Số trang đọc được trong ngày 1 là \(200\cdot\dfrac{2}{5}=80\left(trang\right)\)

Số trang đọc được trong ngày 2 là \(80\cdot\dfrac{7}{10}=56\left(trang\right)\)

Số trang đọc được trong ngày 3 là:

200-80-56=64(trang)

Số trang đọc được trong ngày 3 chiếm:

\(\dfrac{64}{200}=\dfrac{8}{25}\)(cuốn sách)

A=1x3+2x4+3x5+...+99x100

A=(1x3+3x5+...+99x101)+(2x4+4x6+...+98x100)

đặt S=1x3+3x5+...+99x101

=>6S=6x(1x3+3x5+...+99x101)

=1x3x(5+1)+3x5x(7-1)+...+97x99x(101-95)+99x101x(103-97)

=1x3x5+1x3x1+3x5x7-1x3x5+....+97x99x101-95x97x99+99x101x103-97x99x101

=1x3x1+99x101x103

=>S=(3+99x101x103):6=171650

=>C=171650+(2x4+4x6+...+98x100)

đặt A=2x4+4x6+...+98x100

=>6A=6x(2x4+4x6+...+98x100)

=>6A=2x4x6+4x6x(8-2)+...+96x98x(100-94)+98x100x(102-96)

=2x4x6+4x6x8-2x4x6+...+96x98x100-94x96x98+98x100x102-96x98x100

=98x100x102

=>A=98x100x102:6=166600

=>A=166600+171650

=>A=338250

ủa đây là lớp 6 rồii

\(\dfrac{7}{11}-\dfrac{6}{13}-\left(\dfrac{4}{7}+\dfrac{20}{13}-3\right)\)

\(=\dfrac{7}{11}-\dfrac{6}{13}-\dfrac{4}{7}-\dfrac{20}{13}+3\)

\(=\dfrac{7}{11}-\dfrac{4}{7}+3-2=\dfrac{7}{11}-\dfrac{4}{7}+1\)

\(=\dfrac{49-44+77}{77}=\dfrac{82}{77}\)

xy+12=x+y

=>xy-x-y+12=0

=>x(y-1)-y+1+11=0

=>(x-1)(y-1)=-11

=>\(\left(x-1\right)\left(y-1\right)=1\cdot\left(-11\right)=\left(-1\right)\cdot11=11\cdot\left(-1\right)=\left(-11\right)\cdot1\)

=>\(\left(x-1;y-1\right)\in\left\{\left(1;-11\right);\left(-1;11\right);\left(11;-1\right);\left(-11;1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(2;-10\right);\left(0;12\right);\left(12;0\right);\left(-10;2\right)\right\}\)

1+mấy=bao nhiêu(bao nhiêu <0)

---------------- HẾT ----------------

Tôi ko biết

1: \(\dfrac{x}{5}=-\dfrac{12}{10}\)

=>\(\dfrac{x}{5}=-\dfrac{6}{5}\)

=>x=-6

2: \(\dfrac{-1}{3}=\dfrac{3}{x}\)

=>\(x=\dfrac{3\cdot3}{-1}=-9\)

3: \(\dfrac{-x}{-12}=\dfrac{9}{4}\)

=>\(\dfrac{x}{12}=\dfrac{9}{4}\)

=>\(\dfrac{x}{12}=\dfrac{27}{12}\)

=>x=27

4: \(\dfrac{8}{7}=\dfrac{16}{-x}\)

=>\(\dfrac{16}{-x}=\dfrac{16}{14}\)

=>-x=14

=>x=-14

5:

ĐKXĐ: x<>0

 \(\dfrac{3}{5}=-\dfrac{9}{x}\)

=>\(x=\dfrac{-5\cdot9}{3}=-5\cdot3=-15\)(nhận)

6: 

ĐKXĐ: x<>9

\(\dfrac{4}{5}=\dfrac{-12}{x-9}\)

=>\(x-9=\dfrac{5\cdot\left(-12\right)}{4}=\dfrac{-60}{4}=-15\)

=>x=-15+9=-6(nhận)

7: \(\dfrac{1}{6}=\dfrac{x+3}{18}\)

=>\(x+3=18\cdot\dfrac{1}{6}=3\)

=>x=3-3=0

8: ĐKXĐ: x<>7

\(\dfrac{x-7}{25}=\dfrac{9}{x-7}\)

=>\(\left(x-7\right)\left(x-7\right)=9\cdot25\)

=>\(\left(x-7\right)^2=15^2\)

=>\(\left[{}\begin{matrix}x-7=15\\x-7=-15\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=22\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)

9: \(\dfrac{x}{6}=\dfrac{x-1}{5}\)

=>6(x-1)=5x

=>6x-6=5x

=>6x-5x=6

=>x=6

10: ĐKXĐ \(x\notin\left\{0;-1\right\}\)

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

=>\(3x\cdot1=2\left(x+1\right)\)

=>3x=2x+2

=>3x-2x=2

=>x=2(nhận)