(\Leftrightarrow\left|\dfrac{2}{5}x-\dfrac{1}{5}\right|\cdot\dfrac{5}{4}=\dfrac{3}{2}+2=\dfrac{7}{2}\)

\(\Leftrightarrow\left|\dfrac{2}{5}x-\dfrac{1}{5}\right|=\dfrac{7}{2}:\dfrac{5}{4}=\dfrac{7}{2}\cdot\dfrac{4}{5}=\dfrac{28}{10}=\dfrac{14}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}x-\dfrac{1}{5}=\dfrac{14}{5}\\\dfrac{2}{5}x-\dfrac{1}{5}=-\dfrac{14}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{2}{5}=\dfrac{15}{2}\\x=-\dfrac{13}{5}:\dfrac{2}{5}=\dfrac{-13}{2}\end{matrix}\right.\)