1. It's the first time she has studied Japanese

2. It is two years since she last wrote to me

3. I last attended the club in 1988

4. The last time I saw him was 2 months ago

5. The last time she swam in the river was 3 years ago

6. It is 5 years since I last visited the city museum

7. I last went swimming 2 years ago

8. This is the first time I have eaten this kind of food

9. The last time we went to the park was 2 days ago

10. We last met each other five years ago

\(AH=HC.tanC\Rightarrow HC=\dfrac{AH}{tanC}=\dfrac{8}{tan30độ}=8\sqrt{3}\left(cm\right)\)

\(AH^2=BH.HC\Rightarrow BH=\dfrac{AH^2}{HC}=\dfrac{8^2}{8\sqrt{3}}=\dfrac{8\sqrt{3}}{3}\left(cm\right)\)

\(\Rightarrow BC=BH+HC=8\sqrt{3}+\dfrac{8\sqrt{3}}{3}=\dfrac{32\sqrt{3}}{3}\left(cm\right)\)

\(AB=BC.sinC=\dfrac{32\sqrt{3}}{3}.sin30độ=\dfrac{16\sqrt{3}}{3}\left(cm\right)\)

\(AC=BC.cosC=\dfrac{32\sqrt{3}}{3}.cos30độ=16\left(cm\right)\)

\(a)BH=AB.cosB\Rightarrow AB=\dfrac{BH}{cos60độ}=\dfrac{12.5}{cos60độ}=25\left(cm\right)\) (hệ thức về cạnh và góc trong tam giác vuông)

Theo hệ thức lượng trong tam giác vuông ta có:

\(+)AB^2=BH.BC\Leftrightarrow BC=\dfrac{AB^2}{BH}=\dfrac{25^2}{12,5}=50\left(cm\right)\)

\(\Rightarrow HC=BC-BH=50-12,5=37,5\left(cm\right)\)

\(+)AH^2=BH.HC=37,5.12,5=468,75\)

\(\Leftrightarrow AH=\dfrac{25\sqrt{3}}{2}\left(cm\right)\)

Áp dụng đlí pytago vào tam giác ABC vuông tại A ta có:

\(AB^2+AC^2=BC^2\)

\(\Leftrightarrow AC^2=BC^2-AB^2=50^2-25^2=1875\)

\(\Leftrightarrow AC=25\sqrt{3}\left(cm\right)\)

\(x^4+4x^3+6x^2+4x+\sqrt{x^2+2x+17}=3\)

Ta có: \(x^2+2x+17=(x^2+2x+1)+16=\left(x+1\right)^2+16\ge16\)

\(\Rightarrow\sqrt{x^2+2x+17}\ge\sqrt{16}=4\)

\(\Rightarrow x^4+4x^3+6x^2+4x+\sqrt{x^2+2x+17}=3\ge x^4+4x^3+6x^2+4x+4\)

\(\Leftrightarrow x^4+4x^3+6x^2+4x+1\le0\)

\(\Leftrightarrow\left(x+1\right)^4\le0\)

Mà \(\left(x+1\right)^4\ge0\Rightarrow(x+1)^4=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Thử lại ta thấy x=-1 thỏa mãn bài toán

Vậy, pt có nghiệm duy nhất là x=-1

1. I have worked the job for six months

2. I haven't seen her for 10 years

3. I haven't seen him since June

4. This is the first time I have eaten this kind of food

5. It's the first time she has studied Japanese

6. She hasn't had a holiday for two years

7. This is the first time I have driven a car

8. He hasn't come to the cinema for three weeks

9. The last time we went to the park was 2 days ago

10. We haven't met each other for five years

\(\Leftrightarrow x=-1\left(tm\right)\)

Vậy, pt có nghiệm duy nhất là x=-1

\(x^2+4x+5=2\sqrt{2x+3}\)

\(ĐK:x\ge-\dfrac{3}{2}\)

\(pt\Leftrightarrow(2x+3-2\sqrt{2x+3}+1)+x^2+2x+1=0\)

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

Vì \(\left(\sqrt{2x+3}-1\right)^2\ge0;-\left(x+1\right)^2\le0\forall x\)

\(\Rightarrow\left\{{}\begin{matrix}(\sqrt{2x+3}-1)^2=0\\\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2x+3}-1=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2x+3}=1\\x=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3=1\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\left(tm\right)}\)

Kẻ đường cao AH của tam giác ABC

\(sinC=\dfrac{AH}{AC}\)

\(sinB=\dfrac{AH}{AB}\)

\(\Rightarrow\dfrac{sinB}{sinC}=\dfrac{\dfrac{AH}{AB}}{\dfrac{AH}{AC}}=\dfrac{AC}{AB}=\dfrac{b}{c}\Rightarrow\dfrac{b}{sinB}=\dfrac{c}{sinC}\left(1\right)\)

Kẻ đường cao CE của tam giác ABC rồi CMTT ta được:

\(\dfrac{a}{sinA}=\dfrac{b}{sinB}\left(2\right)\)

Từ (1) và (2) suy ra đpcm