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Đáp án :
\(x_0=^3\sqrt{38-17}\sqrt{5}+^3\sqrt{38+17}.\sqrt{5}\)
\(=x_0=38-17\sqrt{5}+38+17\sqrt{5}-3^3\sqrt{\left(38-17\sqrt{5}\right)\left(38+17\sqrt{5}\right).x_0}\)
\(=76-3^3\sqrt{-1}.x_0=76+3x_0\)
\(=x_0^3\)\(-3x_0-76=0\)
\(=\left(x_0-4\right)\left(x_0^2+4x_0+19\right)=0\)
\(=x_0=4\)
Thay x0 = 4 vào phương trình x3 - 3x2 - 2x - 8 = 0 ta có đẳng thức đúng là:
43 - 3.42 - 2.4 - 8 = 0
Vậy x0 là nghiệm của phương trình x3 - 3x2 - 2x - 8 = 0
\(x^3=10+3x\sqrt[3]{\left(5-\sqrt{17}\right)\left(5+\sqrt{17}\right)}=10+6x\)
Thay vào -> dpcm
\(x=\sqrt[3]{5-\sqrt{17}}+\sqrt[3]{5+\sqrt{17}}\)
\(\Leftrightarrow x^3=5-\sqrt{17}+5+\sqrt{17}\)
\(+3\left(\sqrt[3]{5-\sqrt{17}}+\sqrt[3]{5+\sqrt{17}}\right)\sqrt[3]{5-\sqrt{17}}\sqrt[3]{5+\sqrt{17}}\)
\(\Leftrightarrow x^3=10+3x\sqrt[3]{\left(5-\sqrt{17}\right)\left(5+\sqrt{17}\right)}\)
\(\Leftrightarrow x^3=10+3x\sqrt[3]{8}\Leftrightarrow x^3=10+6x\)
\(\Leftrightarrow x^3-6x-10=0\)
\(\Rightarrow\) Đpcm
Chúc bạn học tốt !!!
cho x = \(\sqrt[3]{38+17\sqrt{5}}+\sqrt[3]{38-17\sqrt{5}}\)
Tính C= \(\left(x^3+3x+1935\right)2018\)
\(x=\sqrt[3]{38+17\sqrt{5}}+\sqrt[3]{38-17\sqrt{5}}=\sqrt[3]{5\sqrt{5}+3.5.2+3.\sqrt{5}.4+8}+\sqrt[3]{8-3.4.\sqrt{5}+3.2.5-5\sqrt{5}}=\sqrt[3]{\left(2+\sqrt{5}\right)^3}+\sqrt[3]{\left(2-\sqrt{5}\right)^3}=2+\sqrt{5}+2-\sqrt{5}=4\)Vậy C=(43+3.4+1935)2018=2011.2018=4058198
\(x=\sqrt[3]{5-\sqrt{17}}+\sqrt[3]{5+\sqrt{17}}\)
\(\Leftrightarrow x^3=5-\sqrt{17}+5+\sqrt{17}+3\left(\sqrt[3]{5-\sqrt{17}}+\sqrt[3]{5+\sqrt{17}}\right)\sqrt[3]{5-\sqrt{17}}\sqrt[3]{5+\sqrt{17}}\)
\(\Leftrightarrow x^3=10+3x\sqrt[3]{\left(5-\sqrt{17}\right)\left(5+\sqrt{17}\right)}\)
\(\Leftrightarrow x^3=10+3x\sqrt[3]{8}\)\(\Leftrightarrow x^3=10+6x\)
\(\Leftrightarrow x^3-6x-10=0\)
Hay ta co DPCM
\(x=\sqrt[3]{17\sqrt{5}+38}-\sqrt[3]{17\sqrt{5}-38}\)
\(=\sqrt[3]{\left(\sqrt{5}+2\right)^3}+\sqrt[3]{\left(\sqrt{5}-2\right)^3}\)
\(=\sqrt{5}+2+\sqrt{5}-2\)
\(=2\sqrt{5}\)
Dùng cách phổ thông hơn bạn nhé!
\(x^3=17\sqrt{5}+38-17\sqrt{5}+38-3\sqrt[3]{\left(17\sqrt{5}+38\right)\left(17\sqrt{5}-38\right)}x\)
\(=76-3x\sqrt[3]{1445-1444}\)
\(=76-3x\)
\(\Rightarrow x^3+3x-76=0\)
\(\Leftrightarrow x^3-16x+19x-76=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)+19\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2+4x+19\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left[\left(x+2\right)^2+15\right]=0\)
Vì [...] > 0
Nên x - 4 = 0
=> x = 4
a) ĐKXĐ: \(x\ge0\)
Ta có: \(3\sqrt{18x}-5\sqrt{8x}+4\sqrt{50x}=38\)
\(\Leftrightarrow9\sqrt{2x}-10\sqrt{2x}+20\sqrt{2x}=38\)
\(\Leftrightarrow19\sqrt{2x}=38\)
\(\Leftrightarrow\sqrt{2x}=2\)
\(\Leftrightarrow2x=4\)
hay x=2(thỏa ĐK)
b) ĐKXĐ: \(x\ge0\)
Ta có: \(3\sqrt{12x}-2\sqrt{27x}+4\sqrt{3x}=8\)
\(\Leftrightarrow6\sqrt{3x}-6\sqrt{3x}+4\sqrt{3x}=8\)
\(\Leftrightarrow\sqrt{3x}=2\)
\(\Leftrightarrow3x=4\)
hay \(x=\dfrac{4}{3}\)
c) ĐKXĐ: \(x\ge5\)
Ta có: \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\)
hay x=9
a)
\(3.3\sqrt{2x}-5.2\sqrt{2x}+4.5.\sqrt{2x}=38\\ \Leftrightarrow19\sqrt{2x}=38\\ \Leftrightarrow\sqrt{2x}=2\\ \Leftrightarrow x=2\)
b)
\(3.2.\sqrt{3x}-2.3.\sqrt{3x}+4.\sqrt{3x}=8\\ \Leftrightarrow4\sqrt{3x}=8\\ \Leftrightarrow\sqrt{3x}=2\\\Leftrightarrow x=\dfrac{2^2}{3}=\dfrac{4}{3} \)
c)
\(\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\)
\(x^3=76+3\sqrt[3]{\left(38-17\sqrt{5}\right)\left(38+17\sqrt{5}\right)}\left(\sqrt[3]{38-17\sqrt{5}}+\sqrt[3]{38+17\sqrt{5}}\right)\)
\(\Leftrightarrow x^3=76-3x\)
\(\Leftrightarrow x^3+3x-76=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2+4x+19\right)=0\)
\(\Leftrightarrow x=4\)
\(\Rightarrow x^3-3x^2-2x-8=0\)