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1)
\(M=\frac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{6+4\sqrt{2}}}+\frac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{6-4\sqrt{2}}}\)
\(=\frac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{4+2.2.\sqrt{2}+2}}+\frac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{4-2.2.\sqrt{2}+2}}\)
\(=\frac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{\left(2+\sqrt{2}\right)^2}}+\frac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{\left(2-\sqrt{2}\right)^2}}\)
\(=\frac{6+4\sqrt{2}}{2+2\sqrt{2}}+\frac{6-4\sqrt{2}}{-2+2\sqrt{2}}\)
\(=\frac{2.\left(3+2\sqrt{2}\right)}{2.\left(1+\sqrt{2}\right)}+\frac{2.\left(3-2\sqrt{2}\right)}{2.\left(\sqrt{2}-1\right)}\)
\(=\frac{3+2\sqrt{2}}{\sqrt{2}+1}+\frac{3-2\sqrt{2}}{\sqrt{2}-1}\)
\(=\frac{\left(3+2\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\left(3-2\sqrt{2}\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\)
\(=1+\sqrt{2}+\sqrt{2}-1=2\sqrt{2}\)
B3: \(\sqrt{x^4-4x^3+2x^2+4x+1}=3x-1\)
\(pt\Leftrightarrow x^4-4x^3+2x^2+4x+1=\left(3x-1\right)^2\)
\(\Leftrightarrow x^4-4x^3+2x^2+4x+1=9x^2-6x+1\)
\(\Leftrightarrow x^4-4x^3-7x^2+10x=0\)
\(\Leftrightarrow x\left(x^3-4x^2-7x+10\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=5\end{cases}}\) (thỏa mãn (mấy cái kia loại hết))
+Xét 2 riêng trường hợp x = 0 và y = 0.
+Xét x, y đều khác 0
Hệ \(\Leftrightarrow\int^{\frac{1}{4}+\frac{2\sqrt{x}+\sqrt{y}}{x+y}=\frac{2}{\sqrt[4]{x}}}_{\frac{1}{4}-\frac{2\sqrt{x}+\sqrt{y}}{x+y}=\frac{1}{\sqrt[4]{y}}}\Leftrightarrow\frac{1}{2}=\frac{2}{\sqrt[4]{x}}+\frac{1}{\sqrt[4]{y}}\text{ }\&\text{ }2.\frac{2\sqrt{x}+\sqrt{y}}{x+y}=\frac{2}{\sqrt[4]{x}}-\frac{1}{\sqrt[4]{y}}\)
\(\Rightarrow\frac{2\sqrt{x}+\sqrt{y}}{x+y}=\left(\frac{2}{\sqrt[4]{x}}+\frac{1}{\sqrt[4]{y}}\right)\left(\frac{2}{\sqrt[4]{x}}-\frac{1}{\sqrt[4]{y}}\right)=\frac{4}{\sqrt{x}}-\frac{1}{\sqrt{y}}\)
Đặt \(\sqrt{y}=t.\sqrt{x}\text{ }\left(t>0\right)\)
Suy ra: \(\frac{2+t}{1+t^2}=4-\frac{1}{t}\Leftrightarrow\left(2t-1\right)\left(2t^2+1\right)=0\Leftrightarrow t=\frac{1}{2}\)
\(\Rightarrow\sqrt{x}=2\sqrt{y}\)
Thay vào phương trình đầu của hệ ban đầu:
\(\sqrt{2\sqrt{y}}\left(\frac{1}{4}+\frac{5\sqrt{y}}{5y}\right)=2\Leftrightarrow\frac{1}{4}+\frac{1}{\sqrt{y}}=\frac{2}{\sqrt{2\sqrt{y}}}\)
\(\Leftrightarrow\frac{1}{4}+2t^2=2t\text{ với }t=\frac{1}{\sqrt{2\sqrt{y}}}\)
Tới đây dễ rồi.
rút gọn giúp mình nha mình quên ghi
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)
\(A=\frac{\sqrt{x}.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}-\frac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)
\(A=\frac{\sqrt{x}.\left(\sqrt{x}+1\right)+3.\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)
\(A=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)