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![](https://rs.olm.vn/images/avt/0.png?1311)
a) Để tính giá trị của biểu thức P=(x^3+12x−9)^{2005}=(√3+12√−9)^{2005} với x=3√4(√5+1)−3√4(√5−1). Đầu tiên, ta thay x bằng giá trị đã cho vào biểu thức P: P=(3√4(√5+1)−3√4(√5−1))^3+12(3√4(√5+1)−3√4(√5−1))−9)^{2005} Tiếp theo, ta thực hiện các phép tính để đơn giản hóa biểu thức: P=(4(5+1)^{1/2}−4(5−1)^{1/2})^3+12(4(5+1)^{1/2}−4(5−1)^{1/2})−9)^{2005} =(4√6−4√4)^3+12(4√6−4√4)−9)^{2005} =(4√6−8)^3+12(4√6−8)−9)^{2005} =(64√6−192+96√6−96−9)^{2005} =(160√6−297)^{2005} ≈ 1.332 × 10^3975
b) Để tính giá trị của biểu thức Q=x^3+ax+b=√3+√a+√b^2+√a^3+√3+√a−√b^2+√a^3 với x=3√−b^2+√b^2/4+a^3/(27+3√−b^2−√b^2/4+a^3/27). Tương tự như trên, ta thay x bằng giá trị đã cho vào biểu thức Q: Q=(3√−b^2+√b^2/4+a^3/(27+3√−b^2−√b^2/4+a^3/27))^3+a(3√−b^2+√b^2/4+a^3/(27+3√−b^2−√b^2/4+a^3/27))+b Tiếp theo, ta thực hiện các phép tính để đơn giản hóa biểu thức: Q=(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))^3+a(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))+b =−b^3+3√b^2/4+a^3/(27−3b√b^2/4+a^3/(27))+a(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))+b =−b^3+3√b^2/4+a^3/(27−3b√b^2/4+a^3/(27))+a(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))+b =−b^3+3√b^2/4+a^3/(27−3b√b^2/4+a^3/(27))+a(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))+b =−b^3+3√b^2/4+a^3/(27−3b√b^2/4+a^3/(27))+a(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))+b ≈ −b^3+3√b^2/4+a^3/(27−3b√b^2/4+a^3/(27))+a(−b+√b^2/4+a^3/(27−b+√b^2/4+a^3/27))+b
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x=\dfrac{\sqrt[3]{\left(2+\sqrt{3}\right)^3}\left(2-\sqrt{3}\right)}{\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}}=\dfrac{1}{\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}}\)
Đặt \(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)\(\Leftrightarrow A^3=18+3\sqrt[3]{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\\ \Leftrightarrow A^3=18+3A\sqrt[3]{1}\\ \Leftrightarrow A^3-3A-18=0\\ \Leftrightarrow A=3\\ \Leftrightarrow X=\dfrac{1}{3}\\ \Leftrightarrow Q=\left[3\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{3}\right)^2-1\right]^{2021}=\left(\dfrac{1}{9}-\dfrac{1}{9}-1\right)^{2021}=\left(-1\right)^{2021}=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\dfrac{\sqrt{5}\left(\sqrt{3}-\sqrt{7}\right)}{\sqrt{3}-\sqrt{7}}\right).\left(\sqrt{2}+\sqrt{5}\right)\)
\(=\left(\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}+\sqrt{5}\right)=2-5=-3\)
\(B=\dfrac{12\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}-\dfrac{2\sqrt{3}.\sqrt{3}}{\sqrt{3}}+\dfrac{3}{\sqrt{2}}-\dfrac{3}{\sqrt{3}}\)
\(=\dfrac{12\left(3-\sqrt{3}\right)}{6}-2\sqrt{3}+\dfrac{3\sqrt{2}}{2}-\sqrt{3}\)
\(=2\left(3-\sqrt{3}\right)-3\sqrt{3}+\dfrac{3\sqrt{2}}{2}=6-5\sqrt{3}+\dfrac{3\sqrt{2}}{2}\) (câu này khả năng đề sai, dấu \(\sqrt{3}.\sqrt{2}\) ở mẫu cuối cùng là dấu trừ mới hợp lý)
\(C=\left(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\dfrac{3}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)^2}\)
Dấu giữa 2 dấu ngoặc là dấu chia sẽ hợp lý hơn
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}\cdot\dfrac{\sqrt{x}-2+3}{3}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{x-9}\cdot\dfrac{\sqrt{x}+1}{3}=\dfrac{-\sqrt{x}-1}{\sqrt{x}-3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a:
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-1\right):\dfrac{9-x+x-9-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)^2}=\dfrac{3}{\sqrt{x}-2}\)
b: Khi x=7-4căn 3 thì
\(A=\dfrac{3}{2-\sqrt{3}-2}=\dfrac{3}{-\sqrt{3}}=-\sqrt{3}\)
c: A=3
=>căn x-2=1
=>x=9(loại)
\(a,A=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\left(dkxd:x\ne4,x\ge0,x\ne9\right)\)
\(=\dfrac{x-3\sqrt{x}-x+9}{x-9}:\dfrac{9-x+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{9-x+x-9-x+4\sqrt{x}-4}\)
\(=\dfrac{-3\left(\sqrt{x}-3\right)}{\sqrt{x}-3}.\dfrac{\sqrt{x}-2}{4\sqrt{x}-4-x}\)
\(=\dfrac{-3\left(\sqrt{x}-2\right)}{-\left(x-4\sqrt{x}+4\right)}\)
\(=\dfrac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)^2}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
\(b,x=7-4\sqrt{3}\Rightarrow A=\dfrac{3}{\sqrt{7-4\sqrt{3}}-2}=\dfrac{3}{\sqrt{\left(\sqrt{3}-2\right)^2}-2}=\dfrac{3}{\left|\sqrt{3}-2\right|-2}=\dfrac{3}{-\sqrt{3}+2-2}=\dfrac{\sqrt{3^2}}{-\sqrt{3}}=-\sqrt{3}\)
\(c,A=3\Rightarrow\dfrac{3}{\sqrt{x}-2}=3\\ \Rightarrow\dfrac{3-3\left(\sqrt{x}-2\right)}{\sqrt{x}-2}=0\\ \Rightarrow3-3\sqrt{x}+6=0\\ \Rightarrow-3\sqrt{x}=-9\\ \Rightarrow\sqrt{x}=3\\ \Rightarrow x=9\left(ktm\right)\)
Vậy không có giá trị x thỏa mãn đề bài.
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=3+\left(-2\right)\cdot\sqrt{3}+3\cdot\sqrt{3}-2-\sqrt{3}\)
\(=3-2=1\)
\(B=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b: B<A
=>B-1<0
=>\(\dfrac{\sqrt{x}-1-\sqrt{x}}{\sqrt{x}}< 0\)
=>-1/căn x<0
=>căn x>0
=>x>0 và x<>1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\left(\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right)\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}+4}\)
\(=\dfrac{-3}{2\sqrt{x}+4}\)
Để \(C< -\dfrac{1}{3}\) thì \(\dfrac{-3}{2\sqrt{x}+4}+\dfrac{1}{3}< 0\)
\(\Leftrightarrow-9+2\sqrt{x}+4< 0\)
\(\Leftrightarrow\sqrt{x}< \dfrac{5}{2}\)
hay \(0\le x< \dfrac{25}{4}\)