f(1)=3

f(-1)=3

\(\text{f(1)=}2.1^2+1=3\)

\(\text{f(-1)=}2.\left(-1\right)^2+1=3\)

\(\text{f(2)=}2.2^2+1=9\)

\(\text{f(0)=}2.0^2+1=1\)

\(\text{f(-3)=}=2.\left(-3\right)^2+1=19\)

a) Ta có: \(\widehat{xOy}+\widehat{yOz}=180^0\)(hai góc kề bù)

\(\Leftrightarrow\widehat{zOy}+140^0=180^0\)

hay \(\widehat{yOz}=40^0\)

Vậy: \(\widehat{yOz}=40^0\)

a) Ta có: xOy+yOz=180⁰xOy+yOz=180⁰(hai góc kề bù)

              ⇔zOy+140⁰=180⁰

     hay yOz=40⁰yOz=40⁰

     Vậy: yOz=40⁰

1 were - would you play

2 weren't studying - would have 

3 had taken - wouldn't have got

4 would you go - could

5 will you give - is

6 recycle - won't be

7 had heard - wouldn't have gone

8 would you buy - had

9 don't hurry - will miss

10 had phoned - would have given

11 were - wouldn't eat

12 will go - rains

13 had known - would have sent

14 won't feel - swims

15 hadn't freezed - would have gone

Ngày thứ hai nhà máy sản xuất được số tấn đường là:

1,17-0,12= 1,05 (tấn)

Cả hai ngày thứ nhất và thứ hai, nhà máy sản xuất được số tấn đương là:

1,17+1,05=2,22 (tấn)

Ngày thứ ba nhà máy sản xuất được số tấn đường là:

4,14-2,22=1,92 (tấn)

"Chúc em học tốt."

Bn cần bài nào trong 2 bài nhỉ?

e tách câu hỏi ra nhe tạm thời cj giúp mụt câu nhe

a: \(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1+\dfrac{5}{7}=\dfrac{-5}{7}+1+\dfrac{5}{7}=1\)

b: \(=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{16}\cdot4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{13}{56}\)

c: \(=\dfrac{2}{3}+\dfrac{1}{3}\cdot\dfrac{-24+45}{54}\cdot\dfrac{12}{7}\)

\(=\dfrac{2}{3}+\dfrac{1}{3}\cdot\dfrac{21}{54}\cdot\dfrac{12}{7}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{6+2}{9}=\dfrac{8}{9}\)

d: \(=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{107}-\dfrac{1}{111}\)

\(=\dfrac{1}{3}-\dfrac{1}{111}=\dfrac{108}{333}=\dfrac{12}{37}\)

giúp cái quái gì

giúp gì vậy bn

TL

Giúp gì vậy trời

Hok tốt

Bài 4: 

\(x=130^0\)

`@` `\text {Ans}`

`\downarrow`

`4,`

Vì `\text {MN // BC}`

`=>` $\widehat {B} = \widehat {BMN} = 114^0 (\text {2 góc đối đỉnh})$

Ta có: \(\left\{{}\begin{matrix}\widehat{\text{BMN}}+\widehat{\text{AMN}}=180^0\left(\text{2 góc kề bù}\right)\\\widehat{\text{CNM}}+\widehat{\text{ANM}}=180^0\left(\text{2 góc kề bù}\right)\end{matrix}\right.\)

`=>`\(\left\{{}\begin{matrix}\widehat{\text{AMN}}=180^0-114^0=66^0\\\widehat{\text{ANM}}=180^0-130^0=50^0\end{matrix}\right.\)

Xét `\Delta AMN`:

\(\widehat{\text{A}}+\widehat{\text{M}}+\widehat{\text{N}}=180^0\left(\text{định lý tổng 3 góc trong 1 tgiac}\right)\)

`=>`\(\widehat{\text{A}}+66^0+50^0=180^0\)

`=>`\(\widehat{\text{A}}=180^0-66^0-50^0=64^0\)

Mà \(\widehat{\text{A}}=\widehat{\text{x}}\)

`=>`\(\widehat{\text{x}}=64^0\)

Vậy, số đo của góc `x = 64^0.`

Đề bài thấy hơi sao sao ấy

Dễ nhưng dài lắm :)) cô cho lm đầy 

\(a,2\dfrac{1}{3}.5\dfrac{4}{7}=\dfrac{7}{3}.\dfrac{39}{7}=13\)

\(b,\dfrac{-1}{2}+\dfrac{4}{9}+\dfrac{3}{2}+\dfrac{5}{9}=-1+1=0\)

\(c,0,75+\left(-1,25\right)=-1,50\)

\(d,\dfrac{2}{5}\cdot\dfrac{3}{8}-\dfrac{2}{5}\cdot\dfrac{6}{11}=\dfrac{2}{5}\cdot\left(\dfrac{3}{8}-\dfrac{6}{11}\right)=\dfrac{2}{5}.\dfrac{-15}{88}=\dfrac{-3}{44}\)