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\(1.A=\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=\dfrac{1}{2}\sqrt{\dfrac{1}{3}.144}-2\sqrt{\dfrac{1}{3}.225}-\sqrt{\dfrac{1}{3}.9}+5\sqrt{\dfrac{4}{3}}=6\sqrt{\dfrac{1}{3}}-30\sqrt{\dfrac{1}{3}}-3\sqrt{\dfrac{1}{3}}+10\sqrt{\dfrac{1}{3}}=-17\sqrt{\dfrac{1}{3}}\) \(2.B=\left(2\sqrt{27}-3\sqrt{48}+3\sqrt{75}-\sqrt{192}\right)\left(1-\sqrt{3}\right)=\left(6\sqrt{3}-12\sqrt{3}+15\sqrt{3}-8\sqrt{3}\right)\left(1-\sqrt{3}\right)=\sqrt{3}\left(1-\sqrt{3}\right)=\sqrt{3}-3\) \(3.C=\left(2\sqrt{7}-2\sqrt{6}\right).\sqrt{6}-\sqrt{168}=2\sqrt{42}-12-2\sqrt{42}=-12\) \(4.D=\left(\sqrt{28}-2\sqrt{8}+\sqrt{7}\right).\sqrt{7}+4\sqrt{14}=\left(3\sqrt{7}-4\sqrt{2}\right)\sqrt{7}=21-4\sqrt{14}+4\sqrt{14}=21\)
ai nay dung kinh nghiem la chinh
cau a)
ta thay \(10+6\sqrt{3}=\left(1+\sqrt{3}\right)^3\)
\(6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)
khi do \(x=\frac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{5}}\)
\(x=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{1+\sqrt{5}-\sqrt{5}}\)
\(x=\frac{3-1}{1}=2\)
suy ra
x^3-4x+1=1
A=1^2018
A=1
b)
ta thay
\(7+5\sqrt{2}=\left(1+\sqrt{2}\right)^3\)
khi do
\(x=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\frac{1}{\sqrt[3]{\left(1+\sqrt{2}\right)^3}}\)
\(x=1+\sqrt{2}-\frac{1}{1+\sqrt{2}}=\frac{\left(1+\sqrt{2}\right)^2-1}{1+\sqrt{2}}=\frac{2+2\sqrt{2}}{1+\sqrt{2}}\)
x=2
thay vao
x^3+3x-14=0
B=0^2018
B=0
Bài 3: \(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)
\(\Leftrightarrow\left(3-8x\right)\sqrt{2x^2+1}=3x^2+x+3\)
\(\Rightarrow\left(3-8x\right)^2\left(2x^2+1\right)=\left(3x^2+x+3\right)^2\)
\(\Leftrightarrow119x^4-102x^3+63x^2-54x=0\)
\(\Leftrightarrow x\left(7x-6\right)\left(17x^2+9\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{6}{7}\end{cases}}\)
Thử lại, ta nhận được \(x=0\)là nghiệm duy nhất của phương trình
a, \(2\sqrt{3}-\sqrt{27}+\sqrt{75}=6,92820323\)
a) \(2\sqrt{3}-\sqrt{27}+\sqrt{75}\)
\(=2\sqrt{3}-3\sqrt{3}+5\sqrt{3}\)
\(=\sqrt{3}\left(2-3+5\right)\)
\(=4\sqrt{3}\)
b)\(\sqrt{\left(1-\sqrt{3}\right)^2}+\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(1-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=\left(1-\sqrt{3}\right)+\left(1+\sqrt{3}\right)\)
\(=2\)