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Do \(12=\sqrt{144}>\sqrt{135}\) nên \(x>0\)
Đặt \(a=\sqrt[3]{\dfrac{12+\sqrt{135}}{3}}+\sqrt[3]{\dfrac{12-\sqrt{135}}{3}}\) \(\Rightarrow x=\dfrac{1}{3}\left(a+1\right)\)
\(a^3=8+3\left(\sqrt[3]{\dfrac{12+\sqrt{135}}{3}}+\sqrt[3]{\dfrac{12-\sqrt{135}}{3}}\right)=8+3a\)
Ta có: \(x=\dfrac{1}{3}\left(a+1\right)\Rightarrow3x=a+1\Rightarrow9x^2=a^2+2a+1\)
Lại có: \(x^3=\dfrac{1}{27}\left(a+1\right)^3\Leftrightarrow9x^3=\dfrac{1}{3}\left(a^3+3a^2+3a+1\right)\)
\(\Leftrightarrow9x^3=\dfrac{1}{3}\left(8+3a+3a^2+3a+1\right)=a^2+2a+3\)
\(\Rightarrow M=\left(a^2+2a+3-a^2-2a-1-3\right)^2=\left(-1\right)^2=1\)
Đặt \(a=\sqrt[3]{\dfrac{12+\sqrt{135}}{3}}+\sqrt[3]{\dfrac{12-\sqrt{135}}{3}}\) \(\Rightarrow x=\dfrac{1}{3}\left(a+1\right)\)
\(\Rightarrow3x=a+1\Rightarrow9x^2=a^2+2a+1\) (1)
\(x^3=\dfrac{1}{27}\left(a+1\right)^3=\dfrac{1}{27}\left(a^3+3a^2+3a+1\right)\)
Ta có:
\(a^3=\left(\sqrt[3]{\dfrac{12+\sqrt{135}}{3}}+\sqrt[3]{\dfrac{12-\sqrt{135}}{3}}\right)^3\)
\(\Rightarrow a^3=\dfrac{24}{3}+3\sqrt[3]{\dfrac{\left(12+\sqrt{135}\right)\left(12-\sqrt{135}\right)}{9}}.\left(\sqrt[3]{\dfrac{12+\sqrt{135}}{3}}+\sqrt[3]{\dfrac{12-\sqrt{135}}{3}}\right)\)
\(\Rightarrow a^3=8+3a\)
\(\Rightarrow x^3=\dfrac{1}{27}\left(8+3a+3a^2+3a+1\right)=\dfrac{1}{9}\left(a^2+2a+3\right)\)
\(\Rightarrow9x^3=a^2+2a+3\) (2)
Thay (1), (2) vào M ta được:
\(M=\left(9x^3-9x^2-3\right)^2=\left(a^2+2a+3-\left(a^2+2a+1\right)-3\right)^2\)
\(\Rightarrow M=\left(-1\right)^2=1\)
Đặt \(a=\sqrt[3]{\frac{12+\sqrt{135}}{3}}+\sqrt[3]{\frac{12-\sqrt{135}}{3}}\)
\(\Rightarrow a^3=\left(\sqrt[3]{4+\sqrt{15}}+\sqrt[3]{4-\sqrt{15}}\right)^3\)
Có (a+b)^3=a^3+b^3+3ab(a+b)
\(\Rightarrow a^3=4+\sqrt{15}+4-\sqrt{15}+3\sqrt[3]{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}a\)
\(\Rightarrow a^3=8+3a\Rightarrow a^3-3a-8=0\)-> khó
Từ \(x=\frac{1}{3}\left(1+\sqrt[3]{\frac{12+\sqrt{135}}{3}}+\sqrt[3]{\frac{12-\sqrt{135}}{3}}\right)\)
\(\Rightarrow3x-1=\left(\sqrt[3]{\frac{12+\sqrt{135}}{3}}+\sqrt[3]{\frac{12-\sqrt{135}}{3}}\right)\)
\(\Leftrightarrow\left(3x-1\right)^3=\left(\sqrt[3]{\frac{12+\sqrt{135}}{3}}+\sqrt[3]{\frac{12-\sqrt{135}}{3}}\right)^3\)
\(\Rightarrow\left(3x-1\right)^3=8+3\left(3x+1\right)\)
\(\Leftrightarrow9x^3-9x^2-2=0\)
\(\Rightarrow M=-1\)
\(a.\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\dfrac{3}{3-\sqrt{6}}=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{\sqrt{3}.\sqrt{3}}{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}=\sqrt{6}-\dfrac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{3\sqrt{2}-3\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{-3\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}=-3\) \(b.\left(2\sqrt{2}-\sqrt{3}\right)^2-2\sqrt{3}\left(\sqrt{3}-2\sqrt{2}\right)=\left(2\sqrt{2}-\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)=8-3=5\) \(c.\left(\dfrac{1}{3-\sqrt{5}}-\dfrac{1}{3+\sqrt{5}}\right):\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\dfrac{3+\sqrt{5}-3+\sqrt{5}}{9-5}:\sqrt{5}=\dfrac{2\sqrt{5}}{4}.\dfrac{1}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}.\dfrac{1}{\sqrt{5}}=\dfrac{1}{2}\) \(d.\left(3-\dfrac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3+\dfrac{\sqrt{ab}-3\sqrt{a}}{\sqrt{b}-3}\right)=\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)=9-a\)
\(b1:=\sqrt{2}\left(\sqrt{3}+1\right).\sqrt{2-\sqrt{3}}\\ =\left(\sqrt{3}+1\right).\sqrt{4-2\sqrt{3}}\\ =\left(\sqrt{3}+1\right).\left(\sqrt{3}-1\right)\\ =2\\ \\ b2:a,=\sqrt{\dfrac{\left(3\sqrt{5}+1\right)\left(2\sqrt{5}-3\right)}{\left(2\sqrt{5}-3\right)^2}}.\left(\sqrt{10}-\sqrt{2}\right)\\ =\dfrac{\sqrt{27-7\sqrt{5}}}{2\sqrt{5}-3}.\left(\sqrt{10}-\sqrt{2}\right)\\ =\dfrac{\sqrt{2}}{\sqrt{2}}.\dfrac{\sqrt{27-7\sqrt{5}}}{2\sqrt{5}-3}.\left(\sqrt{10}-\sqrt{2}\right)\\ =\dfrac{\sqrt{54-14\sqrt{5}}}{2\sqrt{10}-3\sqrt{2}} .\left(\sqrt{10}-\sqrt{2}\right)\\ \)\(=\dfrac{\sqrt{\left(7-\sqrt{5}\right)^2}}{2\sqrt{10}-3\sqrt{2}}.\left(\sqrt{10}-\sqrt{2}\right)\)\(\\ =\dfrac{8\sqrt{10}-12\sqrt{2}}{2\sqrt{10}-3\sqrt{2}}\\ =4\)
1)
a) \(3\sqrt{\dfrac{1}{3}}-\dfrac{1}{\sqrt{3}+\sqrt{2}}\)
\(=\sqrt{3}-\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\sqrt{3}-\sqrt{3}+\sqrt{2}\)
\(=\sqrt{2}\)
b) \(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=\left|\sqrt{3}+1\right|-\left|1-\sqrt{3}\right|\)
\(=\sqrt{3}+1-\sqrt{3}+1\)
\(=2\)
a. Ta có:A = \(\dfrac{\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
= \(\dfrac{\sqrt{3-\sqrt{5}}.\left(\sqrt{3+\sqrt{5}}\right)^2}{\sqrt{2}.\left(\sqrt{5}-1\right)}\)
= \(\dfrac{\sqrt{9-5}.\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{2}.\left(\sqrt{5}-1\right)}\)
= \(\dfrac{2\sqrt{3+\sqrt{5}}}{\sqrt{2}.\left(\sqrt{5+1}\right)}\)
=\(\dfrac{\sqrt{2}.\sqrt{3+\sqrt{5}}}{\sqrt{5}+1}\)
⇒A2 = \(\dfrac{2.\left(3+\sqrt{5}\right)}{5+2\sqrt{5}+1}\)=\(\dfrac{6+2\sqrt{5}}{6+2\sqrt{5}}\)=1
⇒A=\(\sqrt{1}\)=1
2, \(\dfrac{1}{2}\sqrt{x+1}-\dfrac{3}{2}\sqrt{9x+9}+24\sqrt{\dfrac{x+1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x+1}-\dfrac{9}{2}\sqrt{x+1}+3\sqrt{x+1}=-17\)
\(\Leftrightarrow-\sqrt{x+1}=-17\)
\(\Leftrightarrow x+1=289\left(x>0\right)\)
\(\Leftrightarrow x=288\)
Vậy x = 288
3, \(-5x+7\sqrt{x}+12=0\)
\(\Leftrightarrow-5x+12\sqrt{x}-5\sqrt{x}+12=0\)
\(\Leftrightarrow\sqrt{x}\left(12-5\sqrt{x}\right)+\left(12-5\sqrt{x}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(12-5\sqrt{x}\right)=0\)
Do \(\sqrt{x}+1>0\)
\(\Rightarrow12-5\sqrt{x}=0\Leftrightarrow x=\dfrac{144}{25}\)
Vậy...
1. (Đề có chút sai sai nên mình sửa lại nhé) \(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)
(ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{36\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{4\left(x-1\right)}=16-\sqrt{x-1}\)
\(\Leftrightarrow6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)
\(\Leftrightarrow2\sqrt{x-1}=16\)
\(\Leftrightarrow\sqrt{x-1}=8\)
\(\Leftrightarrow x-1=64\)
\(\Leftrightarrow x=65\left(tm\right)\)
Vậy pt đã cho có nghiệm x=65.
2. \(\dfrac{1}{2}\sqrt{x+1}-\dfrac{3}{2}\sqrt{9x+9}+24\sqrt{\dfrac{x+1}{64}}=-17\)
(ĐK: \(x\ge-1\))
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x+1}-\dfrac{3}{2}\sqrt{9\left(x+1\right)}+3\sqrt{x+1}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x+1}-\dfrac{9}{2}\sqrt{x+1}+3\sqrt{x+1}=-17\)
\(\Leftrightarrow-\sqrt{x+1}=-17\)
\(\Leftrightarrow\sqrt{x+1}=17\)
\(\Leftrightarrow x+1=289\)
\(\Leftrightarrow x=288\left(tm\right)\)
Vậy \(S=\left\{288\right\}\)
3. \(-5x+7\sqrt{x}+12=0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow5x-7\sqrt{x}-12=0\)
\(\Leftrightarrow5x+5\sqrt{x}-12\sqrt{x}-12=0\)
\(\Leftrightarrow5\sqrt{x}\left(\sqrt{x}+1\right)-12\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(5\sqrt{x}-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=0\\5\sqrt{x}-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-1\left(vô.lý\right)\\5\sqrt{x}=12\end{matrix}\right.\Leftrightarrow\sqrt{x}=\dfrac{12}{5}\Leftrightarrow x=\dfrac{144}{25}\left(tm\right)\)
Vậy pt có nghiệm \(x=\dfrac{144}{25}\)