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\(x=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(\sqrt{5}-2\right)^3}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}=\dfrac{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}=\dfrac{5-4}{\sqrt{5}+3-\sqrt{5}}=\dfrac{1}{3}\)A=\(\left(3\left(\dfrac{1}{3}\right)^3+8\left(\dfrac{1}{3}\right)^2+2\right)^{2009}-3^{2009}=3^{2009}-3^{2009}=0\)
\(x=\dfrac{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}{\sqrt{5}+3-\sqrt{5}}=\dfrac{3}{3}=1\)
\(A=\left(3\cdot1+8\cdot1+2\right)^{2018}=13^{2018}\)
Mẫu của x
\(\sqrt{5}+\sqrt{3^2-2.3.\sqrt{5}+5}=\sqrt{5}+\left|3-\sqrt{5}\right|=3\)
Tử của x
\(\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}=\left(\sqrt{5}+2\right)\sqrt[3]{\left(5\sqrt{5}\right)-3.\left(\sqrt{5}\right)^2.2+3.\sqrt{5}.2^2-2^3}=\left(\sqrt{5}+2\right)\sqrt{\left(\sqrt{5}-2\right)^3}=\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)=5-4=1\)
=> \(x=\dfrac{1}{3}\)
\(A=\left(\dfrac{3}{3^3}+\dfrac{8}{3^2}+2\right)^{1998}=\left(\dfrac{1+8+9}{3^2}\right)^{1998}=2^{1998}\)
\(x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3+2\sqrt{2}}\)
Ta có: Đặt \(A=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\)=> \(A^2=\frac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{\sqrt{5}+1}\)
=> \(A^2=\frac{2\sqrt{5}+2\sqrt{5-4}}{\sqrt{5}+1}=\frac{2\left(\sqrt{5}+1\right)}{\sqrt{5}+1}=2\)=> \(A=\sqrt{2}\)
\(\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
==> \(x=\sqrt{2}-\left(\sqrt{2}+1\right)=-1\)
Do đó: N = (-1)2019 + 3.(-1)2020 - 2.(-1)2021 = -1 + 3 + 2 = 4
5.
ĐKXĐ: ...
\(\Leftrightarrow3x^2-14x-5+\sqrt{3x+1}-4+1-\sqrt{6-x}=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x-5\right)+\frac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\frac{x-5}{1+\sqrt{6-x}}=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x+1+\frac{3}{\sqrt{3x+1}+4}+\frac{1}{1+\sqrt{6-x}}\right)=0\)
\(\Leftrightarrow x=5\)
6.
ĐKXĐ: \(-4\le x\le4\)
\(\Leftrightarrow\frac{\left(\sqrt{x+4}-2\right)\left(\sqrt{x+4}+2\right)\left(\sqrt{4-x}+2\right)}{\sqrt{x+4}+2}=2x\)
\(\Leftrightarrow\frac{x\left(\sqrt{4-x}+2\right)}{\sqrt{x+4}+2}=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{\sqrt{4-x}+2}{\sqrt{x+4}+2}=2\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{4-x}+2=2\sqrt{x+4}+4\)
\(\Leftrightarrow2\sqrt{x+4}-\frac{4}{5}+\frac{14}{5}-\sqrt{4-x}=0\)
\(\Leftrightarrow\frac{2\left(x+4-\frac{4}{25}\right)}{\sqrt{x+4}+\frac{2}{5}}+\frac{\frac{196}{25}-4+x}{\frac{14}{5}+\sqrt{4-x}}=0\)
\(\Leftrightarrow\left(x-\frac{96}{25}\right)\left(\frac{2}{\sqrt{x+4}+\frac{2}{5}}+\frac{1}{\frac{14}{5}+\sqrt{4-x}}\right)=0\)
\(\Rightarrow x=\frac{96}{25}\)
1.
Bạn coi lại đề
2.
ĐKXĐ: \(1\le x\le2\)
Nhận thấy \(\sqrt{x+2}+\sqrt{x-1}>0;\forall x\) , nhân 2 vế của pt với nó:
\(\left(\sqrt{x+2}+\sqrt{x-1}\right)\left(\sqrt{x+2}-\sqrt{x-1}\right)\left(\sqrt{2-x}+1\right)=\sqrt{x+2}+\sqrt{x-1}\)
\(\Leftrightarrow3\left(\sqrt{2-x}+1\right)=\sqrt{x+2}+\sqrt{x-1}\)
\(\Leftrightarrow3\sqrt{2-x}+3=\sqrt{x+2}+\sqrt{x-1}\)
\(\Leftrightarrow3\sqrt{2-x}+2-\sqrt{x+2}+1-\sqrt{x-1}=0\)
\(\Leftrightarrow3\sqrt{2-x}+\frac{2-x}{2+\sqrt{x+2}}+\frac{2-x}{1+\sqrt{x-1}}=0\)
\(\Leftrightarrow\sqrt{2-x}\left(3+\frac{\sqrt{2-x}}{2+\sqrt{x+2}}+\frac{\sqrt{2-x}}{1+\sqrt{x-1}}\right)=0\)
\(\Leftrightarrow\sqrt{2-x}=0\Rightarrow x=2\)
Sửa đề:
\(x=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}\)
\(=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+12\sqrt{5}-8}}{\sqrt{5}+\sqrt{9-6\sqrt{5}+5}}\)
\(=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(\sqrt{5}-2\right)^3}}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}\)
\(=\dfrac{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}{\sqrt{5}+\left(3-\sqrt{5}\right)}=\dfrac{1}{3}\)
Thế vô A ta được
\(A=\left(3.\dfrac{1}{3^3}+8.\dfrac{1}{3^2}+2\right)^{2018}=3^{2018}\)
\(x=\dfrac{\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}.\left(\sqrt{5}+2\right)=\dfrac{\sqrt[3]{5\sqrt{5}-3.5.2+3\sqrt{5}.4-8}}{\sqrt{5}+\sqrt{9-2.3\sqrt{5}+5}}.\left(\sqrt{5}+2\right)=\dfrac{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}{3}=\dfrac{5-4}{3}=\dfrac{1}{3}\) Thay : \(x=\dfrac{1}{3}\) vào A , ta được :
\(A=\left(\dfrac{3}{27}+\dfrac{8}{9}-\dfrac{3}{3}+1\right)^{2012}=1^{2012}=1\)
Vậy ,...