a: =-2,4x1,25=-3

b: =-5,85:2,25=-2,6

c: =4,2-0,126-2,18=1,894

d: =-45,3+7,3-1,5=-38-1,5=-39,5

a) \(\left(-2,4\right)\cdot1,25=-\left(2,4\cdot1,25\right)=-3\)
b) \(-5,85:2,25=\)\(-\left(5,85:2,25\right)=-2,6\)
c) \(4,2-\left(0,126+2,18\right)=4,2-2,306=1,894\)
d) \(-45,3+\left[7,3+\left(-1,5\right)\right]=-45,3+5,8=-39,5\)

2 C có so far chia HTHT

4 b 

5 d 

6 a ăn => đau bụng

2. C
so far = HTHT
4. B
put on weight: tăng cân
5. D
make a different: tạo nên sự khác bọt
6. A
ate: ăn -> liên quan đến "stomachache": đau bụng

1 A => the girl

2 B => should be kept

3 A => had

4 C => mine

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

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

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\x\ne2\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{2}{x\left(x-2\right)}=\dfrac{x-2}{x\left(x-2\right)}\)

\(\Leftrightarrow x^2+2x-2=x-2\)

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

\(\Rightarrow\left\{{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\left(kTM\right)\\x=-1\left(TM\right)\end{matrix}\right.\)

Vậy phương trình đã cho có nghiệm \(x=-1\)

Ta có: \(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\)

\(\Leftrightarrow x\left(x+2\right)-2=x-2\)

\(\Leftrightarrow x^2+2x-x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

hay x=-1

bằng 5

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

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

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

\(\Leftrightarrow-\left(x^2-x+2x-2\right)=2\)

\(\Leftrightarrow x^2+x-2=-2\)

\(\Leftrightarrow x\left(x+1\right)=0\)

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

Vậy: S={0}

?

e gửi ảnh mà nó lỗi r

a) Ta có : 51ⁿ=\(\overline{...1}\)

                47¹⁰²=47².(47⁴)²⁵=\(\left(\overline{...9}\right).\left(\overline{...1}\right)=\overline{...9}\)

\(\Rightarrow51^n+47^{102}=\left(\overline{...1}\right)+\left(\overline{...9}\right)=\overline{...0}⋮10\)

Vậy 51ⁿ+47¹⁰²\(⋮\)10.

b) Ta có : \(17^5=17.17^4=17.\left(\overline{...1}\right)=\overline{...7}\)

                \(24^4=\overline{...6}\)

                 \(13^{21}=13.\left(13^4\right)^5=13.\left(\overline{...1}\right)=\overline{...3}\)

\(\Rightarrow17^5+24^4-13^{21}=\left(\overline{...7}\right)+\left(\overline{...6}\right)-\left(\overline{...3}\right)=\overline{...0}⋮10\)

Vậy 17⁵+24⁴+13²¹\(⋮\)10

Tóm tắt

\(F=200N\)

\(s=600m\)

\(\text{℘}=300W\)

===========

\(t=?s\)

Công thực hiện được:

\(A=F.s=200.600=120000J\)

Thời gian đạp xe:

\(\text{℘}=\dfrac{A}{t}\Rightarrow t=\dfrac{A}{\text{℘}}=\dfrac{120000}{300}=400s\)

Đề có thiếu không vậy? 

Tóm tắt:

\(F=200N\\ s=600m\\ P\left(hoa\right)=300W\\ --------\\ t=?s\) 

Giải:

Công của người ấy đạp xe: \(A=F.s\\ =200.600=120000\left(J\right)\) 

Thời gain người ấy đạp xe: \(P\left(hoa\right)=\dfrac{A}{t}\Rightarrow t=\dfrac{A}{P\left(hoa\right)}\\ =\dfrac{120000}{300}=400\left(s\right).\)