Question

[Linh Nhật]

Answer 1

[Nguyễn Lê Phước Thịnh]

c: \(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{200}\left(1+2+...+200\right)\)

\(=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{200}\cdot\dfrac{200\cdot201}{2}\)

\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{201}{2}\)

\(=\dfrac{2+3+...+201}{2}=\dfrac{\left(201-2+1\right)\cdot\dfrac{\left(201+2\right)}{2}}{2}\)

\(=\dfrac{200\cdot203}{4}=50\cdot203=10150\)

d: \(\dfrac{2^7\cdot3+2^{10}}{13\cdot2^7-14\cdot2^6}=\dfrac{2^7\left(3+2^3\right)}{2^6\left(13\cdot2-14\right)}=2\cdot\dfrac{3+8}{26-14}\)

\(=2\cdot\dfrac{11}{12}=\dfrac{11}{6}\)

Bài 2:

1: 

a: \(\left(x-\dfrac{1}{2}\right):\dfrac{2}{3}+0,75=-3\dfrac{3}{4}\)

=>\(\left(x-\dfrac{1}{2}\right):\dfrac{2}{3}=-3-\dfrac{3}{4}-\dfrac{3}{4}=-3-\dfrac{3}{2}=-\dfrac{9}{2}\)

=>\(x-\dfrac{1}{2}=-\dfrac{9}{2}\cdot\dfrac{2}{3}=-\dfrac{9}{3}=-3\)

=>\(x=-3+\dfrac{1}{2}=-\dfrac{5}{2}\)

b: \(\left(2x-15\right)^3=\left(2^2\cdot3^3-2^3\cdot3^2\right):\left(-36\right)\)

=>\(\left(2x-15\right)^3=\left(4\cdot27-8\cdot9\right):\left(-36\right)\)

=>\(\left(2x-15\right)^3=\dfrac{\left(108-72\right)}{-36}=\dfrac{36}{-36}=-1\)

=>2x-15=-1

=>2x=14

=>x=7

c: \(\left(4x+1\right)^4=\left(4x+1\right)^6\)

=>\(\left(4x+1\right)^6-\left(4x+1\right)^4=0\)

=>\(\left(4x+1\right)^4\cdot\left[\left(4x+1\right)^2-1\right]=0\)

=>\(\left(4x+1\right)^4\left(4x+1-1\right)\left(4x+1+1\right)=0\)

=>\(4x\left(4x+2\right)\left(4x+1\right)^4=0\)

=>\(\left[{}\begin{matrix}x=0\\4x+2=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

d: \(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\cdot x^2=99\)

=>\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\cdot x^2=99\)

=>\(\left(1-\dfrac{1}{100}\right)\cdot x^2=99\)

=>\(x^2=99:\dfrac{99}{100}=100\)

=>\(x=\pm10\)

2: 2xy-x-y=2

=>\(x\left(2y-1\right)-y+\dfrac{1}{2}=2+\dfrac{1}{2}\)

=>\(2x\left(y-\dfrac{1}{2}\right)-\left(y-\dfrac{1}{2}\right)=\dfrac{5}{2}\)

=>\(\left(2x-1\right)\left(y-\dfrac{1}{2}\right)=\dfrac{5}{2}\)

=>(2x-1)(2y-1)=5

=>\(\left(2x-1;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(0;-2\right);\left(-2;0\right)\right\}\)

Answer 2

[Nguyễn Thị Thương Hoài]

Bài 1:

a; 56.712 - 7.50.80 - 7.12.8

= 56.712 - (7.8).(50.10) - (7.8).12

= 56.712 - 56. 500 - 56.12

= 56.(712 - 500 - 12)

= 56.(212 - 12)

= 56.200

= 11200

b; 5\(^0\): \(\frac{3}{\left(-1\right)^{2024}}\) + (6\(\frac16\) - 2\(\frac49\) .2\(\frac{5}{11}\))

= 1:\(\frac31\) + (\(6+\frac16\) - \(\frac{22}{9}\).\(\frac{27}{11}\))

= \(\frac13\) + (\(6+\frac16\) - 6)

=(\(\frac26\) + \(\frac16\)) + ( 6 - 6)

= \(\frac12\)