ĐK: \(x\ge-7\)

PT \(\Leftrightarrow\left(\sqrt[3]{x-8}-\left(x-8\right)\right)+\left[\sqrt{x+7}-4\right]+\left(x-9\right)\left(x^2+x+2\right)=0\)

\(\Leftrightarrow\frac{-\left(x-9\right)\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}+\frac{x-9}{\sqrt{x+7}+4}+\left(x-9\right)\left(x^2+x+2\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left[x^2+x+2+\frac{1}{\sqrt{x+7}+4}-\frac{\left(x-7\right)\left(x-8\right)}{\left(\sqrt[3]{x-8}\right)^2+\left(x-8\right)\sqrt[3]{x-8}+\left(x-8\right)^2}\right]=0\)

\(\Leftrightarrow x=9\) 

P/s:em chả biết đánh giá cái ngoặc to thế nào nữa:((((

6-2x=0:(x mu 2+x+1)

6-2x=0

2x=6

x=3

Dung thi k cho mih nha

Cam on bn nhieu luong!!!

\(\left(x^2+x+1\right)\left(6-2x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2+x+1=0\\6-2x=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x\left(x+1\right)+1=0\\2x=6\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x\left(x+1\right)=-1\\x=3\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x\text{ là tích của hai số liên tiếp nên khác 1 }\Rightarrow\text{ }x\in\varnothing\\x=3\end{cases}}\)

\(\text{Vậy }x=3\)

=\(\sqrt{27+2.3\sqrt{3}.\sqrt{2}+2}-\sqrt{27-2.3\sqrt{3}.\sqrt{5}+5}\)

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

=\(3\sqrt{3}+\sqrt{2}-|3\sqrt{3}-\sqrt{5}|\)

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

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

Chuc ban hoc tot

\(DK:x\ge1\)

\(\Leftrightarrow\left(3\sqrt{x-1}-3\right)+\left(\sqrt{x+2}-2\right)-\left(10x-20\right)-\left(6\sqrt{x^2+x-2}-12\right)=0\)

\(\Leftrightarrow3\left(\sqrt{x-1}-1\right)+\left(\sqrt{x+2}-2\right)-10\left(x-2\right)-6\left(\sqrt{x^2+x-2}-2\right)=0\)

\(\Leftrightarrow\frac{3\left(x-2\right)}{\sqrt{x-1}+1}+\frac{x-2}{\sqrt{x+2}+2}-10\left(x-2\right)-\frac{6\left(x^2+x-6\right)}{\sqrt{x^2+x-2}+2}=0\)

\(\Leftrightarrow\frac{3\left(x-2\right)}{\sqrt{x-1}+1}+\frac{x-2}{\sqrt{x+2}+2}-10\left(x-2\right)-\frac{6\left(x-2\right)\left(x+3\right)}{\sqrt{x^2+x-2}+2}=0\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{3}{\sqrt{x-1}+1}+\frac{1}{\sqrt{x+2}}-10-\frac{6x+18}{\sqrt{x^2+x-2}+2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\\frac{3}{\sqrt{x-1}+1}+\frac{1}{\sqrt{x+2}}=10+\frac{6x+18}{\sqrt{x^2+x-2}+2}\end{cases}}\)

Cái PT 2 nó vô nghiệm,chắc la quy dong lên là duoc

Vay PT co nghiem la \(x=2\)

Vẫn là liên hợp nhưng em có cách khác:D Nó sẽ nhanh hơn ở chỗ xử lý cái ngoặc to đấy:)

\(ĐK:x\ge1\)

\(PT\Leftrightarrow6\left(\sqrt{x^2+x-2}-x\right)+12x-24+3\left[\left(x-1\right)-\sqrt{x-1}\right]+x-\sqrt{x+2}=0\)

\(\Leftrightarrow\frac{6\left(x-2\right)}{\sqrt{x^2+x-2}+x}+12\left(x-2\right)+\frac{3\left(x-2\right)\left(x-1\right)}{\left(x-1\right)+\sqrt{x-1}}+\frac{\left(x-2\right)\left(x+1\right)}{x+\sqrt{x+2}}=0\)

\(\Leftrightarrow\left(x-2\right)\left[\frac{6}{\sqrt{x^2+x-2}+x}+12+\frac{3\left(x-1\right)}{\left(x-1\right)+\sqrt{x-1}}+\frac{\left(x+1\right)}{x+\sqrt{x+2}}\right]=0\)

Cái ngoặc to không cần đánh giá cũng >0 :D. Vậy x = 2 (TM)

P/s: Em có tính sai chỗ nào không nhỉ:))

8x3-8-8-8-8-8-8=-24

học tốt

8 . 3 - 8 - 8 - 8 - 8 - 8 - 8 

= 24 - ( 8 . 6 )

= 24 - 48 

= -24