ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne4\end{matrix}\right.\)

\(P=\left(\dfrac{2}{x-4}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{1}{\sqrt{x}+2}\)

\(=\dfrac{2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{1}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

Để P=3/2 thì \(\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{3}{2}\)

=>\(3\left(\sqrt{x}-2\right)=2\sqrt{x}\)

=>\(3\sqrt{x}-2\sqrt{x}=6\)

=>\(\sqrt{x}=6\)

=>x=36(nhận)

a: Khi x=707228 thì x+87002=707228+87002=794230

b: Khi x=100 thì 2035xX=2035x100=203500

c: Khi x=84560 thì x:2=84560:2=42280

c: Khi x=304110 thì 564320-x=564320-304110=260210

Do đều dư 16 nên:

1012 - 16 và 1178 - 16 ⋮ a

⇒ 996 và 1162 ⋮ a

Do dư 16 nên a > 16.

Ta có:

996 = 2² x 3 x 83

1162 = 2 x 7 x 83

Các số a có thể ϵ {83; 166}

4051100

40 511 000.

Xét ΔEDI có \(\widehat{EIF}\) là góc ngoài

nên \(\widehat{EIF}=\widehat{IED}+\widehat{IDE}\)

=>\(\widehat{IED}=110^0-90^0=20^0\)

EI là phân giác của góc DEF

=>\(\widehat{DEF}=2\cdot\widehat{DEI}=40^0\)

ΔDEF vuông tại D

=>\(\widehat{DEF}+\widehat{DFE}=90^0\)

=>\(\widehat{DFE}=90^0-40^0=50^0\)

 cảm ơn bạn nhiều nha

Sửa đề: \(\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{2^{12}\cdot9^6+8\cdot9^5}\)

\(=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^{12}+2^3\cdot3^{10}}\)

\(=\dfrac{2^{12}\cdot3^4\left(3-1\right)}{2^3\cdot3^{10}\left(2^9\cdot3^2+1\right)}\)

\(=\dfrac{2^9}{3^6}\cdot\dfrac{2}{1028\cdot9+1}=\dfrac{2^{10}}{729\left(1028\cdot9+1\right)}\)

sorry mình ấn nhầm nha

Trong tập hợp T, những phần tử có số ngày 31 là

\(T=\) { tháng 10; tháng 12 }

giúp với...

Bài 2:

a: \(\left|x+\dfrac{1}{5}\right|-\dfrac{1}{2}=\dfrac{9}{10}\)

=>\(\left|x+\dfrac{1}{5}\right|=\dfrac{1}{2}+\dfrac{9}{10}=\dfrac{14}{10}=\dfrac{7}{5}\)

=>\(\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{7}{5}\\x+\dfrac{1}{5}=-\dfrac{7}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=-\dfrac{8}{5}\end{matrix}\right.\)

b: \(\dfrac{5}{4}-3\left|2x+5\right|=\dfrac{3}{4}\)

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

=>\(\left|2x+5\right|=\dfrac{1}{6}\)

=>\(\left[{}\begin{matrix}2x+5=\dfrac{1}{6}\\2x+5=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}-5=-\dfrac{29}{6}\\2x=-\dfrac{1}{6}-5=-\dfrac{31}{6}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{29}{12}\\x=-\dfrac{31}{12}\end{matrix}\right.\)

c: \(\left(\dfrac{3}{5}x+\dfrac{1}{2}\right)^2=\dfrac{25}{16}\)

=>\(\left[{}\begin{matrix}\dfrac{3}{5}x+\dfrac{1}{2}=\dfrac{5}{4}\\\dfrac{3}{5}x+\dfrac{1}{2}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{5}x=\dfrac{5}{4}-\dfrac{1}{2}=\dfrac{3}{4}\\\dfrac{3}{5}x=-\dfrac{5}{4}-\dfrac{1}{2}=-\dfrac{7}{4}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{3}{4}:\dfrac{3}{5}=\dfrac{5}{4}\\x=-\dfrac{7}{4}:\dfrac{3}{5}=-\dfrac{7}{4}\cdot\dfrac{5}{3}=-\dfrac{35}{12}\end{matrix}\right.\)

d: \(3-\left(2x+1\right)^2=2\)

=>\(\left(2x+1\right)^2=3-2=1\)

=>\(\left[{}\begin{matrix}2x+1=1\\2x+1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Bài 1:

a: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{9}{16}-\sqrt{\dfrac{4}{81}}:\dfrac{16}{9}+\left|-0,25\right|\)

\(=\dfrac{4}{9}\cdot\dfrac{9}{16}-\dfrac{2}{9}\cdot\dfrac{9}{16}+\dfrac{1}{4}\)

\(=\dfrac{4}{16}-\dfrac{2}{16}+\dfrac{1}{4}=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)

b: \(\left(-2\right)^3+\dfrac{1}{2}:\dfrac{1}{8}-\sqrt{25}+\left|-8\right|\)

\(=-8+\dfrac{1}{2}\cdot8-5+8\)

=4-5=-1

c: \(\left(\dfrac{4}{3}-\dfrac{3}{2}\right)^2-2:\left|-\dfrac{1}{9}\right|+\dfrac{-5}{18}\)

\(=\left(\dfrac{8}{6}-\dfrac{9}{6}\right)^2-2:\dfrac{1}{9}-\dfrac{5}{18}\)

\(=\dfrac{1}{36}-18-\dfrac{5}{18}=\dfrac{1}{36}-\dfrac{10}{36}-18=-\dfrac{9}{36}-18\)

\(=-18-\dfrac{1}{4}=-18,25\)

d: \(\left(-\dfrac{3}{4}\right)^2:\left(-\dfrac{1}{4}\right)^2+9\left(\dfrac{1}{3}\right)^2+\left|-\dfrac{3}{2}\right|\)

\(=\left(-\dfrac{3}{4}:\dfrac{-1}{4}\right)^2+9\cdot\dfrac{1}{9}+\dfrac{3}{2}\)

\(=3^2+1+\dfrac{3}{2}=9+1+\dfrac{3}{2}=10+\dfrac{3}{2}=11,5\)