\(ĐK:x\ge\frac{3}{2}\)

Đặt : \(\sqrt{4x^2+9}=a;\sqrt{2x-3}=b\); a lớn hơn  0; b lớn hơn hoặc bằng 0

ta có: \(a^2-b^2=4x^2+9-2x+3=2\left(2x^2-x+6\right)\)

Ta có phương trình:

\(\frac{a^2-b^2}{2x}=a+b\Leftrightarrow\frac{\left(a-b\right)\left(a+b\right)}{2x}=a+b\)

mà a+b lớn hơn 0

phương trình trên <=> \(\frac{a-b}{2x}=1\Leftrightarrow a-b=2x\)( chia hai vế cho a+b)

Khi đó ta có phương trình ẩn x

\(\sqrt{4x^2+9}-\sqrt{2x-3}=2x\)

=> \(4x^2+9+2x-3-2\sqrt{\left(4x^2+9\right)\left(2x-3\right)}=4x^2\)

<=> \(3+x=\sqrt{8x^3-12x^2+18x-27}\)

<=> \(8x^3-13x^2+12x-36=0\)

<=> \(\left(x-2\right)\left(8x^2+3x+18\right)\)=0

<=> x=2  (tmđk)

thử lại vào phương trình ban đầu thấy thỏa mãn

Vậy x=2

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

\(=4\sin^2\alpha-2\sin^2\alpha+2\cos^2\alpha=2\left(\sin^2\alpha+\cos^2\alpha\right)=2\)

\(B=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\)

\(=\left(\sin^2\alpha+\cos^2\alpha\right)^2-1=0\)

\(C=3\left(\sin^4\alpha+\cos^4\alpha\right)-2\sin^2\alpha.\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=3\left(\sin^4\alpha+\cos^4\alpha\right)-2\sin^2\alpha.\cos^2\alpha\)

\(=3\left(\sin^2\alpha+\cos^2\alpha-\frac{1}{9}\right)^2-\frac{1}{9}=\frac{61}{27}\)

ĐK x >0

\(PT\Leftrightarrow2x+2\sqrt{x^2-\frac{1}{x^4}}=\frac{4}{x^2}.\)

\(\Leftrightarrow2\sqrt{x^2-\frac{1}{x^4}}=\frac{4}{x^2}-2x\)

\(\Leftrightarrow x^2-\frac{1}{x^4}=\frac{4}{x^4}-\frac{4}{x}+x^2\)(chia cả 2 vế cho 2)

\(\Leftrightarrow\frac{5}{x^4}-\frac{4}{x}=0\Leftrightarrow5-4x^3=0\Leftrightarrow4x^3=5\)

\(\Leftrightarrow x^3=\frac{5}{4}\Leftrightarrow x=\sqrt[3]{\frac{5}{4}}\)

Vậy................................

\(A=\sqrt{4+\sqrt{4+\sqrt{4+....}}}\)vô số dấu căn

\(\Leftrightarrow A^2=4+\sqrt{4+\sqrt{4+\sqrt{4+....}}}\)

\(\Leftrightarrow A^2-A-4=0\)

\(\Leftrightarrow\orbr{\begin{cases}A=\frac{1-\sqrt{17}}{2}\left(l\right)\\A=\frac{1+\sqrt{17}}{2}=2,56< 3\end{cases}}\)

Từ đây ta có \(\sqrt{4+\sqrt{4+\sqrt{4+....}}}< 3\)

\(A=\sqrt{4+\sqrt{4}+\sqrt{4+.....}}\)vô số dấu căn

\(\Leftrightarrow A^2=4+\sqrt{4+\sqrt{4+\sqrt{4+...}}}\)

\(\Leftrightarrow A^2-A-A=0\)

\(\Leftrightarrow\orbr{\begin{cases}A=\frac{1-\sqrt{17}}{2}\\A=\frac{1+\sqrt{17}}{2}=2,56< 3\end{cases}}\)

Từ đây ta có: \(\sqrt{4+\sqrt{4}+\sqrt{4+.....}}< 3\)

Rất vui vì giúp đc bạn <3

a) \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}+\frac{2b}{a-b}\)

\(=\frac{a+b+2\sqrt{ab}}{2\left(a-b\right)}-\frac{a+b-2\sqrt{ab}}{2\left(a-b\right)}+\frac{4b}{2\left(a-b\right)}=\frac{a+b+2\sqrt{ab}-a-b+2\sqrt{ab}+4b}{2\left(a-b\right)}\)

\(=\frac{4\sqrt{ab}+4b}{2\left(a-b\right)}=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)}\)

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

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)-\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\frac{4\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)\(=\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(A=\frac{x^2}{x^2+1}=\frac{x^2+1-1}{x^2+1}=1-\frac{1}{x^2+1}\)

A lớn nhất\(\Leftrightarrow\frac{1}{x^2+1}\)nhỏ nhất\(\Leftrightarrow x^2+1\)lớn nhất

Dễ chứng minh được \(x^2+1\)không có GTLN

Vậy A không có GTLN

\(\sqrt{x^2-\frac{x^2}{7}}=\sqrt{\frac{7x^2}{7}-\frac{x^2}{7}}=\sqrt{\frac{6x^2}{7}}=\frac{\sqrt{6x^2}}{\sqrt{7}}=\frac{\sqrt{6x^2}.\sqrt{7}}{\sqrt{7}.\sqrt{7}}=\frac{x\sqrt{6.7}}{7}=\frac{x\sqrt{42}}{7}\)