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\(x=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
\(\Rightarrow x^3=5+2\sqrt{13}+5-2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}.x\)
\(=10+3x\sqrt[3]{25-52}\)
\(=10+3x\sqrt[3]{-27}\)
\(=10-9x\)
\(\Rightarrow x^3+9x-10=0\)
\(\Leftrightarrow x^3-x+10x-10=0\)
\(\Leftrightarrow x\left(x^2-1\right)+10\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)+10\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+10\right)=0\)
Vì \(x^2+x+10=\left(x+\frac{1}{2}\right)^2+\frac{39}{4}>0\forall x\)
=> x - 1 = 0
=> x = 1
Thay vào A = 12015 - 12016 = 0
Vậy A = 0
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
a, Ta có : \(x=\sqrt{3+2\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}=4\)
Thay x = 4 => \(\sqrt{x}=2\) vào B ta được :
\(B=\frac{2+5}{2-3}=-7\)
b, Ta có : Với \(x\ge0;x\ne9\)
\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13-\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)
\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}=\frac{x-25}{x-9}\)
Lại có \(P=\frac{A}{B}\Rightarrow P=\frac{\frac{x-25}{x-9}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)
\(C=\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\)
\(C^2=\left(\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\right)^2\)
\(C^2=x^2+2\sqrt{x^2-1}-2\sqrt{\left(x^2+2\sqrt{x^2-1}\right)\left(x^2-2\sqrt{x^2-1}\right)}+x^2-2\sqrt{x^2-1}\)
\(C^2=2x^2-2\sqrt{x^4-2x^2\sqrt{x^2-1}+2x^2\sqrt{x^2-1}-\left(2\sqrt{x^2-1}\right)^2}\)
\(C^2=2x^2-2\sqrt{x^4-4\left(x^2-1\right)}\)
\(C^2=2x^2-2\sqrt{x^4-4x^2+4}\)
\(C=\sqrt{2x^2-2\sqrt{x^4-4x^2+4}}\)
Thay: \(x=\sqrt{5}\) vào C, ta có:
\(C=\sqrt{2\sqrt{5}^2-2\sqrt{\sqrt{5}^4-4\sqrt{5}^2+4}}\)
\(C=\sqrt{10-2\sqrt{25-20+4}}\)
\(C=\sqrt{10-2\sqrt{9}}\)
\(C=\sqrt{10-6}\)
\(C=\orbr{\begin{cases}-2\\2\end{cases}}\)
Mà theo bài ra: \(\sqrt{x^2+2\sqrt{x^2-1}}>\sqrt{x^2-2\sqrt{x^2-1}}\)
\(\Rightarrow\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}>0\)
\(\Rightarrow C=2\)
mọi ng ơi mk viết thiếu dấu ngoặc nha.thiếu ngoặc lownns nha. đóng ngoắc ở trước dấu chia