Viet Nam is in the Soul - East Asia . It has lots of beautiful mountains rivers and beaches . There are tow long river the mekong rivers is the south rive in the soul east asia and of couse it is longer than the red rivers . The mekong river starts in tibet and flows into to the bien dong phanxipang is the hingest mountian in Viet Nam . It 3,14 meters Viet Nam aslo has many nice beaches such as Sam Son , Nha Trang , Vung tau .

Học tốt

1. lots

2. there

3. river

4. south

5. longest

6. than

7. flows into

8. the 

9. heigh

10. has

Sửa lại tí: + (-3^x) là + 3^x vì + số lớn như vậy mà \(\frac{9^{1006}-1}{4}\)> 0 nên sửa như vậy và \(\frac{9^{1006}-1}{4}\)đổi thành \(\frac{9^{1006}+1}{4}\)

Cái đó lát nữa biết.

Bg

Ta có: 1 - 3 + 3² - 3³ +...+ 3^x = \(\frac{9^{1006}+1}{4}\)

Đặt B = 1 - 3 + 3² - 3³ +...+ 3^x:

=> B = 1 - 3 + 3² - 3³ +...+ 3^x

=> 3B = 3 - 3² + 3³ - 3⁴ +...+ 3^{x + 1}  

=> 3B + B = 3 - 3² + 3³ - 3⁴ +...+ 3^{x + 1} + (1 - 3 + 3² - 3³ +...+ 3^x)

=> 4B = 3 - 3² + 3³ - 3⁴ +...+ 3^{x + 1} + 1 - 3 + 3² - 3³ +...+ 3^x 

=> 4B = (3 - 3) + (3² - 3²) + (3³ - 3³) +...+ (3^x - 3^x) + 3^{x + 1} + 1

=> 4B = 3^{x + 1} + 1

=> B = \(\frac{3^{x+1}+1}{4}\)

=> \(\frac{3^{x+1}+1}{4}=\frac{9^{1006}+1}{4}\)

=> 3^{x + 1} + 1 = 9¹⁰⁰⁶ + 1

=> 3^{x + 1} = 9¹⁰⁰⁶

=> 3^{x + 1} = 9¹⁰⁰⁶ 

=> 3^{x + 1} = (3²)¹⁰⁰⁶

=> 3^{x + 1} = 3^2.1006 

=> 3^{x + 1} = 3²⁰¹² 

=> x + 1 = 2012

=> x       = 2012 - 1

=> x       = 2011

đề mình ra sai hay sao hả bạn

Trả lời:

\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{12+4\sqrt{3}+1}}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{6}+\sqrt{2}}\)

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

\(A=\frac{\sqrt{2}.\sqrt{4+2\sqrt{3}}}{\sqrt{2}.\left(\sqrt{3}+1\right)}\)

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

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

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

\(A=1\)

không có cây trả lời

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

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

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

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

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

Dễ thấy theo AM - GM ta có:

\(P\ge3\sqrt[3]{\sqrt{\frac{a+b}{c+ab}\cdot\sqrt{\frac{b+c}{a+bc}}\cdot\sqrt{\frac{c+a}{b+ca}}}}\)

Ta cần chứng minh \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(c+ab\right)\left(a+bc\right)\left(b+ca\right)\)

Mặt khác theo AM - GM:

\(\left(c+ab\right)\left(a+bc\right)\le\frac{\left(c+ab+a+bc\right)^2}{4}=\frac{\left(b+1\right)^2\left(a+c\right)^2}{4}\)

Tương tự thì:

\(\left(c+ab\right)\left(a+bc\right)\left(b+ca\right)\le\frac{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(a+b\right)\left(b+c\right)\left(c+a\right)}{8}\)

Ta cần chứng minh:\(\left(a+1\right)\left(b+1\right)\left(c+1\right)\le8\)

Áp dụng tiếp AM - GM:

\(\left(a+1\right)\left(b+1\right)\left(c+1\right)\le\frac{\left(a+1+b+1+c+1\right)^3}{27}=8\)

Vậy ta có đpcm

Chuyên Phan năm nay :))

j) \(\frac{8}{3}.\frac{2}{5}.\frac{3}{8}.10.\frac{19}{92}=\left(\frac{8}{3}.\frac{3}{8}\right).\left(\frac{2}{5}.10\right).\frac{19}{92}=1.4.\frac{19}{92}\)

\(=\frac{19}{23}\)

k)\(\frac{-5}{7}.\frac{2}{11}.\frac{-13}{17}.\frac{19}{12}.\frac{17}{13}=\left(\frac{2}{11}.\frac{19}{12}\right).\left(\frac{-13}{17}.\frac{17}{13}\right).\frac{-5}{7}\)

\(=\frac{-19}{66}.\frac{-5}{7}=\frac{95}{462}\)

l)\(\frac{12}{19}.\frac{7}{15}.\frac{-13}{17}.\frac{19}{12}.\frac{17}{13}=\left(\frac{12}{19}.\frac{19}{12}\right).\left(\frac{-13}{17}.\frac{17}{13}\right).\frac{7}{15}\)

\(=\frac{-7}{15}\)

cậu tham khảo trên này ạ, nếu đúng cho mk 1 t.i.c.k ạ, thank nhiều

mình nghĩ đề như này mới đúng chứ :))

\(\sqrt{2x-5}=\sqrt{5}\)

<=> 2x - 5 = 5

<=> 2x = 5 + 5

<=> 2x = 10

<=> x = 5

\(\sqrt{2x}-5=\sqrt{5}\left(dk:x\ge0\right)\)

\(< =>\sqrt{2x}=\sqrt{5}\left(1+\sqrt{5}\right)\)

\(< =>\sqrt{x}=\frac{\sqrt{5}\left(1+\sqrt{5}\right)}{\sqrt{2}}\)

\(< =>x=\frac{5\left(1+\sqrt{5}\right)^2}{2}\)