Chà sao lại đặt trong lớp 1 vậy em
1.35
\(x=\dfrac{\sqrt{4+2\sqrt{3}}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt{9-4\sqrt{5}}-2}=\dfrac{\sqrt{3+2\sqrt{3}+1}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt{5-2.2.\sqrt{5}+4}-2}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt{\left(\sqrt{5}-2\right)^2}-2}=\dfrac{\left|\sqrt{3}+1\right|-\sqrt{3}}{\left(\sqrt{5}+2\right)\left|\sqrt{5}-2\right|-2}\)
\(=\dfrac{\sqrt{3}+1-\sqrt{3}}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)-2}=\dfrac{1}{5-4-2}=\dfrac{1}{-1}=-1\)
\(\Rightarrow P=\left(\left(-1\right)^2-1+1\right)^{2017}=1^{2017}=1\)
1.36.
\(A=\left(\dfrac{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\right)\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}+3\right)}\right)\)
\(=\left(\dfrac{x+2\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\left(\dfrac{1}{2\sqrt{x}+3}\right)\)
\(=\dfrac{\left(x+2\sqrt{x}+1\right)}{\sqrt{x}+1}.\dfrac{1}{2\sqrt{x}+3}=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}\)
b.
\(\dfrac{x}{4}=\sqrt{\dfrac{1009+\sqrt{2017}}{2}}-\sqrt{\dfrac{1009-\sqrt{2017}}{2}}\)
\(x=4\sqrt{\dfrac{1009+\sqrt{2017}}{2}}-4\sqrt{\dfrac{1009-\sqrt{2017}}{2}}\)
\(x=2.\sqrt{2018+2\sqrt{2017}}-2\sqrt{2018-2\sqrt{2017}}\)
\(x=2\sqrt{2017+2\sqrt{2017}+1}-2\sqrt{2017-2\sqrt{2017}+1}\)
\(x=2\sqrt{\left(\sqrt{2017}+1\right)^2}-2\sqrt{\left(\sqrt{2017}-1\right)^2}\)
\(x=2\left|\sqrt{2017}+1\right|-2\left|\sqrt{2017}-1\right|\)
\(x=2\left(\sqrt{2017}+1\right)-2\left(\sqrt{2017}-1\right)=4\)
\(\Rightarrow A=\dfrac{\sqrt{4}+1}{2\sqrt{4}+3}=\dfrac{3}{7}\)
1.37
\(P=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{5\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(3x-6\sqrt{x}\right)+\left(x+3\sqrt{x}+2\right)-\left(5\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{4x-8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{4\sqrt{x}}{\sqrt{x}+2}\)
b.
\(x=\sqrt[3]{1+\dfrac{\sqrt[]{84}}{9}}+\sqrt[3]{1-\dfrac{\sqrt[]{84}}{9}}\)
\(\Rightarrow x^3=\left(\sqrt[3]{1+\dfrac{\sqrt[]{84}}{9}}+\sqrt[3]{1-\dfrac{\sqrt[]{84}}{9}}\right)^3\)
Áp dụng HĐT: \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)
\(\Rightarrow x^3=2+3\sqrt[3]{\left(1+\dfrac{\sqrt[]{84}}{9}\right)\left(1-\dfrac{\sqrt[]{84}}{9}\right)}.\left(\sqrt[3]{1+\dfrac{\sqrt[]{84}}{9}}+\sqrt[3]{1-\dfrac{\sqrt[]{84}}{9}}\right)\)
\(\Rightarrow x^3=2+3.\sqrt[3]{1-\dfrac{84}{81}}.x\)
\(\Rightarrow x^3=2-x\)
\(\Rightarrow x^3+x-2=0\)
\(\Rightarrow\left(x-1\right)\left(x^2+x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x^2+x+2=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=1\)
\(\Rightarrow P=\dfrac{4\sqrt[]{1}}{\sqrt[]{1}+2}=\dfrac{4}{3}\)
cái này mà lớp 1 á t lớp 6 còn chưa thấy
