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

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

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

f(1)=3

f(-1)=3

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

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

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

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

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

1.A
2.A
3.B
4.C
5.B
6.C
7.A
8.A
9.B
10.A
11.B
12.A
13.C
14.B
15.B
16.A
17.A
18.A
19.A
20.C

1 C

2 B

3 B

4 C

5 B

6 B

7 D

8 B

9 A

10 B

Bài 4

1 for

2 of

3 about

a) Xét \(\Delta ABC:\)

\(BC^2=10^2=100.\\ AB^2+AC^2=6^2+8^2=100.\\ \Rightarrow BC^2=AB^2+AC^2.\)

\(\Rightarrow\Delta ABC\) vuông tại A (Pytago đảo).

\(\Rightarrow\widehat{BAC}=90^o.\)

b) Ta có: \(\widehat{B}+\widehat{D}=90^o.\\ \widehat{B}+\widehat{C}=90^o.\)

\(\Rightarrow\widehat{D}=\widehat{C}.\)

Xét \(\Delta ABC\) và \(\Delta AED:\)

\(\widehat{D}=\widehat{C}\left(cmt\right).\)

\(AC=AD\left(=8cm\right).\)

\(\widehat{BAC}=\widehat{EAD}\left(=90^o\right).\)

\(\Rightarrow\) \(\Delta ABC\) \(=\Delta AED\left(g-c-g\right).\)

c) Xét \(\Delta BDC:\)

DK là đường cao \(\left(DK\perp BC\right).\)

CA là đường cao \(\left(CA\perp AB\right).\)

Mà E là giao điểm của DK; CA (gt).

b: \(x+1657-34524=1289\)

\(\Leftrightarrow x+1657=35813\)

hay \(x=34156\)

c) x - 54 683 = 32 596

   x                = 32 956 + 54 863

   x                =     87 819

a) x + 27 569 = 35 986

    x                = 35986 - 27569

   x                 =  8417

b) x + 1657 - 34524 = 1289

x  + 1657      =    1289 + 34524

x + 1657       =      35813

x                   =     35813 - 1657

x                  =          34156

Kẻ đường cao AH cho tam giác ABC 

sinB = AH/AB => \(\dfrac{\sqrt{3}}{2}=\dfrac{AH}{6}\Rightarrow AH=3\sqrt{3}\)cm 

cosB = BH/AB => \(\dfrac{1}{2}=\dfrac{BH}{6}\Rightarrow BH=3cm\)

=> CH = BC - BH = 1 cm 

Theo Pytago tam giác AHC vuông tại H

\(AC=\sqrt{AH^2+HC^2}=2\sqrt{7}cm\)

-> chọn A

17. The fire isn't very hot. It won't boil a kettle.

\(\Rightarrow\) The fire isn't hot enough to boil a kettle.
18. You are quite thin. You could slip between the bars.

\(\Rightarrow\) You are thin enough to slip between the bars.
19. He is very ill. He can't eat anything.

\(\Rightarrow\) He is too ill to eat anything.
20. Our new car is very wide. It won't get through those gates.

\(\Rightarrow\) Our new car is too wide to get through those gates.

Đề bài là sd enuogh or too nhé