c. Vì AD.AB = AE.AC (chứng minh trên) \(\Rightarrow\dfrac{AD}{AE}=\dfrac{AC}{AB}\)

Xét \(\Delta ADE\) và \(\Delta ACB\) có:

\(\dfrac{AD}{AE}=\dfrac{AC}{AB}\) (chứng minh trên)

\(\widehat{BAC}\) chung

Do đó: \(\Delta ADE\) đồng dạng với \(\Delta ACB\) (cạnh.góc.cạnh)

\(\Rightarrow\widehat{ADE}=\widehat{ACB}\) hay \(\widehat{ADK}=\widehat{ACB}\) (1)

Xét \(\Delta ABC\) vuông tại \(A\) có: \(AI\) là đường trung tuyến ứng với cạnh huyền \(BC\) (do \(I\) là trung điểm \(BC\))

\(\Rightarrow AI=BI=CI=\dfrac{1}{2}BC\)

Xét \(\Delta ABI\) có: \(AI=BI\) (chứng minh trên)

\(\Rightarrow\Delta ABI\) cân tại \(I\Rightarrow\widehat{ABI}=\widehat{BAI}\) hay \(\widehat{ABC}=\widehat{DAK}\) (2)

Từ (1) và (2) \(\Rightarrow\widehat{ABC}+\widehat{ACB}=\widehat{DAK}+\widehat{ADK}\) hay \(\widehat{BAC}=\widehat{AKD}\)

mà \(\widehat{BAC}=90^o\) (vì \(\Delta ABC\) vuông tại \(A\)) \(\Rightarrow\widehat{AKD}=90^o\)

\(\Rightarrow AK\) là đường cao của \(\Delta ADE\)  vuông tại \(A.\)

\(\Rightarrow\dfrac{1}{AK^2}=\dfrac{1}{AD^2}+\dfrac{1}{AE^2}\) (đpcm)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{\left(a+b-c\right)+\left(b+c-a\right)+\left(c+a-b\right)}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)

\(\Rightarrow\left\{{}\begin{matrix}a+b-c=c\\b+c-a=a\\c+a-b=b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)

Do đó: \(P=\left(1+\dfrac{b}{a}\right).\left(1+\dfrac{c}{b}\right).\left(1+\dfrac{a}{c}\right)=\dfrac{a+b}{a}.\dfrac{b+c}{b}.\dfrac{c+a}{c}=\dfrac{2c}{a}.\dfrac{2a}{b}.\dfrac{2b}{c}=\dfrac{8abc}{abc}=8\)

1. B
2. C
3. D
4. C
5. B
6. A
7. B
8. C
9. D
10. A

6. C
7. D
8. A
Bài 4:
1. Incorrect
2. Correct
3. Correct
4. Incorrect
5. Incorrect
6. Correct
7. Incorrect
8. Correct

Bài 5:

- \(A=\dfrac{2^2}{3}.\dfrac{3^2}{8}.\dfrac{4^2}{15}.\dfrac{5^2}{24}.\dfrac{6^2}{35}.\dfrac{7^2}{48}.\dfrac{8^2}{63}.\dfrac{9^2}{80}\)

\(=\dfrac{2^2}{3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}.\dfrac{6^2}{5.7}.\dfrac{7^2}{6.8}.\dfrac{8^2}{7.9}.\dfrac{9^2}{8.10}\)

\(=\dfrac{2^2.3^2.4^2.5^2.6^2.7^2.8^2.9^2}{3.2.4.3.5.4.6.5.7.6.8.7.9.8.10}\)

\(=\dfrac{2^2.3^2.4^2.5^2.6^2.7^2.8^2.9^2}{2.3^2.4^2.5^2.6^2.7^2.8^2.9.10}\)

\(=\dfrac{9}{5}\)

- \(B=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}...\dfrac{2499}{2500}\)

\(=\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.\dfrac{4.6}{5^2}...\dfrac{49.51}{50^2}\)

\(=\dfrac{2.4.3.5.4.6...49.51}{3^2.4^2.5^2...50^2}\)

\(=\dfrac{2.51}{3.50}\)

\(=\dfrac{17}{25}\)

- \(C=\left(1+\dfrac{8}{10}\right)\left(1+\dfrac{8}{22}\right)\left(1+\dfrac{8}{36}\right)...\left(1+\dfrac{8}{8352}\right)\)

\(=\dfrac{18}{10}.\dfrac{30}{22}.\dfrac{44}{36}...\dfrac{8360}{8352}\)

\(=\dfrac{2.9}{10}.\dfrac{3.10}{2.11}.\dfrac{4.11}{3.12}...\dfrac{88.95}{87.96}\)

\(=\dfrac{2.9.3.10.4.11...88.95}{10.2.11.3.12...87.96}\)

\(=\dfrac{9.88}{96}\)

\(=\dfrac{3}{4}\)

Câu 2:

\(\dfrac{3}{2\sqrt{6}+5}+\dfrac{3}{2\sqrt{6}-5}\)

\(=\dfrac{3\left(2\sqrt{6}-5\right)+3\left(2\sqrt{6}+5\right)}{\left(2\sqrt{6}-5\right)\left(2\sqrt{6}+5\right)}\)

\(=\dfrac{6\sqrt{6}-15+6\sqrt{6}+15}{24-25}\)

\(=-12\sqrt{6}\)

Câu 3:

a) Điều kiện: \(x\ge2\)

\(\sqrt{x-2}+1=4\)

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

\(\Leftrightarrow x-2=9\)

\(\Leftrightarrow x=11\) (nhận)

Vậy phương trình có tập nghiệm là: \(S=\left\{11\right\}.\)

b) \(\sqrt{4x^2}=3\)

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

\(\Leftrightarrow\left|2x\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};-\dfrac{3}{2}\right\}.\)

Câu 4:

a) \(B=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{1}{\sqrt{x}-1}\right]:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)

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

b) Yêu cầu bài toán: \(\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{1}{3}\)

\(\Rightarrow3\left(\sqrt{x}-1\right)=\sqrt{x}\)

\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}\)

\(\Leftrightarrow2\sqrt{x}=3\)

\(\Leftrightarrow\sqrt{x}=\dfrac{3}{2}\)

\(\Leftrightarrow x=\dfrac{9}{4}\) (nhận)

Vậy với \(x=\dfrac{9}{4}\) thì \(B=\dfrac{1}{3}.\)

Đề phải là \(A=1-\left(\dfrac{2x^2+x-1}{1-x^2}+\dfrac{2x^3-x+x^2}{1+x^3}\right).\dfrac{\left(1-x\right)\left(x^2-x\right)}{2x-1}\) thì mới làm đc.

Điều kiện xác định: \(\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne\dfrac{1}{2}\end{matrix}\right.\)

\(A=1-\left[\dfrac{2x^2+x-1}{\left(1-x\right)\left(1+x\right)}+\dfrac{x\left(2x^2+x-1\right)}{\left(1+x\right)\left(1-x+x^2\right)}\right].\dfrac{\left(1-x\right).x\left(x-1\right)}{2x-1}\)

\(=1-\left\{\dfrac{\left(2x^2-x\right)+\left(2x-1\right)}{\left(1-x\right)\left(1+x\right)}+\dfrac{x\left[\left(2x^2-x\right)+\left(2x-1\right)\right]}{\left(1+x\right)\left(1-x+x^2\right)}\right\}.\dfrac{\left(1-x\right).x\left(x-1\right)}{2x-1}\)

\(=1-\left\{\dfrac{x\left(2x-1\right)+\left(2x-1\right)}{\left(1-x\right)\left(1+x\right)}+\dfrac{x\left[x\left(2x-1\right)+\left(2x-1\right)\right]}{\left(1+x\right)\left(1-x+x^2\right)}\right\}.\dfrac{\left(1-x\right).x\left(x-1\right)}{2x-1}\)

\(=1-\left[\dfrac{\left(x+1\right)\left(2x-1\right)}{\left(1-x\right)\left(x+1\right)}+\dfrac{x\left(x+1\right)\left(2x-1\right)}{\left(x+1\right)\left(1-x+x^2\right)}\right].\dfrac{\left(1-x\right).x\left(x-1\right)}{2x-1}\)

\(=1-\left[\dfrac{2x-1}{1-x}+\dfrac{x\left(2x-1\right)}{1-x+x^2}\right].\dfrac{\left(1-x\right).x\left(x-1\right)}{2x-1}\)

\(=1-\dfrac{\left(2x-1\right)\left(1-x+x^2\right)+x\left(2x-1\right)\left(1-x\right)}{\left(1-x\right)\left(1-x+x^2\right)}.\dfrac{\left(1-x\right).x\left(x-1\right)}{2x-1}\)

\(=1-\dfrac{\left(2x-1\right)\left[\left(1-x+x^2\right)+x\left(1-x\right)\right]}{\left(1-x\right)\left(1-x+x^2\right)}.\dfrac{\left(1-x\right).x\left(x-1\right)}{2x-1}\)

\(=1-\dfrac{\left(2x-1\right)\left[1-x+x^2+x-x^2\right]}{\left(1-x\right)\left(1-x+x^2\right)}.\dfrac{\left(1-x\right).x\left(x-1\right)}{2x-1}\)

\(=1-\dfrac{\left(2x-1\right)}{\left(1-x\right)\left(1-x+x^2\right)}.\dfrac{\left(1-x\right).x\left(x-1\right)}{2x-1}\)

\(=1-\dfrac{x\left(2x-1\right)\left(1-x\right)\left(x-1\right)}{\left(2x-1\right)\left(1-x\right)\left(1-x+x^2\right)}\)

\(=1-\dfrac{x\left(x-1\right)}{1-x+x^2}\)

\(=\dfrac{1-x+x^2-x\left(x-1\right)}{1-x+x^2}\)

\(=\dfrac{1-x+x^2-x^2+x}{1-x+x^2}\)

\(=\dfrac{1}{1-x+x^2}\)

đề có bị lỗi ko vậy?

hết rồi

Đề phải là \(H=\left(\dfrac{1}{x+1}-\dfrac{3}{x^3+1}+\dfrac{3}{x^2-x+1}\right).\dfrac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\dfrac{2x-2}{x^2+2x}\) thì mới làm đc.

Điều kiện xác định: \(\left\{{}\begin{matrix}x\ne-2\\x\ne-1\\x\ne0\end{matrix}\right.\)

\(H=\left[\dfrac{1}{x+1}-\dfrac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{3}{x^2-x+1}\right].\dfrac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\dfrac{2x-2}{x\left(x+2\right)}\)

\(=\dfrac{x^2-x+1-3+3\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\dfrac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\dfrac{2x-2}{x\left(x+2\right)}\)

\(=\dfrac{x^2-x+1-3+3x+3}{\left(x+1\right)\left(x^2-x+1\right)}.\dfrac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\dfrac{2x-2}{x\left(x+2\right)}\)

\(=\dfrac{x^2+2x+1}{\left(x+1\right)\left(x^2-x+1\right)}.\dfrac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\dfrac{2x-2}{x\left(x+2\right)}\)

\(=\dfrac{\left(x+1\right)^2}{\left(x+1\right)\left(x^2-x+1\right)}.\dfrac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\dfrac{2x-2}{x\left(x+2\right)}\)

\(=\dfrac{3\left(x+1\right)^2.\left(x^2-x+1\right)}{\left(x+1\right)^2.\left(x^2-x+1\right)\left(x+2\right)}-\dfrac{2x-2}{x\left(x+2\right)}\)

\(=\dfrac{3}{x+2}-\dfrac{2x-2}{x\left(x+2\right)}\)

\(=\dfrac{3x-\left(2x-2\right)}{x\left(x+2\right)}\)

\(=\dfrac{3x-2x+2}{x\left(x+2\right)}\)

\(=\dfrac{x+2}{x\left(x+2\right)}\)

\(=\dfrac{1}{x}\)