\(\dfrac{\dfrac{4}{5}:\left(\dfrac{4}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{27}{25}-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(\dfrac{59}{9}-\dfrac{13}{4}\right)\cdot\dfrac{36}{17}}+\left(\dfrac{6}{5}\cdot\dfrac{1}{2}\right):\dfrac{4}{5}\)

\(=\dfrac{4}{5}:\dfrac{3}{5}+\dfrac{7}{4}:7+\dfrac{3}{5}:\dfrac{4}{5}\)

\(=\dfrac{4}{3}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\dfrac{7}{3}\)

1: \(\left(3x-\dfrac{1}{5}\right)^2=\left(-\dfrac{3}{25}\right)^2\)

=>3x-1/5=3/25 hoặc 3x-1/5=-3/25

=>3x=8/25 hoặc 3x=2/25

=>x=8/75 hoặc x=2/75

2: \(\left(2x-\dfrac{1}{3}\right)^2=\left(-\dfrac{2}{9}\right)^2\)

=>2x-1/3=2/9 hoặc 2x-1/3=-2/9

=>2x=5/9 hoặc 2x=1/9

=>x=5/18 hoặc x=1/18

log\(_5\)(\(\dfrac{1}{25}=log_5\left(5^{-2}\right)=-2\)

log\(_{27}9\)=log\(_{3^3}3^2\)=\(\dfrac{2}{3}\)

\(\Rightarrow\) log\(_5\dfrac{1}{25}\).\(log_{27}9\)=\(\dfrac{-4}{3}\)

\(log_24=log_22^2=2\)

\(log_{\dfrac{1}{4}}2=log_{2^{-2}}2=\dfrac{-1}{2}\)

\(\Rightarrow log_24.log_{\dfrac{1}{4}}2=-1\)

Có: \(\dfrac{\left(-3\right)^{10}x15^5}{25^3x\left(-9\right)^7}=\dfrac{3^{10}.\left(3.5\right)^5x}{-\left(3^2\right)^7\left(5^2\right)^3x}\)

\(=\dfrac{3^{15}.5^5x}{-3^{14}.5^6x}\)\(=\dfrac{3^{14}.5^5\left(3x\right)}{3^{14}.5^5\left(-5x\right)}=\dfrac{3x}{-5x}=-\dfrac{3}{5}\)

Vậy...

\(PT\Leftrightarrow\dfrac{5}{2}\sqrt{2x+1}-\sqrt{\dfrac{\dfrac{2x+1}{2}}{2}}=\dfrac{3}{2}\\ \Leftrightarrow\dfrac{5}{2}\sqrt{2x+1}-\dfrac{1}{2}\sqrt{2x+1}=\dfrac{3}{2}\\ \Leftrightarrow2\sqrt{2x+1}=\dfrac{3}{2}\\ \Leftrightarrow\sqrt{2x+1}=\dfrac{3}{4}\\ \Leftrightarrow2x+1=\dfrac{9}{16}\\ \Leftrightarrow2x=-\dfrac{7}{16}\\ \Leftrightarrow x=-\dfrac{7}{32}\\ \Leftrightarrow a=-\dfrac{7}{32}\\ \Leftrightarrow1-36a=1+36\cdot\dfrac{7}{32}=...\)

Chọn \(f\left(x\right)=5x+5\)

Khi đó: \(\lim\limits_{x\rightarrow1}\dfrac{5x-5}{\left(\sqrt{x}-1\right)\left(\sqrt{20x+29}+3\right)}=\lim\limits_{x\rightarrow1}\dfrac{5\left(\sqrt{x}+1\right)}{\sqrt{20x+29}+3}=\dfrac{10}{7+3}=1\)

1.\(\left(\dfrac{1}{3}-x\right)^2=\dfrac{9}{25}\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}-x=\dfrac{3}{5}\\\dfrac{1}{3}-x=-\dfrac{3}{5}\end{matrix}\right.\)  \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{15}\\x=\dfrac{14}{15}\end{matrix}\right.\)

2.\(\left(5-x\right)^2=25\Leftrightarrow\left[{}\begin{matrix}5-x=5\\5-x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)

\(=\sqrt{\dfrac{4}{25}+\dfrac{4}{5}-\dfrac{9}{5}\cdot\dfrac{1}{9}+\dfrac{3}{4}}=\sqrt{\dfrac{4}{25}+\dfrac{4}{5}-\dfrac{1}{5}+\dfrac{3}{4}}\)

\(=\sqrt{\dfrac{4}{25}+\dfrac{3}{5}+\dfrac{3}{4}}\)

\(=\sqrt{\dfrac{16+60+75}{100}}=\dfrac{\sqrt{151}}{10}\)

bạn viết đúng đề chứ ạ

\(\sqrt{\dfrac{4}{25}+\left|-\dfrac{4}{5}\right|-\dfrac{9}{5}.\left(\dfrac{-1}{3}\right)^2+0,75}\)

\(=\sqrt{\dfrac{4}{25}+\dfrac{4}{5}-\dfrac{9}{5}.\dfrac{1}{9}+\dfrac{3}{4}}=\sqrt{\dfrac{4}{25}+\dfrac{20}{25}-\dfrac{9}{45}+\dfrac{3}{4}}=\sqrt{\dfrac{24}{25}-\dfrac{9}{45}+\dfrac{3}{4}}\)

\(=\sqrt{\dfrac{19}{25}+\dfrac{3}{4}}=\sqrt{\dfrac{76}{100}+\dfrac{75}{100}}=\sqrt{\dfrac{151}{100}}=\dfrac{\sqrt{151}}{10}\)

a: \(\left(\dfrac{5}{9}-\dfrac{\sqrt{9}}{12}\right):\dfrac{3}{4}+\dfrac{11}{3}:\dfrac{3}{4}\)

\(=\left(\dfrac{5}{9}-\dfrac{3}{12}\right)\cdot\dfrac{4}{3}+\dfrac{11}{3}\cdot\dfrac{4}{3}\)

\(=\left(\dfrac{5}{9}-\dfrac{1}{4}+\dfrac{11}{3}\right)\cdot\dfrac{4}{3}\)

\(=\dfrac{20-9+132}{36}\cdot\dfrac{4}{3}\)

\(=\dfrac{143}{3}\cdot\dfrac{1}{9}=\dfrac{143}{27}\)

b: \(\left(0.\left(3\right)+\dfrac{\left|-2\right|}{3}\right):\dfrac{\sqrt{25}}{4}-\left(2^3+3^2\right)^0\)

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

\(=\dfrac{4}{5}-1=-\dfrac{1}{5}\)