Mình chịu thua thôi!!!

Bài làm:

a) \(\sqrt{4-\sqrt{7}}=\frac{\sqrt{2\left(4-\sqrt{7}\right)}}{\sqrt{2}}=\sqrt{\frac{8-2\sqrt{7}}{2}}=\sqrt{\frac{7-2\sqrt{7}+1}{2}}\)

\(=\sqrt{\frac{\left(\sqrt{7}-1\right)^2}{2}}=\frac{\left(\sqrt{7}-1\right)\sqrt{2}}{2}=\frac{\sqrt{14}-\sqrt{2}}{2}\)

b) \(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\) (đề vậy chứ)

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

\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=\sqrt{3}+1-\sqrt{3}+1\)

\(=2\)

c) \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)

\(=\sqrt{5-4\sqrt{5}+4}-\sqrt{5+4\sqrt{5}+4}\)

\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)

\(=\sqrt{5}-2-\sqrt{5}-2\)

\(=-4\)

d) \(\sqrt{x-2\sqrt{x-1}}\)

\(=\sqrt{\left(x-1\right)-2\sqrt{x-1}+1}\)

\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}\)

\(=\left|\sqrt{x-1}-1\right|\)

tả nời

B = 1

Mik chỉ bt làm thế này thôi bạn áp dụng vào bài nhá

cos75 = sin(90-75) = sin15

cos18 = sin(90-18) = sin72

Vì 15 < 65 < 70 < 72 < 79

Nên sin15 < sin 65 < sin70 < sin72 < sin79

Tít cho mik

\(P=\left(\frac{1}{x+1}+\frac{1}{x-1}\right):\frac{2x}{x-1}\)

a) Điều kiện xác định:

\(\hept{\begin{cases}x+1\ne0\\x-1\ne0\\2x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne0-1\\x\ne0+1\\x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne-1\\x\ne1\\x\ne0\end{cases}}\)

Vậy để P có nghĩa thì \(x\ne-1;x\ne1\) và \(x\ne0.\)

b) Rút gọn:

\(P=\left(\frac{1}{x+1}+\frac{1}{x-1}\right):\frac{2x}{x-1}\)

\(P=\left(\frac{1.\left(x-1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{1.\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}\right):\frac{2x}{x-1}\)

\(P=\left(\frac{x-1}{\left(x-1\right).\left(x+1\right)}+\frac{x+1}{\left(x-1\right).\left(x+1\right)}\right):\frac{2x}{x-1}\)

\(P=\left(\frac{x-1+x+1}{\left(x-1\right).\left(x+1\right)}\right):\frac{2x}{x-1}\)

\(P=\frac{2x}{\left(x-1\right).\left(x+1\right)}:\frac{2x}{x-1}\)

\(P=\frac{2x}{\left(x-1\right).\left(x+1\right)}.\frac{x-1}{2x}\)

\(P=\frac{2x.\left(x-1\right)}{2x.\left(x-1\right).\left(x+1\right)}\)

\(P=\frac{1}{x+1}.\)