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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Áp dụng định lí Pytago vào ΔABC vuông tại A, ta được:
\(BC^2=AB^2+AC^2\)
\(\Leftrightarrow BC^2=6^2+8^2=100\)
hay BC=10(cm)
Áp dụng hệ thức lượng trong tam giác vuông vào ΔABC vuông tại A có AH là đường cao ứng với cạnh huyền BC, ta được:
\(AH\cdot BC=AB\cdot AC\)
\(\Leftrightarrow AH\cdot10=6\cdot8=48\)
hay AH=4,8(cm)
![](https://rs.olm.vn/images/avt/0.png?1311)
IV
1 to have
2 making
3 leaving
4 seeing
5 to get
6 arguing - working
7 to have
8 to seeing
9 not touching
10 to disappoint
V
1 on - on
2 at - at
3 in - in
4 at
5 at
6 in
7 in - in
8 at - in
9 in - at
10 in
VI
1 are - reach
2 comes
3 flies
4 have just decided - will undertake
5 would take
6 was
8 am attending - was attending
9 arrived - was waiting
10 had lived
VII
1 send - will receive
2 will - improve - do
3 will - has
4 doesn't phone - will leave
tờ 2
5 don't study - won't oas
VIII
1 had - would learn
2 told - would be
3 lived - would do
4 would help - knew
5 would buy - had
IX
1 went
2 were
3 wrote
4 could
5 bought
6 studied
7 went
8 would stop
9 were
10 lead
X
1 He opened the window in order to let fresh air in
2 I took my camera so that I could take some phôt
3 He studied really hard in order to get better marks
4 Jason learns Chinese to work in China
5 I've collected money in order that I will buy a new car
XI
1 A new museum has been built in the city center by the council
2The explosion had been caused by a bomb
3 Their flat was broken into last month
4 Jane won't be invited to his birthday party by him
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1.
Tờ vé số có dạng \(\overline{a_1a_2a_3a_4a_5a_6}\in A=\left\{0;1;2;3;4;5;6;7;8;9\right\}\)
\(;a_i\ne a_j\)
Chọn \(a_1\ne0\) nên \(a_1\) có 9 cách chọn.
5 số còn lại là chỉnh hợp chập 5 của 8 số còn lại \(\in A\backslash\left\{a_1\right\}\)
\(\Rightarrow\)Có \(A_8^5\) cách.
Vậy có tất cả \(A_8^5\cdot9=60480\) vé số.
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 9:
a: Ta có: \(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow x-5=0\)
hay x=5
b: ta có: \(4x^2-4x=-1\)
\(\Leftrightarrow4x^2-4x+1=0\)
\(\Leftrightarrow2x-1=0\)
hay \(x=\dfrac{1}{2}\)
c: Ta có: \(\left(2x-1\right)^2=\left(3x-2\right)^2\)
\(\Leftrightarrow\left(3x-2\right)^2-\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left(3x-2-2x+1\right)\left(3x-2+2x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{5}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 8:
a: \(73^2-27^2=\left(73-27\right)\left(73+27\right)=4600\)
b: \(63^2-27^2+72^2-18^2\)
\(=\left(63-18\right)\left(63+18\right)+\left(72-27\right)\left(72+27\right)\)
\(=45\cdot\left(63+18+72+27\right)\)
\(=45\cdot180=8100\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: Ta có: \(\left(x+y\right)^2-x^2+4xy-4y^2\)
\(=\left(x+y\right)^2-\left(x-2y\right)^2\)
\(=\left(x+y-x+2y\right)\left(x+y+x-2y\right)\)
\(=3y\cdot\left(2x-y\right)\)
c: Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)
\(=2y^3+6x^2y\)
\(=2y\left(3x^2+y^2\right)\)