đk: \(x>0;x\ne1\)

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

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

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

\(\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\right)=\frac{x-1}{\sqrt{x}}:[\left(\sqrt{x}-1\right).\left(\frac{1}{\sqrt{x}}+\frac{-1}{\sqrt{x}\left(1+\sqrt{x}\right)}\right)\)

\(=[\frac{x-1}{\sqrt{x}}:\left(\sqrt{x}-1\right)]:\left(\frac{1}{\sqrt{x}}+\frac{-1}{\sqrt{x}\left(1+\sqrt{x}\right)}\right)=\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{\sqrt{x}+1-1}{\sqrt{x}\left(1+\sqrt{x}\right)}\)

\(\frac{\left(\sqrt{x}+1\right)\sqrt{x}\left(1+\sqrt{x}\right)}{\sqrt{x}.\sqrt{x}}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}=\sqrt{x}+2+\frac{1}{\sqrt{x}}\)

chị hiểu ko ạ, nếu chị hiểu k dùm em ạ

câu này chủ yếu tập trung vào công thức nhé bạn 

cos bình cộng sin bình bằng 1

thế cos vào tính sin 

tan bằng sin chia cos

cot a bằng cos chia sin 

thế nào ra nhé cẩn thận bạn có thể thiếu trường hợp nhé cám ơn nhiều 

cần hõi gì cứ nhắn THẰNG THẦY LỢI YOUTUBE

ta có : \(sina=\sqrt{1-cos^2a}=\sqrt{1-0.4^2}=\frac{\sqrt{21}}{5}\)

ta có : \(\hept{\begin{cases}tana=\frac{sina}{cosa}=\frac{\sqrt{21}}{2}\\cota=\frac{1}{tana}=\frac{2}{\sqrt{21}}\end{cases}}\)

\(ĐKXĐ:x\ge\sqrt[3]{\frac{-7}{2}}\)

\(\left(6x\sqrt{2x^3+7}-18\right)=6x^3+2x-8-\left(4\sqrt{2x^3+7}-12\right)\)

\(\frac{36x^2\left(2x^3+7\right)-324}{6x\sqrt{2x^3+7}+18}=2\left(3x^3+x-4\right)-\frac{16\left(2x^3+7\right)-144}{4\sqrt{2x^3+7}+12}\)

\(\frac{72x^5+252x^2-324}{6x\sqrt{2x^3+7}+18}=2\left(x-1\right)\left(3x^2+3x+4\right)-\frac{32x^3-32}{4\sqrt{2x^3+7}+12}\)

\(\frac{36\left(2x^5+7x^2-9\right)}{6x\sqrt{2x^3+7}+18}=2\left(x-1\right)\left(3x^2+3x+4\right)-\frac{32\left(x-1\right)\left(x^2+x+1\right)}{4\sqrt{2x^3+7}+12}\)

\(\frac{36\left(x-1\right)\left(2x^4+2x^3+2x^2+9x+9\right)}{6x\sqrt{2x^3+7}+18}=2\left(x-1\right)\left(3x^2+3x+4\right)-\frac{32\left(x-1\right)\left(x^2+x+1\right)}{4\sqrt{2x^3+7}+12}\)

\(\left(x-1\right)\left[\frac{36\left(2x^4+2x^3+2x^2+9x+9\right)}{6x\sqrt{2x^3+7}+18}+\frac{32\left(x^2+x+1\right)}{4\sqrt{2x^3+7}+12}-2\left(3x^2+3x+4\right)\right]=0\)

\(\orbr{\begin{cases}x=1\left(TM\right)\\\frac{36\left(2x^4+2x^3+2x^2+9x+9\right)}{6x\sqrt{2x^3+7}+18}+\frac{32\left(x^2+x+1\right)}{4\sqrt{2x^3+7}+12}-2\left(3x^2+3x+4\right)=0\end{cases}}\)

chưa biết cm câu cuối thế nào :v

Cảm ơn cậu nhaa

a. Điều kiện xác định: \(x^2-4\ne0\Leftrightarrow x\ne\pm2\)

b. ta có :\(\frac{x^2-4x+4}{x^2-4}=\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{x-2}{x+2}=A\)

c.\(\left|x\right|=3\Leftrightarrow\orbr{\begin{cases}x=3\Rightarrow A=\frac{3-2}{3+2}=\frac{1}{5}\\x=-3\Rightarrow A=\frac{-3-2}{-3+2}=5\end{cases}}\)

a, $ĐKXĐ:x^2-4\ne 0\Rightarrow x\ne \pm 2$

b,Đặt  $A=\frac{x^2-4x+4}{x^2-4}$

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

c, $\left|x\right|=3\Rightarrow \text{orbr}\{\begin{array}{c}x=3\\ x=-3\end{array}$ (thỏa mãn ĐKXĐ)

Với x = 3 thì $A=\frac{3-2}{3+2}=\frac{1}{5}$

Với x = -3 thì $A=\frac{-3-2}{-3+2}=5$

d, $A<2\Rightarrow \frac{x-2}{x+2}<2\Rightarrow x-2<2x+4\Rightarrow -2-4<2x-x\Rightarrow x$

a. \(A=\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)\left(\frac{x-\sqrt{x}}{\sqrt{x}+1}-\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)=\left(\frac{x-1}{2\sqrt{x}}\right).\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{x-1}\right)\)

\(=\left(\frac{x-1}{2\sqrt{x}}\right).\left(\frac{-4x}{x-1}\right)=-2\sqrt{x}\)

Để \(A>-6\Leftrightarrow-2\sqrt{x}>-6\Leftrightarrow\sqrt{x}< 3\Leftrightarrow0\le x< 9\)

để \(y=\left(\sqrt{3}-\sqrt{5}\right)x+\sqrt{5}+\sqrt{3}=1\)

thì \(\left(\sqrt{3}-\sqrt{5}\right)x=1-\sqrt{5}-\sqrt{3}\)

\(\Leftrightarrow x=\frac{1-\sqrt{3}-\sqrt{5}}{\sqrt{3}-\sqrt{5}}\)

b.\(f^2\left(x\right)=\left[\left(\sqrt{3}-\sqrt{5}\right)x+\sqrt{5}+\sqrt{3}\right]^2=8+2\sqrt{15}=\left(\sqrt{5}+\sqrt{3}\right)^2\)

\(\Leftrightarrow\left[\left(\sqrt{3}-\sqrt{5}\right)x+2\sqrt{5}+2\sqrt{3}\right]\left(\sqrt{3}-\sqrt{5}\right)x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{2\left(\sqrt{3}+\sqrt{5}\right)x}{\left(\sqrt{3}-\sqrt{5}\right)x}\end{cases}}\)

a, Thay x = 36 vào B ta được : \(B=\frac{6}{6-3}=\frac{6}{3}=2\)

b, \(B< \frac{1}{2}\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{1}{2}< 0\Leftrightarrow\frac{\sqrt{x}+3}{2\left(\sqrt{x}-3\right)}< 0\)Với \(x>0;x\ne9\)

\(\Rightarrow\sqrt{x}-3< 0\Leftrightarrow x< 9\Rightarrow0< x< 9\)

c, Với \(x>0;x\ne1\)

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

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

d, \(P=AB=\frac{\sqrt{x}+2}{\sqrt{x}-3}=\frac{\sqrt{x}-3+5}{\sqrt{x}-3}=1+\frac{5}{\sqrt{x}-3}\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)