a: \(\left[\left(-21,8\right)+4,125\right]+\left[11,8+\left(-2,125\right)\right]\)

=-21,8+4,125+11,8-2,125

=(4,125-2,125)+(-21,8+11,8)

=2-10

=-8

b: \(\left(-124,5\right)+\left(-6,24+124,5\right)\)

\(=-124,5-6,24+124,5\)

=-6,24

\(a,\left[\left(-21,8\right)+4,125\right]+\left[11,8+\left(-2,125\right)\right]\)

\(=-21,8+4,125+11,8-2,125\)

\(=\left[\left(-21,8\right)+11,8\right]+\left(4,125-2,125\right)\)

\(=-10+2\)

\(=-8\)

\(b,\left(-124,5\right)+\left(-6,24+124,5\right)\)

\(=-124,5-6,24+124,5\)

\(=\left[\left(-124,5\right)+124,5\right]-6,24\)

\(=0-6,24\)

\(=-6,24\)

$\color{#90EE90}{\text{4}}$  $\color{#B0E0E6}{\text{56}}$

a: \(\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\\left(x+y\right)+2\left(x-y\right)=5\end{matrix}\right.\)

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

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

=>\(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=3x-5=-\dfrac{3}{2}-5=-\dfrac{13}{2}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\left(x+1\right)\left(y-1\right)=xy-1\\\left(x-3\right)\left(y+3\right)=xy-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}xy-x+y-1=xy-1\\xy+3x-3y-9=xy-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x+y=0\\3x-3y=-3+9=6\end{matrix}\right.\)

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

vậy: Hệ vô nghiệm

a) 

\(\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\\left(x+y\right)+2\left(x-y\right)=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+2y+3x-3y=4\\x+y+2x-2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5x-y=4\\3x-y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-1\\3x-y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\-\dfrac{3}{2}-y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{3}{2}-5=-\dfrac{13}{2}\end{matrix}\right.\)

b) 

\(\left\{{}\begin{matrix}\left(x+1\right)\left(y-1\right)=xy-1\\\left(x-3\right)\left(y+3\right)=xy-3\end{matrix}\right. \Leftrightarrow\left\{{}\begin{matrix}xy-x+y-1=xy-1\\xy+3x-3y-9=xy-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-x+y=0\\3x-3y=6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-x+y=0\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-y=2\end{matrix}\right.\)

mà: 2 khác 0

=> Hpt vô nghiệm 

ĐKXĐ: \(x\notin\left\{0;2\right\}\)
\(\dfrac{x-1}{2x^2-4x}-\dfrac{7}{8x}=\dfrac{5-x}{4x^2-8x}-\dfrac{1}{8x-16}\)

=>\(\dfrac{x-1}{2x\left(x-2\right)}-\dfrac{7}{8x}=\dfrac{5-x}{4x\left(x-2\right)}-\dfrac{1}{8\left(x-2\right)}\)

=>\(\dfrac{4\left(x-1\right)}{8x\left(x-2\right)}-\dfrac{7\left(x-2\right)}{8x\left(x-2\right)}=\dfrac{2\left(5-x\right)}{8x\left(x-2\right)}-\dfrac{x}{8x\left(x-2\right)}\)

=>4(x-1)-7(x-2)=2(5-x)-x

=>4x-4-7x+14=10-2x-x

=>-3x+10=-3x+10

=>0x=0(luôn đúng)

Vậy: \(x\in R\backslash\left\{0;2\right\}\)

\(\dfrac{x-1}{2x^2-4x}-\dfrac{7}{8x}=\dfrac{5-x}{4x^2-8x}-\dfrac{1}{8x-16}\left(x\ne0;x\ne2\right)\\ \Leftrightarrow\dfrac{x-1}{2x\left(x-2\right)}-\dfrac{7}{8x}=\dfrac{5-x}{4x\left(x-2\right)}-\dfrac{1}{8\left(x-2\right)}\\ \Leftrightarrow\dfrac{4\left(x-1\right)}{8x\left(x-2\right)}-\dfrac{7\left(x-2\right)}{8x\left(x-2\right)}=\dfrac{2\left(5-x\right)}{8x\left(x-2\right)}-\dfrac{x}{8x\left(x-2\right)}\\ \Leftrightarrow4\left(x-1\right)-7\left(x-2\right)=2\left(5-x\right)-x\\ \Leftrightarrow4x-4-7x+14=10-2x-x\\ \Leftrightarrow-3x+10=10-3x\\ \Leftrightarrow0x=0\)

=> Pt vô số nghiệm khi x khác 0 và 2 

Số phần quyển sách còn lại sau ngày thứ nhất là:

\(1-40\%=60\%=\dfrac{3}{5}\)

Số phần quyển sách còn lại sau ngày thứ hai là:

\(\dfrac{3}{5}\times\left(1-60\%\right)=\dfrac{3}{5}\times\dfrac{2}{5}=\dfrac{6}{25}\)

Số phần quyển sách còn lại sau ngày thứ ba là:

\(\dfrac{6}{25}\times\left(1-80\%\right)=\dfrac{6}{25}\times\dfrac{1}{5}=\dfrac{6}{125}\)

Số trang của quyển sách là:\(30:\dfrac{6}{125}=30\times\dfrac{125}{6}=625\left(trang\right)\)

giúp ik mik tích choa =))

Cậu chia nhỏ bài ra để được hỗ trợ nhanh hơn nhé!

$\color{#0000CD}{\text{A}}$   $\color{#0000FF}{\text{n}}$
$\color{#8A2BE2}{\text{nn}}$

\(\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\left(x\ne1\right)\\ \Leftrightarrow\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ \Leftrightarrow x^2+x+1-3x^2=2x\left(x-1\right)\\ \Leftrightarrow-2x^2+x+1=2x^2-2x\\ \Leftrightarrow4x^2-3x-1\\ \Leftrightarrow4x^2-4x+x-1=0\\ \Leftrightarrow4x\left(x-1\right)+\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(ktm\right)\\x=-\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)

a: ĐKXĐ: \(x\ne0;y\ne0\)

Đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\)

Hệ phương trình sẽ trở thành: \(\left\{{}\begin{matrix}a-2b=-1\\2a+b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-2b=-1\\4a+2b=6\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a-2b+4a+2b=-1+6\\2a+b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5a=5\\b=3-2a\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a=1\\b=3-2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=1\\\dfrac{1}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\left(nhận\right)\)

b: ĐKXĐ: \(\left\{{}\begin{matrix}x\ne y\\x\ne-\dfrac{y}{2}\end{matrix}\right.\)

Đặt \(\dfrac{1}{x-y}=a;\dfrac{1}{2x+y}=b\)

Hệ phương trình sẽ trở thành:

\(\left\{{}\begin{matrix}a+b=2\\3a-2b=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a+2b=4\\3a-2b=-2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2a+2b+3a-2b=4-2\\a+b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5a=2\\b=2-a\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a=\dfrac{2}{5}\\b=2-\dfrac{2}{5}=\dfrac{10}{5}-\dfrac{2}{5}=\dfrac{8}{5}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{x-y}=\dfrac{2}{5}\\\dfrac{1}{2x+y}=\dfrac{8}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=\dfrac{5}{2}\\2x+y=\dfrac{5}{8}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-y+2x+y=\dfrac{5}{2}+\dfrac{5}{8}\\x-y=\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=\dfrac{25}{8}\\y=x-\dfrac{5}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{25}{8}:3=\dfrac{25}{24}\\y=\dfrac{25}{24}-\dfrac{5}{2}=\dfrac{25}{24}-\dfrac{60}{24}=-\dfrac{35}{24}\end{matrix}\right.\left(nhận\right)\)

c: ĐKXĐ: \(x\ne1;y\ne-3\)

Đặt \(\dfrac{x}{x-1}=a;\dfrac{1}{y+3}=b\)

Hệ phương trình sẽ trở thành:

\(\left\{{}\begin{matrix}3a-2b=3\\4a+b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a-2b=3\\8a+2b=10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3a-2b+8a+2b=13\\4a+b=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11a=13\\b=5-4a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{13}{11}\\b=5-4\cdot\dfrac{13}{11}=\dfrac{55}{11}-\dfrac{52}{11}=\dfrac{3}{11}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{x}{x-1}=\dfrac{13}{11}\\\dfrac{1}{y+3}=\dfrac{3}{11}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13\left(x-1\right)=11x\\y+3=\dfrac{11}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x-13=11x\\y=\dfrac{11}{3}-3=\dfrac{2}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{13}{2}\\y=\dfrac{2}{3}\end{matrix}\right.\left(nhận\right)\)

ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

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

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

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

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

=>\(2+3\left(x-1\right)+\left(x-1\right)^2=0\)

=>\(2+3x-3+x^2-2x+1=0\)

=>\(x^2+x=0\)

=>x(x+1)=0

=>\(\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)

1 must/ have to

2 have to/ must

3 must not

4 do not have to

5 ought

6 should not

7 Should

8 does not have to

Câu 1 mình nghĩ dùng " should " cũm đc :vv nhưng mà nếu để cho câu 1 thì các câu còn lại nó bị rối tè le

a) 

\(\left(-12,5\right)+3.4+12,5+\left(-3,4\right)\\ =\left(12,5-12,5\right)+\left(3,4-3,4\right)\\ =0+0=0\)

b) 

\(32,8+4,2+\left(-4,3\right)+\left(-32,8\right)+4,3\\ =\left(32,8-32,8\right)+\left(4,3-4,3\right)+4,2\\ =0+0+4,2\\ =4,2\)

c) 

\(-\left(42,5+150\right)\cdot2,5-7,5\cdot2,5\\ =2,5\left(-42,5-150-7,5\right)\\ =2,5\cdot\left(-50-150\right)\\ =2,5\cdot-200\\ =-500\) 

d) 

\(\left(-2,45\right)\cdot2,6+2,6\cdot\left(-7,55\right)\\ =2,6\cdot\left(-2,45-7,55\right)\\ =2,6\cdot-10\\ =-26\)

a: \(\left(-12,5\right)+3,4+12,5+\left(-3,4\right)\)

\(=\left(-12,5+12,5\right)+\left(3,4-3,4\right)\)

=0+0=0

b: \(32,8+4,2+\left(-4,3\right)+\left(-32,8\right)+4,3\)

\(=\left(32,8-32,8\right)+\left(4,3-4,3\right)+4,2\)

=0+0+4,2

=4,2

c: \(-\left(42,5+150\right)\cdot2,5-7,5\cdot2,5\)

\(=2,5\cdot\left(-42,5-150-7,5\right)\)

\(=2,5\cdot\left(-200\right)=-500\)

d: Sửa đề: \(\left(-2,45\right)\cdot2,6+2,6\cdot\left(-7,55\right)\)

\(=2,6\left(-2,45-7,55\right)\)

\(=2,6\cdot\left(-10\right)=-26\)