Ex6

1 was

2 had been

3 found

4 had talked

5 had spoken

6 snowed

7 were

8 painted

9 had

10 had known

Ex7

2 had got

3 had hurt

4 were

5 were

6 was

7 wanted

8 ?

9 stopped

10 had eaten

9. is/ tells

10. are

11. has

12. isn't

--------

B3

1. is

2. teaches

3. learn

4. comes

5. has

6. goes

7. meets

8. drinks

------------

B4

1. plays

2. drink

3. starts

4. live

5. do

6. speak

7. watches

------------

B5

1. does -> do

2. Do -> Does

3. watch -> watches

4. goes -> go

5. carries -> carry

6. don't -> doesn't

7. plays -> play

8. are -> is

bài nào vs ak

có refer mà sao xoá bài em

đọc lại đề bài hộ tớ vs.

Bài 5:

1: \(A=\frac{1}{2\cdot15}+\frac{3}{11\cdot2}+\frac{4}{1\cdot11}+\frac{5}{2\cdot1}\)

\(=7\left(\frac{1}{14\cdot15}+\frac{3}{11\cdot14}+\frac{4}{7\cdot11}+\frac{5}{2\cdot7}\right)\)

\(=7\left(\frac12-\frac17+\frac17-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}\right)\)

\(=7\left(\frac12-\frac{1}{15}\right)=7\cdot\frac{13}{30}=\frac{91}{30}\)

Bài 4:

1: Trên cùng một nửa mặt phẳng bờ chứa tia Ox, ta có: \(\hat{xOy}<\hat{xOt}\left(65^0<130^0\right)\)

nên tia Oy nằm giữa hai tia Ox và Ot

2: tia Oy nằm giữa hai tia Ox và Ot

=>\(\hat{xOy}+\hat{yOt}=\hat{xOt}\)

=>\(\hat{yOt}=130^0-65^0=65^0\)

3: Ta có: tia Oy nằm giữa hai tia Ox và Ot

mà \(\hat{xOy}=\hat{yOt}\left(=65^0\right)\)

nên Oy là phân giác của góc xOt

4: Ta có: \(\hat{xOt}+\hat{mOt}=180^0\) (hai góc kề bù)

=>\(\hat{mOt}=180^0-130^0=50^0\)

BÀi 2:

1: \(x+\frac72=\frac{15}{4}\)

=>\(x+\frac{14}{4}=\frac{15}{4}\)

=>\(x=\frac{15}{4}-\frac{14}{4}=\frac14\)

2: \(0,8+\left|x-\frac12\right|=1\)

=>\(\left|x-\frac12\right|=1-0,8=0,2=\frac15\)

=>\(\left[\begin{array}{l}x-\frac12=\frac15\\ x-\frac12=-\frac15\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac15+\frac12=\frac{7}{10}\\ x=\frac12-\frac15=\frac{3}{10}\end{array}\right.\)

3: \(\left(\frac13-x\right)^2-1\frac39=1\frac49\)

=>\(\left(x-\frac13\right)^2=1+\frac49+1+\frac39=2+\frac79=\frac{25}{9}\)

=>\(\left[\begin{array}{l}x-\frac13=\frac53\\ x-\frac13=-\frac53\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac53+\frac13=\frac63=2\\ x=-\frac53+\frac13=-\frac43\end{array}\right.\)

4: \(-1\le\frac{x}{5}<\frac15\)

=>\(-\frac55\le\frac{x}{5}<\frac15\)

=>-5<=x<1

mà x nguyên

nên x∈{-5;-4;-3;-2;1;0}

66 * ( 12-12)*(132*6+66) = 66 x 0 x *(132*6+66) = 0

   66 x ( 12 - 12 ) x ( 132 x 6 + 66 )

= 66 x 0 x ( 132 x 6 + 66 )

= 0

Vì tam giác ABH vuông tại H

\(\Rightarrow AH^2+x^2=AB^2\)

mà AH = 4,8 cm; AB = 6 cm

\(\Rightarrow4,8^2+x^2=6^2\) \(\Rightarrow23,04+x^2=36\) \(\Rightarrow x^2=36-23,04=12,96\) \(\Rightarrow x=3,6\left(cm\right)\)

Vì tam giác ACH vuông tại H

\(\Rightarrow AH^2+y^2=AC^2\)

mà AH = 4,8 cm; AC = 8 cm

\(\Rightarrow4,8^2+y^2=8^2\) \(\Rightarrow23,04+y^2=64\) \(\Rightarrow y^2=64-23,04=40,96\) \(\Rightarrow y=6,4\left(cm\right)\)

Vậy x = 3,6 cm; y = 6,4 cm

~~ Chúc bạn học tốt ~~

áp dụng định lí Py-ta-go vào \(\Delta\) vuông AHC, ta có:

\(AC^2=AH^2+HC^2\)

\(\Rightarrow\)\(HC^2=AC^2-AH^2\)

    \(HC^2=8^2-4,8^2\)

    \(HC^2=64-23,04\)

     \(HC^2=40,96\)

    \(\Rightarrow\)\(HC=\sqrt{40,96}=6,4\)

vậy \(y\)\(=\)\(6,4\)