no no no

ai trả lời được ko

\(=\left(\sin^2\alpha+\cos^2\alpha\right)^2=1\)

=100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

khổ thân cái máy tính

\(P=\left(\frac{2\left(\sqrt{x}+2\right)+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right).\frac{x+2\sqrt{x}}{2\sqrt{x}}\) điều kiện x >0

\(P=\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}.\frac{x+2\sqrt{x}}{2\sqrt{x}}\)

\(P=\frac{2\sqrt{x}+4+x}{2\sqrt{x}}=1+\frac{4+x}{2\sqrt{x}}.\)

b) P = 3

\(\Leftrightarrow1+\frac{4+x}{2\sqrt{x}}=3\Leftrightarrow\frac{4+x}{2\sqrt{x}}=2\)

\(\Leftrightarrow4+x=4\sqrt{x}\Leftrightarrow4+x-4\sqrt{x}=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)

\(\Leftrightarrow\sqrt{x}-2=0\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)

Ngô Văn Tuyên cảm ơn bạn nha. Nhưng cho mình hỏi tí sao bạn lại tách ra thành \(1+\frac{4-x}{2\sqrt{x}}\)

giải thích hộ mình với nhé. Cảm ơn nhiều !!

a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ne3\\x\ge1\end{matrix}\right.\)

Ta có : \(P=\dfrac{x-1-2}{\sqrt{x-1}-\sqrt{2}}=\dfrac{\left(\sqrt{x-1}-\sqrt{2}\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}\)

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

b, Thấy : \(\sqrt{x-1}\ge0\)

\(\Rightarrow P=\sqrt{x-1}+\sqrt{2}\ge\sqrt{2}\)

Vậy \(Min=\sqrt{2}\Leftrightarrow x=1\)

Vậy ,...
 

\(A=\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{\sqrt{a}+1}{a}\)

\(A=\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right).\frac{a}{\sqrt{a}+1}\)

\(A=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right).\frac{a}{\sqrt{a}+1}\)

\(A=\frac{a-1}{\sqrt{a}}.\frac{a}{\sqrt{a}+1}\)

\(A=\left(\sqrt{a}-1\right).\sqrt{a}\)

\(A=a-\sqrt{a}\)

A=\(\left(\frac{\sqrt{a}\left(\sqrt{a}\right)-1}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right):\frac{\sqrt{a}+1}{a}\)= \(\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\):\(\frac{\sqrt{a}+1}{a}\)=

=\(\left(\frac{a-1}{\sqrt{a}}\right)\). \(\frac{a}{\sqrt{a}+1}\)= \(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\)\(\frac{a}{\sqrt{a}+1}\)= \(\frac{\sqrt{a}-1}{\sqrt{a}}\)

a: \(=\left|4-3\sqrt{2}\right|=3\sqrt{2}-4\)

b: \(\sqrt{\left(2+\sqrt{5}\right)^2}=\left|\sqrt{5}+2\right|=\sqrt{5}+2\)

c: \(\sqrt{\left(4+\sqrt{2}\right)^2}=\left|4+\sqrt{2}\right|=4+\sqrt{2}\)

d: \(\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)