Ace nào có nhu cầu lấy các topics và bài essays mẫu của IELTS và TOEFL thì cmt email bên dưới em gửi bài cho ạ :3

1. Despite her dislikes for coffee , she drank it to keep herself warm ( although)

→ Although she didn't like coffee, she drank it to keep herself warm.

2. Mary will take a plane , even though she dislikes flying ( in spite of)

→ In spite of her dislikes flying, Mary will take a plane.

3. In spite of Marcy’s sadness at losing the contest , she managed to smile ( although )

→ Although Marcy was sad at losing the contest, she managed to smile.

4. We took many pictures though the sky was cloudy (despite)

→ Despite the cloudy sky, we took many pictures.

5. Despite her poor memory, the old woman told interesting stories to the children ( even though)

→ The woman told interesting stories to the children, even though her poor memory.

6. Though he has been absent frequently , he has managed to pass the test ( in spite of )

→ In spite of having been absent frequently, he has managed to pass the test.

7. Nancy told me the secret despite having promised not to do so ( though )

→ Nancy told me the secret, though she has promised not to do so.

8. We plan to buy a ticket for the drawing although we know we will not win a prize (even though)

→ We plan to buy a ticket for the drawing, even though we know we will not win a prize.

9. In spite of the high price , my daughters insist on going to the movies every Saturday ( even though)

→ Even though the price was high, my daughter insist on going to the movies every Sunday.

10. He ate the chocolate cake even though he is on a diet ( in spite of)

→ In spite of being on a diet, he ate the chocolate cake.

11. He left school, because his life was hard ( because of)

→ Because of his hard life, he left school.

12. John succeeded in his exam because of his hard and methodical work ( because)

→ John succeeded in his exam, because he worked hard and methodical work.

13. We didn’t want to go out because of the heavy rain (because)

→ Because it rained heavily, we didnt want to go out.

14. Although he had much experience in machinery, he didn’t succeed in repairing this machine ( despite)

→ Despite having much experience in machinery, he didn't succeed in repairing this machine.

15. In spite of the darkness , they continued to work (though)

→ They continued to work, though it was dark.

Bài 1:

a) P= \(\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}-1}{2}\) (x ≥ 0; x ≠ 4)

=\(\left(\frac{x+2}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)\cdot\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{2}{\sqrt{x}-1}\)

= \(\left(\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{2}{\sqrt{x}-1}\)

=\(\left(\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{2}{\sqrt{x}-1}\)

=\(\frac{\left(\sqrt{x}-1\right)^2\cdot2}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\)

=\(\frac{2}{x+\sqrt{x}+1}\)

b) Ta có: x ≥ 0 ⇒ \(\sqrt{x}\) ≥ 0

⇒ \(x+\sqrt{x}+1\) ≥ 1 > 0

mà 2 > 0 ⇒ \(\frac{2}{x+\sqrt{x}+1}\) > 0 ⇒ P > 0

Bài 2:

a) P= \(\left(\frac{2\sqrt{x}+x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\) (x ≥ 0; x ≠ 1)

=\(\left(\frac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right):\left(\frac{x+\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)

=\(\left(\frac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right):\left(\frac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\right)\)

=\(\left(\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right):\left(\frac{x-1}{x+\sqrt{x}+1}\right)\)

=\(\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\cdot\frac{x+\sqrt{x}+1}{x-1}\)

=\(\frac{1}{x-1}\)

b) Ta có: \(\sqrt{P}=\sqrt{\frac{1}{x-1}}\)

= \(\frac{1}{\sqrt{x-1}}\)

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

Thay x vào \(\sqrt{P}\) ta có:

\(\sqrt{P}=\frac{1}{\sqrt{5+2\sqrt{3}-1}}\)

=\(\frac{1}{\sqrt{4+2\sqrt{3}}}\)

=\(\frac{1}{\sqrt{3+2\sqrt{x}+1}}\)

=\(\frac{1}{\sqrt{\left(\sqrt{3}+1\right)^2}}\)

=\(\frac{1}{\left|\sqrt{3}+1\right|}\)

=\(\frac{1}{\sqrt{3}+1}\)

= \(\frac{\sqrt{3}-1}{\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-1\right)}\)

=\(\frac{\sqrt{3}-1}{2}\)

Vậy \(\sqrt{P}=\frac{\sqrt{3}-1}{2}\) khi x = \(5+2\sqrt{3}\)

a) A= \(\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{2}\right)\) (x ≥ 0; x ≠ 4)

= \(\left(\frac{x+2}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)\cdot\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right):\frac{\sqrt{x}-1}{2}\)

=\(\left(\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{2}{\sqrt{x}-1}\)

=\(\left(\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{2}{\sqrt{x}-1}\)

= \(\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\cdot\frac{2}{\sqrt{x}-1}\)

=\(\frac{2}{x+\sqrt{x}+1}\)

b) Ta có: x ≥ 0 ⇒ \(\sqrt{x}\) ≥ 0

⇒x+\(\sqrt{x}\)+1 ≥ 1 > 0

mà 2 > 0

⇒ A > 0 (1)

Ta có:

\(x+\sqrt{x}+1\) ≥ 1

⇒ \(\frac{1}{x+\sqrt{x}+1}\) ≤ 1

⇒\(\frac{2}{x+\sqrt{x}+1}\) ≤ 2

⇒A ≤ 2 (2)

Từ (1) và (2) => 0 < A ≤ 2

1/ \(\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\frac{5-2\sqrt{5}}{2\sqrt{5}-4}\)

=\(\frac{\left(\sqrt{15}-\sqrt{5}\right)\cdot\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+1\right)}+\frac{\left(5-2\sqrt{5}\right)\cdot\left(2\sqrt{5}+4\right)}{\left(2\sqrt{5}-4\right)\cdot\left(2\sqrt{5}+4\right)}\)

=\(\frac{2\sqrt{5}}{2}+\frac{2\sqrt{5}}{4}\)

=\(\sqrt{5}+\frac{\sqrt{5}}{2}\)

=\(\frac{3\sqrt{5}}{2}\)

2/\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)

=\(\frac{\left(\sqrt{15}-\sqrt{12}\right)\cdot\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\cdot\left(\sqrt{5}+2\right)}+\frac{\left(6+2\sqrt{6}\right)\cdot\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}+2\right)\cdot\left(\sqrt{3}-2\right)}\)

=\(\frac{\sqrt{3}}{1}+\frac{2\sqrt{3}}{1}\)

=\(3\sqrt{3}\)

3/\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\)

=\(\frac{\sqrt{3}\cdot\left(3+2\sqrt{3}\right)}{3}+\frac{\left(2+\sqrt{2}\right)\cdot\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\cdot\left(\sqrt{2}-1\right)}-\left(2+\sqrt{3}\right)\)

=\(\frac{6+3\sqrt{3}}{3}+\sqrt{2}-\left(2-\sqrt{3}\right)\)

=\(\frac{3\cdot\left(2+\sqrt{3}\right)}{3}+\sqrt{2}-\left(2+\sqrt{3}\right)\)

=\(2+\sqrt{3}+\sqrt{2}-2-\sqrt{3}\)

=\(\sqrt{2}\)

Câu số 4 bạn có chắc là đúng đề bài không ạ ? Xem lại đề giúp mình nhé, cảm ơn bạn ^^

Hic câu dưới bị giải nhầm nha bạn :<

\(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\)

=\(\sqrt{21-2\sqrt{54}}+\sqrt{6+2\sqrt{18}+3}-\sqrt{24+12\sqrt{3}}\)

=\(\sqrt{18-2\sqrt{54}+3}+\sqrt{6+2\sqrt{18}+3}-\sqrt{24+2\sqrt{108}}\)

=\(\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{3}\right)^2}-\sqrt{18+2\sqrt{108}+6}\)

=\(\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{18}+\sqrt{6}\right)^2}\)

=\(\left|\sqrt{18}-\sqrt{3}\right|+\left|\sqrt{6}+\sqrt{3}\right|-\left|\sqrt{18}+\sqrt{6}\right|\)

=\(\sqrt{18}-\sqrt{3}+\sqrt{6}+\sqrt{3}-\sqrt{18}-\sqrt{6}\)

=0

a) M= \(\left(1-\frac{4\sqrt{x}}{x-1}+\frac{1}{\sqrt{x}-1}\right):\frac{x-2\sqrt{x}}{x-1}\) (x > 0; x ≠ 1; x ≠ 4)

= \(\left(\frac{x-1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}-\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\right)\cdot\)\(\frac{x-1}{x-2\sqrt{x}}\)

= \(\left(\frac{x-1-4\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\right)\cdot\frac{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}{\sqrt{x}\cdot\left(\sqrt{x}-2\right)}\)

=\(\frac{x-3\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\cdot\frac{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}{\sqrt{x}\cdot\left(\sqrt{x}-2\right)}\)

=\(\frac{\sqrt{x}\cdot\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\cdot\frac{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}{\sqrt{x}\cdot\left(\sqrt{x}-2\right)}\)

=\(\frac{\sqrt{x}-3}{\sqrt{x}-2}\)

b)M=\(\frac{1}{2}\)

⇔ \(\frac{\sqrt{x}-3}{\sqrt{x}-2}=\frac{1}{2}\)

⇔ \(2\left(\sqrt{x}-3\right)=\sqrt{x}-2\)

⇔ \(2\sqrt{x}-6=\sqrt{x}-2\)

⇔ \(2\sqrt{x}-6-\sqrt{x}-2=0\)

⇔ \(\sqrt{x}-8=0\)

⇔ \(\sqrt{x}=8\)

⇔ x = 64 (TM)

Vậy x= 64 để M=\(\frac{1}{2}\)

\(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\)

=\(\sqrt{21-2\sqrt{54}}+\sqrt{6+2\sqrt{18}+3}-\sqrt{4\cdot\left(6+3\sqrt{3}\right)}\)

=\(\sqrt{18-2\sqrt{54}+3}+\sqrt{6+2\sqrt{18}+3}-\sqrt{24+12\sqrt{3}}\)

=\(\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{3}\right)^2}-\sqrt{24+2\sqrt{108}}\)

=\(\left|\sqrt{18}-\sqrt{3}\right|+\left|\sqrt{6}+\sqrt{3}\right|-\sqrt{\sqrt{18}+2\sqrt{108}+\sqrt{6}}\)

=\(\left|\sqrt{18}-\sqrt{3}\right|+\left|\sqrt{6}+\sqrt{3}\right|-\sqrt{\left(\sqrt{18}+\sqrt{6}\right)^2}\)

=\(\left|\sqrt{18}-\sqrt{3}\right|+\left|\sqrt{6}+\sqrt{3}\right|-\left|\sqrt{18}-\sqrt{6}\right|\)

=\(\sqrt{18}-\sqrt{3}+\sqrt{6}+\sqrt{3}-\sqrt{18}-\sqrt{6}\)

= 0