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

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

\(=\sqrt{\left(\sqrt{x-1}\right)^2-2\cdot\sqrt{x-1}\cdot1+1^2}\)

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

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

a) \(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)

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

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

\(=2\)

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

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

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

\(=-4\)

a) \(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)

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

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

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

\(=2\)

b) tương tự

Đây nè:

https://vietnamnet.vn/vn/giao-duc/tuyen-sinh/dap-an-de-thi-tuyen-sinh-lop-10-mon-toan-nam-2019-cua-ha-noi-538099.html

Mik chưa học đến lớp 9 nên ko có , sorry bn nha :)))
~ Hok tốt ~
#JH

a) \(\sqrt{9-2\sqrt{14}}\)

\(=\sqrt{7-2\cdot\sqrt{7}\cdot\sqrt{2}+2}\)

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

\(=\sqrt{7}-\sqrt{2}\)

b) chịu 

ĐKXĐ: \(x\ge0;x\ne4.\)

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

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

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

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

b) Để \(A=\frac{5}{4}\)\(\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{5}{4}\Leftrightarrow\frac{4\sqrt{x}}{4\left(\sqrt{x}-2\right)}-\frac{5\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-2\right)}=0\)

\(\Leftrightarrow\frac{4\sqrt{x}-5\sqrt{x}+10}{4\left(\sqrt{x}-2\right)}=0\Leftrightarrow-\sqrt{x}+10=0\)

\(\Leftrightarrow\sqrt{x}=10\Leftrightarrow x=100\left(tmđk\right).\)

Vậy để A=5/4 thì x=100

Tự tìm ĐK nha

a) \(A=\frac{x}{x-4}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)

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

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

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

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

\(A=\frac{\sqrt{x}}{\sqrt{x}-2}\)

b) \(A=\frac{5}{4}\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{5}{4}\)

\(\Leftrightarrow4\sqrt{x}=5\left(\sqrt{x}-2\right)\)

\(\Leftrightarrow4\sqrt{x}=5\sqrt{x}-10\)

\(\Leftrightarrow\sqrt{x}=10\)

\(\Leftrightarrow x=100\)( thỏa mãn )

Vậy...

Cách làm dài bạn thông cảm mình  nghĩ được có zậy thui ak :/

Ta có a, b là các số thực dương 

Từ \(a+3b=ab\Leftrightarrow\frac{1}{b}+\frac{3}{a}=1\ge2\sqrt{\frac{3}{ab}}.\)(bất đẳng thức Cauchy cho 2 số không âm)

\(\Leftrightarrow\frac{12}{ab}\le1\Leftrightarrow ab\ge12\)\(\Leftrightarrow84ab-72ab\ge144\Leftrightarrow84ab\ge72\left(ab+2\right)\)

\(\Leftrightarrow\frac{12ab}{ab+2}\ge\frac{72}{7}\left(1\right)\)

Ta có \(P=\frac{a^2}{1+3b}+\frac{9b^2}{1+a}\ge2\sqrt{\frac{a^2}{1+3b}\frac{9b^2}{1+a}}=\frac{6ab}{\sqrt{\left(1+a\right)\left(1+3b\right)}}\)(Bất đẳng thức Cauchy)

                                                      \(\ge\frac{6ab}{\frac{1+a+1+3b}{2}}=\frac{12ab}{a+3b+2}=\frac{12ab}{ab+2}\)(Bất đẳng thức Cauchy ngược dấu )

Kết hợp với (1) ta được :

\(P\ge\frac{12ab}{ab+2}\ge\frac{72}{7}.\)

Vậy giá trị nhỏ nhất của \(P=\frac{72}{7}\Leftrightarrow\hept{\begin{cases}a=3b\\a+3b=ab\end{cases}\Leftrightarrow\hept{\begin{cases}a=6\\b=2\end{cases}.}}\)

a) \(\sqrt{11-2\sqrt{10}}\)

\(=\sqrt{10-2\sqrt{10}+1}\)

\(=\sqrt{\left(\sqrt{10}-1\right)^2}\)

\(=\sqrt{10}-1\)

b) \(\sqrt{21-6\sqrt{6}}\)

\(=\sqrt{\left(3\sqrt{2}\right)^2-2\cdot3\sqrt{2}\cdot\sqrt{3}+3}\)

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

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

b) có

\(17< 10,25\Rightarrow\sqrt{17}< 4,5\)

\(29< 20,15\Rightarrow\sqrt{19}< 4,5\)

\(\Rightarrow\sqrt{17}+\sqrt{19}< 4,5+4,5=9\)

a) có \(27< 36\)nên \(\sqrt{27}< 6\)

\(\Rightarrow3\sqrt{27}< 18\)(1)

có \(19< 25\Rightarrow\sqrt{19}< 5\Rightarrow23-\sqrt{19}>18\)(2)

từ (1) và (2) suy ra 

\(23-\sqrt{19}>3\sqrt{27}\Rightarrow\frac{23-\sqrt{19}}{3}>\sqrt{27}\)

xin lỗi giờ mình mới nghĩ ra câu a

\(b,\sqrt{\frac{2x-1}{x+3}}\)

\(Đk:\)\(x+3\ne0\Rightarrow x\ne-3\)

Và \(\frac{2x-1}{x+3}\ge0\)

Khi \(\frac{2x-1}{x+3}=0\Rightarrow2x-1=0\)

\(\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)

Khi \(\frac{2x-1}{x+3}>0\)\(\Rightarrow\orbr{\begin{cases}2x-1>0;x+3>0\\2x-1< 0;x+3< 0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x>\frac{1}{2};x>-3\\x< \frac{1}{2};x< -3\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x>\frac{1}{2}\\x< -3\end{cases}}\)

Vậy căn thức xác định khi \(x\ge\frac{1}{2};x< -3\)

\(a,\)\(\sqrt{\frac{1}{\left(x-3\right)^2}}\)

\(đk:\)\(\frac{1}{\left(x-3\right)^3}\ne0\)\(\Rightarrow\left(x-3\right)^3\ne0\)\(\Leftrightarrow x\ne3\)

Và \(\frac{1}{\left(x-3\right)}>0\Rightarrow x-3>0\)\(\Rightarrow x>3\)

Vậy để căn thức xác định thì x > 3

\(\sqrt{8x-x^2-15}\)

\(=\sqrt{-\left(x^2-8x+15\right)}\)

\(=\sqrt{-\left(x^2-8x+16-1\right)}\)

\(=\sqrt{-\left[\left(x^2-8x+16\right)-1\right]}\)

\(=\sqrt{-\left(x-4\right)^2+1}\)

\(đk:\)\(-\left(x-4\right)^2+1\ge0\)

\(\Rightarrow\left(x-4\right)^2\le1\)

\(\Rightarrow\orbr{\begin{cases}\left(x-4\right)^2=1\\\left(x-4\right)^2=0\end{cases}}\)

\(\left(x-4\right)^2=1\Rightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}\)

\(\left(x-4\right)^2=0\Rightarrow x=4\)

Vậy căn thức xác định \(\Leftrightarrow x=\left\{3;4;5\right\}\)