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1: \(\left(2\sqrt{5}-5\right)^2=45-20\sqrt{5}\)
\(\left(\sqrt{5}-3\right)^2=14-6\sqrt{5}\)
mà \(45-20\sqrt{5}< 14-6\sqrt{5}\)
nên \(2\sqrt{5}-5< \sqrt{5}-3\)
3: \(\left(2+\sqrt{3}\right)^2=7+4\sqrt{3}\)
\(\left(\sqrt{2}+\sqrt{5}\right)^2=7+2\sqrt{10}\)
mà 4 căn 3>2 căn 10
nên \(2+\sqrt{3}>\sqrt{2}+\sqrt{5}\)
\(A=2\sqrt{5}-\sqrt{45}+2\sqrt{20}=2\sqrt{5}-\sqrt{3^2.5}+2\sqrt{2^2.5}=2\sqrt{5}-3\sqrt{5}+4\sqrt{5}=3\sqrt{5}\)
\(B=\left(\sqrt{18}-\frac{1}{2}\cdot\sqrt{32}+12\sqrt{2}\right):\sqrt{2}=\left(3\sqrt{2}-\frac{1}{2}\cdot4\sqrt{2}+12\sqrt{2}\right):\sqrt{2}\)
\(=13\sqrt{2}:\sqrt{2}=13\)
\(C=\left(\sqrt{12}+2\sqrt{27}-3\sqrt{3}\right)\cdot\sqrt{3}=\left(2\sqrt{3}+6\sqrt{3}-3\sqrt{3}\right)\cdot\sqrt{3}=5\sqrt{3}\cdot\sqrt{3}=15\)
\(D=\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}=-\sqrt{5}+15\sqrt{2}\)
Đặt \(A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
Áp dụng \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\) ta có:
\(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
\(A^3=2+\sqrt{5}+2-\sqrt{5}+3\sqrt[3]{4-5}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(=4-3A\)
Giải PT:
\(A^3+3A-4=0\Leftrightarrow A^3-1+3A-3=0\)\(\Leftrightarrow\left(A-1\right)\left(A^2+A+1\right)+3\left(A-1\right)=0\)\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}A-1=0\\A^2+A+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}A=1\\A^2+2.\frac{1}{2}A+\frac{1}{4}-\frac{1}{4}+4=0\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}A=1\\\left(A+\frac{1}{2}\right)^2+\frac{15}{4}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}A=1\\\left(A+\frac{1}{2}\right)^2=-\frac{15}{4}\left(L\right)\end{cases}}}\)
Vậy \(A=1\)
a, Điều kiện x ∉ {\(\frac{5}{3};\frac{1}{7}\)}
\(\sqrt{3x-5}=\sqrt{7x-1}\)
\(\left(\sqrt{3x-5}\right)^2=\left(\sqrt{7x-1}\right)^2\)
\(\left|3x-5\right|=\left|7x-1\right|\)
\(3x-5=7x-1\)
\(-4x=4\) => x = -1
Giups gì zậy bạn ?