\(\left(\dfrac{3}{4}x+\dfrac{9}{16}\right)\left(1,5+\dfrac{-3}{5}:x\right)=0\)

=>\(\left[{}\begin{matrix}\dfrac{3}{4}x+\dfrac{9}{16}=0\\1,5+\dfrac{-3}{5}:x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=-\dfrac{9}{16}\\-\dfrac{3}{5}:x=-1,5=-\dfrac{3}{2}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{9}{16}:\dfrac{3}{4}=-\dfrac{9}{16}\cdot\dfrac{4}{3}=-\dfrac{3}{4}\\x=\dfrac{3}{2}:\dfrac{3}{5}=\dfrac{5}{2}\end{matrix}\right.\)

\(\left(\dfrac{-7}{17}\right).\dfrac{5}{12}+\left(-\dfrac{7}{17}\right).\dfrac{7}{12}+\left(-\dfrac{10}{17}\right)\)
\(=\dfrac{-7}{17}.\left(\dfrac{5}{12}+\dfrac{7}{12}\right)+\dfrac{-10}{17}\)
\(=\dfrac{-7}{17}.1+\dfrac{-10}{17}\)
\(=\dfrac{\left(-7\right)+\left(-10\right)}{17}\)
\(=\dfrac{-17}{17}\)
\(=-1\)

\(=\left(-\dfrac{7}{17}\right)\times\left(\dfrac{5}{12}+\dfrac{7}{12}\right)+\left(-\dfrac{10}{17}\right)\)

\(=\left(-\dfrac{7}{17}\right)\times\left(\dfrac{12}{12}\right)+\left(-\dfrac{10}{17}\right)\)

\(=\left(-\dfrac{7}{17}\right)+\left(-\dfrac{10}{17}\right)\)

\(=\dfrac{-17}{17}=-1\)

a: \(\left|x-\dfrac{1}{2}\right|>=0\forall x\)

=>\(2\left|x-\dfrac{1}{2}\right|>=0\forall x\)

=>\(A=2\left|x-\dfrac{1}{2}\right|+\dfrac{1}{2}>=\dfrac{1}{2}\forall x\)

Dấu '=' xảy ra khi \(x-\dfrac{1}{2}=0\)

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

b: \(x^2>=0\forall x;\left|y-2\right|>=0\forall y\)

Do đó: \(x^2+\left|y-2\right|>=0\forall x,y\)

=>\(B=x^2+\left|y-2\right|-5>=-5\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y-2=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)

c: \(C=\left|3x-5\right|+\left|3x-9\right|=\left|3x-5\right|+\left|9-3x\right|\)

=>\(C>=\left|3x-5+9-3x\right|=4\)

Dấu '=' xảy ra khi (3x-5)(3x-9)<=0

=>\(\dfrac{5}{3}< =x< =\dfrac{9}{3}\)

=>\(\dfrac{5}{3}< =x< =3\)

a.

Do \(2\left|x-\dfrac{1}{2}\right|\ge0;\forall x\)

Dấu "=" xảy ra khi \(x=\dfrac{1}{2}\)

Nên \(A\ge\dfrac{1}{2}\)

Vậy \(A_{min}=\dfrac{1}{2}\) khi \(x=\dfrac{1}{2}\)

b.

Do \(\left\{{}\begin{matrix}x^2\ge0\\\left|y-2\right|\ge0\end{matrix}\right.\) ;\(\forall x;y\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)

Nên \(B\ge-5\)

Vậy \(B_{min}=-5\) khi \(\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)

c.

\(C=\left|3x-5\right|+\left|3x-9\right|=\left|3x-5\right|+\left|9-3x\right|\ge\left|3x-5+9-3x\right|=4\)

Dấu "=" xảy ra khi \(\left(3x-5\right)\left(9-3x\right)\ge0\Rightarrow\dfrac{5}{3}\le x\le3\)

Vậy \(C_{min}=4\) khi \(\dfrac{5}{3}\le x\le3\)

a: \(\left|x+\dfrac{3}{4}\right|>=0\forall x;\left|y-1\right|>=0\forall y\)

Do đó: \(\left|x+\dfrac{3}{4}\right|+\left|y-1\right|>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=1\end{matrix}\right.\)

b: \(\left|x-\dfrac{1}{2}\right|>=0\forall x\)

\(\left|y-\dfrac{1}{3}\right|>=0\forall y\)

\(\left|x+y+z\right|>=0\forall x,y,z\)

Do đó: \(\left|x-\dfrac{1}{2}\right|+\left|y-\dfrac{1}{3}\right|+\left|x+y+z\right|>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y-\dfrac{1}{3}=0\\x+y+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{3}\\z=-x-y=-\dfrac{5}{6}\end{matrix}\right.\)

\(2^x-2^y=1920\)

\(\Rightarrow2^y\left(2^{x-y}-1\right)=2^7.3.5\)

\(\Rightarrow\left\{{}\begin{matrix}2^y=2^7\\2^{x-y}-1=15\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y=7\\2^{x-7}=16=2^4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x-7=4\\y=7\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=11\\y=7\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(11;7\right)\)

7a - 21b + 5 = 7 ( a - 3b ) + 5 không chia hết cho 7.

Vậy 7a - 21b + 5 và 7 là hai số nguyên tố cùng nhau.

Vì ( 7a - 2b + 5 ) ( a - 3b + 1 ) chia hết cho 7 nên a - 3b + 1 chia hết cho 7.

Vì 42a + 14b + 14 chia hết cho 7 nên ( 42a + 14b + 14 ) + ( a - 3b + 1 ) chia hết cho 7.

Vậy 43a + 11b + 15 chia hết cho 7.

\(b,\dfrac{15}{37}.\left(\dfrac{38}{41}-\dfrac{74}{45}\right)+\dfrac{38}{41}.\left(\dfrac{82}{76}-\dfrac{15}{37}\right)\)
\(=\dfrac{15}{37}.\dfrac{38}{41}-\dfrac{15}{37}.\dfrac{74}{45}+\dfrac{38}{41}.\dfrac{82}{76}-\dfrac{38}{41}.\dfrac{15}{37}\)
\(=\left(\dfrac{15}{37}.\dfrac{38}{41}-\dfrac{38}{41}.\dfrac{15}{37}\right)-\dfrac{15}{37}.\dfrac{74}{45}+\left(\dfrac{-38}{41}.\dfrac{41}{38}\right)\)
\(=\left[\dfrac{38}{41}.\left(\dfrac{15}{37}-\dfrac{15}{37}\right)\right]-\dfrac{1.2}{3}-1\)
\(=0-\dfrac{2}{3}-1\)
\(=\dfrac{-2}{3}-\dfrac{3}{3}\)
\(=\dfrac{-5}{3}\)
\(c,\dfrac{2020}{2021}.\left(\dfrac{7}{6}-\dfrac{3}{4}\right)+\dfrac{2020}{2021}.\left(\dfrac{5}{6}-\dfrac{1}{4}\right)\)
\(=\dfrac{2020}{2021}.\left(\dfrac{14}{12}-\dfrac{9}{12}\right)+\dfrac{2020}{2021}.\left(\dfrac{10}{12}-\dfrac{3}{12}\right)\)
\(=\dfrac{2020}{2021}.\dfrac{5}{12}+\dfrac{2020}{2021}.\dfrac{7}{12}\)
\(=\dfrac{2020}{2021}.\left(\dfrac{5}{12}+\dfrac{7}{12}\right)\)
\(=\dfrac{2020}{2021}.1\)
\(=\dfrac{2020}{2021}\)
\(e,\dfrac{\dfrac{2}{7}+\dfrac{2}{5}+\dfrac{2}{17}-\dfrac{2}{25}}{\dfrac{3}{14}+\dfrac{3}{10}+\dfrac{3}{34}-\dfrac{3}{50}}\)
\(\dfrac{\dfrac{4}{14}+\dfrac{4}{10}+\dfrac{4}{34}-\dfrac{4}{50}}{\dfrac{3}{14}+\dfrac{3}{10}+\dfrac{3}{34}-\dfrac{3}{50}}\)
\(\dfrac{4.\left(\dfrac{1}{14}+\dfrac{1}{10}+\dfrac{1}{34}-\dfrac{1}{50}\right)}{3.\left(\dfrac{1}{14}+\dfrac{1}{10}+\dfrac{1}{34}-\dfrac{1}{50}\right)}\)
\(=\dfrac{4}{3}\)
\(f,\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}\)
\(=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{\dfrac{-5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}\)
\(=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{\dfrac{-5}{8}-\dfrac{-5}{10}+\dfrac{-5}{11}+\dfrac{-5}{12}}\)
\(=\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}\)
\(=\dfrac{3}{-5}\)
\(=\dfrac{-3}{5}\)

\(b,\dfrac{15}{37}\cdot\left(\dfrac{38}{41}-\dfrac{74}{45}\right)+\dfrac{38}{41}\cdot\left(\dfrac{82}{76}-\dfrac{15}{37}\right)\)

\(=\dfrac{15}{37}\cdot-\dfrac{1324}{1845}+\dfrac{38}{41}\cdot\dfrac{947}{1406}\)

\(=\dfrac{1}{37}\cdot-\dfrac{1324}{123}+\dfrac{1}{41}\cdot\dfrac{947}{37}\)

\(=\dfrac{1}{37}\cdot-\dfrac{1324}{123}+\dfrac{1}{37}\cdot\dfrac{947}{41}\)

\(=\dfrac{1}{37}\cdot\left(-\dfrac{1324}{123}+\dfrac{947}{41}\right)\)

\(=\dfrac{1}{37}\cdot\dfrac{37}{3}\)

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

\(c,\dfrac{2020}{2021}\cdot\left(\dfrac{7}{6}-\dfrac{3}{4}\right)+\dfrac{2020}{2021}\cdot\left(\dfrac{5}{6}-\dfrac{1}{4}\right)\)

\(=\dfrac{2020}{2021}\cdot\left(\dfrac{7}{6}-\dfrac{3}{4}+\dfrac{5}{6}-\dfrac{1}{4}\right)\)

\(=\dfrac{2020}{2021}\cdot\left[\left(\dfrac{7}{6}+\dfrac{5}{6}\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)\right]\)

\(=\dfrac{2020}{2021}\cdot\left[2-1\right]\)

\(=\dfrac{2020}{2021}\cdot1\)

\(=\dfrac{2020}{2021}\)

\(d,\left(1-\dfrac{2}{5}\right)\left(1-\dfrac{2}{7}\right)\left(1-\dfrac{2}{9}\right)...\left(1-\dfrac{2}{99}\right)\)

\(=\dfrac{3}{5}\cdot\dfrac{5}{7}\cdot\dfrac{7}{9}\cdot\cdot\cdot\dfrac{97}{99}\)

\(=\dfrac{3\cdot5\cdot7\cdot\cdot\cdot97}{5\cdot7\cdot9\cdot\cdot\cdot99}\)

\(=\dfrac{1}{33}\)

\(e,\dfrac{\dfrac{2}{7}+\dfrac{2}{5}+\dfrac{2}{17}-\dfrac{2}{25}}{\dfrac{3}{14}+\dfrac{3}{10}+\dfrac{3}{34}-\dfrac{3}{50}}\)

\(=\dfrac{2\cdot\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{25}\right)}{3\cdot\left(\dfrac{1}{14}+\dfrac{1}{10}+\dfrac{1}{34}-\dfrac{1}{50}\right)}\)

\(=2\cdot\dfrac{2}{3}\)

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

\(f,\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}\)

\(=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}\)

\(=\dfrac{3\cdot\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{5\cdot\left(-\dfrac{1}{8}+\dfrac{1}{10}-\dfrac{1}{11}-\dfrac{1}{12}\right)}\)

\(=\dfrac{3}{5}\cdot-1\)

\(=-\dfrac{3}{5}\)

Số tiền lãi của bác Nhi sau kì hạn 1 năm là:

        60 000 000 x 6,5% = 3 900 000 (đồng)

Tổng số tiền trong ngân hàng của bác Nhi sau kì hạn 1 năm là:

       60 000 000 + 3 900 000 = 63 900 000 (đồng)

Số tiền bác Nhi rút ra:

       63 900 000 x \(\dfrac{1}{3}\) = 21 300 000 (đồng)

Số tiền còn lại trong ngân hàng của bác Nhi sau khi rút là:

      63 900 000 - 21 300 000 = 42 600 000 (đồng)

Đáp số: 42 600 000 đồng

Số tiền lãi sau một năm là:
\(60.6,5:100=3,9\) ( triệu đồng)
Số tiền gốc và lãi của Bác Nhi sau một năm là:
\(60+3,9=63,9\) ( triệu đồng)
Số tiền Bác Nhi rút ra được là:
\(\dfrac{1}{3}.63,9=21,3\) ( triệu đồng)
Số tiền còn lại của Bác Nhi trong ngân hàng là:
\(63,9-21,3=42,6\) ( triệu đồng)

Để x là số hữu tỉ dương thì x>0

=>\(\dfrac{-m+2022}{-2024}>0\)

=>\(\dfrac{m-2022}{2024}>0\)

=>m-2022>0

=>m>2022