Chúc mừg số 1 vs số 4 của danh sách `HTGD` nhe <33

vs cả số `6` vs số `9` trong danh sách nhận giải nhe <3

\(12\%\) của \(150\) là số :

\(\dfrac{12}{100}.150=18\)

Đáp số : `18`

\(b,\sqrt{196}-\sqrt{64}+\sqrt{9}\)

\(=14-8+3\)

\(=6+3\)

\(=9\)

`---`

\(d,\sqrt{289}-\sqrt{144}+\sqrt{\left(-3\right)^2}\)

\(=17-12+\sqrt{9}\)

\(=17-12+3\)

\(=5+3\)

`=8`

\(f,\sqrt{1\dfrac{24}{25}}:1,2-2023^0+\left(-\dfrac{1}{3}\right)^2.\sqrt{\left(-3\right)^2}\)

\(=\sqrt{\dfrac{49}{25}}:\dfrac{6}{5}-1+\dfrac{1}{9}.\sqrt{9}\)

\(=\dfrac{7}{5}:\dfrac{6}{5}-1+\dfrac{1}{9}.3\)

\(=\dfrac{7}{6}-1+\dfrac{1}{3}\)

\(=\dfrac{1}{6}+\dfrac{1}{3}\)

\(=\dfrac{1}{2}\)

${color{#B0E0E6}}{\text{Working}}$

P/S: Nếu ai thấy tớ sai ở đâu thì bảo tớ ạ!

\(a,-\dfrac{3}{4}x-\dfrac{1}{2}=\dfrac{1}{6}\)

\(\Rightarrow-\dfrac{3}{4}x=\dfrac{1}{6}+\dfrac{1}{2}\)

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

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

\(\Rightarrow x=-\dfrac{8}{9}\)

Vậy \(x=-\dfrac{8}{9}\)

\(b,\left|x\right|-2,96=\left(-0,2\right)^2\)

\(\Rightarrow\left|x\right|-2,96=0,04\)

\(\Rightarrow\left|x\right|=0,04+2,96\)

\(\Rightarrow\left|x\right|=3\)

\(\Rightarrow x=3\) hoặc \(x=-3\)

Vậy...

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

\(\Rightarrow2x-\dfrac{1}{5}=0\)

\(\Rightarrow2x=0+\dfrac{1}{5}\)

\(\Rightarrow2x=\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{1}{5}:2\)

\(\Rightarrow x=\dfrac{1}{10}\)

\(d,\)\(\left|x+\dfrac{5}{2}\right|-\dfrac{2}{3}=\left(\dfrac{-1}{2}\right)^3:\left(\dfrac{-1}{2}\right)^2\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=\dfrac{1}{6}\\x+\dfrac{5}{2}=-\dfrac{1}{6}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}-\dfrac{5}{2}\\x=-\dfrac{1}{6}-\dfrac{5}{2}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{3}\\x=-\dfrac{8}{3}\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{7}{3};-\dfrac{8}{3}\right\}\)

P/S : Nếu tớ sai ở đâu thì mọi người góp ý giúp ạ!

\(-------\)

\(\dfrac{25}{100}\times30+2,5:10+\dfrac{1}{4}\times29+0,25\times40\)

\(=0,25\times30+0,25\times1+0,25\times29+0,25\times40\)

\(=0,25\times\left(30+1+29+40\right)\)

\(=0,25\times100\)

\(=25\)

\(3,27\times0,25+2,3\times\dfrac{1}{4}+1,75:4+2,68\times25\%\)

\(=3,27\times0,25+2,3\times0,25+1,75\times0,25+2,68\times0,25\)

\(=\left(3,27+2,3+1,75+2,68\right)\times0,25\)

\(=\left(5,57+1,75+2,68\right)\times0,25\)

\(=\left(7,32+2,68\right)\times0,25\)

\(=10\times0,25\)

\(=2,5\)

\(\dfrac{154,8+42,7}{x}+35=74,5\)

\(\Rightarrow\dfrac{197,5}{x}+35=74,5\)

\(\Rightarrow\)\(\dfrac{197,5}{x}=74,5-35\)

\(\Rightarrow\dfrac{197,5}{x}=39,5\)

\(\Rightarrow x=197,5:39,5\)

\(\Rightarrow x=5\)

\(A=1+2^1+2^2+2^3+2^4+2^5\)

\(1,\)

\(2A=2+2^2+2^3+2^4+2^5+2^6\)

\(2A-A=\left(2+2^2+2^3+2^4+2^5+2^6\right)-\left(1+2^1+2^2+2^3+2^4+2^5\right)\)

\(A=2^6-1\)

\(A=64-1=63\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=k\)

Ta có :

\(\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)

\(\Rightarrow xy=54\Leftrightarrow2k.3k=54\)

\(\Rightarrow6k^2=54\)

\(\Rightarrow k^2=54:6\)

\(\Rightarrow k^2=9\)

\(\Rightarrow k^2=\left(\pm3\right)^2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(\pm3\right)=\left(\pm6\right)\\y=3.\left(\pm3\right)=\left(\pm9\right)\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\left(6;9\right),\left(-6;-9\right)\right\}\)

 Mình có thấy cái đề bài đâu nhỉ????????