Question

[Lục Vân Ca]

\(A=\sqrt{80}+\sqrt{45}+\sqrt{5}\)
\(B=\frac{5}{\sqrt{10}}+3,5.\sqrt{40}\)
\(C=\frac{1}{\sqrt{3}-2}+\frac{\sqrt{300}}{10}-\sqrt{12}\)
\(D=4\sqrt{x}+2\sqrt{x^2}-\sqrt{16x}\)( x > hoặc = 0 )

\(E=\sqrt{25x+25}-\sqrt{9x+9}+\sqrt{4x+x}vớix\ge-1\)
\(F=\frac{a-2\sqrt{a}}{\sqrt{a}-2}vớia\ge0,\ne4\)
\(G=\frac{2}{\sqrt{3}+\sqrt{5}}-\frac{2}{\sqrt{5}-\sqrt{7}}\)

Answer 1

[Lục Vân Ca]

Đề bài là Rút gọn biểu thức nha . Mình quên ghi ^^

Answer 2

[Không Tên]

\(A=\sqrt{80}+\sqrt{45}+\sqrt{5}=\sqrt{16.5}+\sqrt{9.5}+\sqrt{5}\)

\(=4\sqrt{5}+3\sqrt{5}+\sqrt{5}=8\sqrt{5}\)

\(B=\frac{5}{\sqrt{10}}+3,5\sqrt{40}=\sqrt{\frac{25}{10}}+3,5\sqrt{16.2,5}\)

\(=\sqrt{2,5}+3,5.4\sqrt{2,5}=15\sqrt{2,5}\)

\(C=\frac{1}{\sqrt{3}-2}+\frac{\sqrt{300}}{10}-\sqrt{12}\)

\(=\frac{\sqrt{3}+2}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+\frac{\sqrt{100.3}}{10}-\sqrt{4.3}\)

\(=-\sqrt{3}-2+\sqrt{3}-2\sqrt{3}=-2\sqrt{3}-2\)

\(D=4\sqrt{x}+2\sqrt{x^2}-\sqrt{16x}=4\sqrt{x}+2x-4\sqrt{x}=2x\) ( do \(x\ge0\))

\(F=\frac{a-2\sqrt{a}}{\sqrt{a}-2}=\frac{\sqrt{a}.\left(\sqrt{a}-2\right)}{\sqrt{a}-2}=\sqrt{a}\)

mk chỉnh đề

\(E=\sqrt{25x+25}-\sqrt{9x+9}+\sqrt{4x+4}\)

\(=\sqrt{25\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}\)

\(=5\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=4\sqrt{x+1}\)

\(G=\frac{2}{\sqrt{3}+\sqrt{5}}-\frac{2}{\sqrt{5}-\sqrt{7}}=\frac{2\left(\sqrt{3}-\sqrt{5}\right)}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{3}-\sqrt{5}\right)}-\frac{2\left(\sqrt{5}+\sqrt{7}\right)}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{5}-\sqrt{7}\right)}\)

\(=\sqrt{3}-\sqrt{5}-\sqrt{5}-\sqrt{7}=\sqrt{3}-\sqrt{7}\)