hình thức đăng vui phương pháp lập phương hai vế sau đó nhẩm nghiệm dùng tiếp sơ đồ hoc-ne :))) là ok

\(x^3=6+\sqrt{\frac{847}{27}}+6-\sqrt{\frac{847}{27}}+3.\sqrt[3]{\left[6^2-\left(\sqrt{\frac{847}{27}}\right)^2\right]}.x\)

\(\Rightarrow x^3=12+3.\sqrt[3]{\frac{125}{27}}x\)

\(\Leftrightarrow x^3-5x-12=0\)

\(\Leftrightarrow x^3-9x+4x-12=0\)

\(\Leftrightarrow x\left(x+3\right)\left(x-3\right)+4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+4\right)=0\).Vì \(x^2+3x+4=x^2+2.\frac{3}{2}.x+\frac{9}{4}+\frac{7}{4}=\left(x+\frac{3}{2}\right)^2+\frac{7}{4}>0\)

\(\Leftrightarrow x=3\)

Sao lại phân tích như thế????? Dòng thứ 4 ko biết dùng sơ đồ hoc-ne à?????

đặt \(a=\sqrt[3]{6+\sqrt{\frac{847}{27}}};b=\sqrt[3]{6-\sqrt{\frac{847}{27}}}\). dễ thấy a> 0; b > 0

=> \(a^3+b^3=6+\sqrt{\frac{847}{27}}+6-\sqrt{\frac{847}{27}}=12\); \(a.b=\sqrt[3]{6+\sqrt{\frac{847}{27}}}.\sqrt[3]{6-\sqrt{\frac{847}{27}}}=\sqrt[3]{36-\frac{847}{27}}=\frac{5}{3}\)

Có: (a+ b)³ = a³ + b³ + 3ab (a+ b)

=> (a + b)³ = 12 + 3. \(\frac{5}{3}\).(a + b) = 12+ 5.(a + b)

=> (a + b)³ - 5.(a +b)  - 12 = 0 

<=> (a + b)³ - 9.(a + b)  + 4.(a + b) - 12 = 0

<=> (a + b). [(a + b)² - 9] + 4.(a + b - 3) = 0 <=> (a + b).(a + b + 3).(a + b- 3) + 4.(a + b - 3) = 0 

<=> (a+ b - 3).[(a + b)(a+ b+ 3) + 4] = 0

<=> a+ b = 3 hoặc (a + b)(a+ b+ 3) + 4 = 0 

tuy nhiên : Vì a > 0; b > 0 nên (a + b)(a+ b+ 3) + 4 > 0 

vậy a + b = 3 => điều phải chứng minh

\(P=\sqrt[3]{6+\sqrt{\frac{847}{27}}}+\sqrt[3]{6+\sqrt{\frac{847}{27}}}\)

Ta áp dụng hằng đẳng thức : 

\(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)

\(\Rightarrow P^3=6+\sqrt{\frac{847}{27}}+6-\sqrt{\frac{847}{27}}+3\sqrt[3]{6+\sqrt{\frac{847}{27}}}.\sqrt[3]{6-\sqrt{\frac{847}{27}}}\left(3\sqrt[3]{6+\sqrt{\frac{847}{27}}}.\sqrt[3]{6-\sqrt{\frac{847}{27}}}\right)\)

\(\Leftrightarrow P^3=12+3.\sqrt[3]{36-\frac{847}{27}}.P=12+5P\)

\(\Leftrightarrow P^3-5P-12=0\)

\(\Leftrightarrow\left(P-3\right)\left(P^2+3P+4\right)=0\)

\(\Leftrightarrow P=3\) hoặc \(P^3+3P+4=0\) vô nghiệm

Vậy \(P=3\)

a)

\(\sqrt{12}-\sqrt{27}+\sqrt{3}=\sqrt{4}.\sqrt{3}-\sqrt{9}.\sqrt{3}+\sqrt{3}=2\sqrt{3}-3\sqrt{3}+\sqrt{3}\)

\(=\sqrt{3}(2-3+1)=0\)

b)

\(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}=\sqrt{4}.\sqrt{63}-\sqrt{4}.\sqrt{175}+\sqrt{4}.\sqrt{252}-\sqrt{4}.\sqrt{112}\)

\(=2(\sqrt{63}-\sqrt{175}+\sqrt{252}-\sqrt{112})\)

\(=2(\sqrt{9}.\sqrt{7}-\sqrt{25}.\sqrt{7}+\sqrt{36}.\sqrt{7}-\sqrt{16}.\sqrt{7})\)

\(=2(3\sqrt{7}-5\sqrt{7}+6\sqrt{7}-4\sqrt{7})=2\sqrt{7}(3-5+6-4)=0\)

------------------

\(\sqrt{3}(\sqrt{12}+\sqrt{27}-\sqrt{3})=\sqrt{36}+\sqrt{81}-\sqrt{9}\)

\(=\sqrt{6^2}+\sqrt{9^2}-\sqrt{3^2}=6+9-3=12\)

c)

\(\frac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}=\frac{\sqrt{2}.\sqrt{3}+\sqrt{2}.\sqrt{5}}{\sqrt{7}.\sqrt{3}+\sqrt{7}.\sqrt{5}}=\frac{\sqrt{2}(\sqrt{3}+\sqrt{5})}{\sqrt{7}(\sqrt{3}+\sqrt{5})}=\frac{\sqrt{2}}{\sqrt{7}}\)

\(\frac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+\sqrt{45}}=\frac{\sqrt{81}.\sqrt{5}+3\sqrt{9}.\sqrt{3}}{3\sqrt{3}+\sqrt{9}.\sqrt{5}}=\frac{9\sqrt{5}+9\sqrt{3}}{3\sqrt{3}+3\sqrt{5}}\)

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

d)

\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}-(\sqrt{6}+\sqrt{9}+\sqrt{12})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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

Bạn không sửa thì m sửa.

Sửa đề: \(P=\sqrt[3]{\sqrt{\frac{2303}{27}}+6}-\sqrt[3]{\sqrt{\frac{2303}{27}}-6}\)

\(P^3=\sqrt{\frac{2303}{27}}+6-\left(\sqrt{\frac{2303}{27}}-6\right)-\frac{3.11.P}{3}\)

\(\Leftrightarrow P^3=12-11P\)

\(\Leftrightarrow P^3+11P-12=0\)

\(\Leftrightarrow\left(P-1\right)\left(P^2+P+12\right)=0\)

Vì \(P^2+P+12>0\) nên ta có

\(P=1\)

Đề bạn chép sai rồi. Sửa lại đi b

theo tớ là cậu chép sai đề rồi cậu chép lại đi

thực hiện phép tính nha cám ơn m.ng