`a)`\(D=\dfrac{x^2-x-1}{3x}+\left(\dfrac{x+2}{3x}+\dfrac{3x+1}{x+1}\right):\dfrac{2-4x}{x+1}\);\(x\ne-1;0\)

\(D=\dfrac{x^2-x-1}{3x}+\left[\dfrac{\left(x+2\right)\left(x+1\right)+3x\left(3x+1\right)}{3x\left(x+1\right)}\right].\dfrac{x+1}{2-4x}\)

\(D=\dfrac{x^2-x-1}{3x}+\dfrac{x^2+3x+2+9x^2+3x}{3x\left(x+1\right)}.\dfrac{x+1}{2-4x}\)

\(D=\dfrac{x^2-x-1}{3x}+\dfrac{10x^2+6x+2}{3x\left(x+1\right)}.\dfrac{x+1}{2-4x}\)

\(D=\dfrac{x^2-x-1}{3x}+\dfrac{10x^2+6x+2}{3x\left(2-4x\right)}\)

\(D=\dfrac{\left(2-4x\right)\left(x^2-x-1\right)+10x^2+6x+2}{3x\left(2-4x\right)}\)

\(D=\dfrac{2x^2-2x-2-4x^3+4x^2+4x+10x^2+6x+2}{3x\left(2-4x\right)}\)

\(D=\dfrac{-4x^3+12x^2+8x}{3x\left(2-4x\right)}\)

\(D=\dfrac{-4x\left(x^2+3x+2\right)}{3x\left(2-4x\right)}\)

\(D=-\dfrac{4\left(x+1\right)\left(x+2\right)}{3\left(2-4x\right)}\)

`b)`\(\left|2x-1\right|=4-x\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4-x;x\ge\dfrac{1}{2}\\1-2x=4-x;x< \dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\left(tm\right)\\x=-3\left(tm\right)\end{matrix}\right.\)

`@`Với `x=5/3` thế vào D, ta được:

\(D=-\dfrac{4\left(\dfrac{5}{3}+1\right)\left(\dfrac{5}{3}+2\right)}{3\left(2-4.\dfrac{5}{3}\right)}=\dfrac{176}{63}\)

`@`Với `x=-3` thế vào D, ta được:

\(D=-\dfrac{4\left(-3+1\right)\left(-3+2\right)}{3\left(2-4.-3\right)}=-\dfrac{4}{21}\)

`c)`\(D=\dfrac{5}{3}\)

`<=>`\(\dfrac{5}{3}=-\dfrac{4\left(x+1\right)\left(x+2\right)}{3\left(2-4x\right)}\)

\(\Leftrightarrow5\left(2-4x\right)=-4\left(x+1\right)\left(x+2\right)\)

\(\Leftrightarrow10-20x=-4x^2-12x-8\)

\(\Leftrightarrow4x^2+8x+2=0\)

\(\Leftrightarrow2x^2+4x+1=0\)

\(\Delta=4^2-4.2=16-8=8>0\)

\(\rightarrow\left[{}\begin{matrix}x=\dfrac{-4+\sqrt{8}}{4}=\dfrac{-2+\sqrt{2}}{2}\left(tm\right)\\x=\dfrac{-4-\sqrt{8}}{4}=\dfrac{-2-\sqrt{2}}{2}\left(tm\right)\end{matrix}\right.\)

`d)`\(D>0\)

`<=>`\(-\dfrac{4\left(x+1\right)\left(x+2\right)}{3\left(2-4x\right)}>0\)

`<=>`\(\dfrac{4\left(x+1\right)\left(x+2\right)}{3\left(2-4x\right)}< 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}4\left(x+1\right)\left(x+2\right)>0\\3\left(2-4x\right)< 0\end{matrix}\right.\) hoặc \(\Leftrightarrow\left\{{}\begin{matrix}4\left(x+1\right)\left(x+2\right)< 0\left(1\right)\\3\left(2-4x\right)>0\left(2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>-1\\x>\dfrac{1}{2}\end{matrix}\right.\)\(\rightarrow x>\dfrac{1}{2}\)        |     \(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x+1>0\\x+2< 0\end{matrix}\right.\) hoặc \(\left[{}\begin{matrix}x+1< 0\\x+2>0\end{matrix}\right.\)

                                                        \(\Leftrightarrow\left[{}\begin{matrix}x>-1\\x< -2\end{matrix}\right.\)    hoặc \(\left[{}\begin{matrix}x< -1\\x>-2\end{matrix}\right.\)  

                                                      \(\left(2\right)\Leftrightarrow2-4x>0\)

                                                          \(\Leftrightarrow x< \dfrac{1}{2}\)

Lời giải:

$\sqrt{7+\sqrt{2x}}=3+\sqrt{5}$

$7+\sqrt{2x}=(3+\sqrt{5})^2=14+6\sqrt{5}$

$\sqrt{2x}=7+6\sqrt{5}$

$2x=(7+6\sqrt{5})^2=229+84\sqrt{5}$

$x=114,5+42\sqrt{5}$

\(\left(2+\sqrt{5}+\sqrt{3}\right)\left(2+\sqrt{5}-\sqrt{3}\right)\)

\(=\left(2+\sqrt{5}\right)^2-\left(\sqrt{3}\right)^2\)

\(=4+4\sqrt{5}+5-3\)

\(=6+4\sqrt{5}\)

tổng kết quả là 14,94 
 

5,96 x 2,5 =14,94 
 mình ko bt giải thích thế nào cho hiểu bạn thong cảm nhé ( tôi cúng lớp 9 )

`a)` A có nghĩa khi \(\left\{{}\begin{matrix}2ab\ge0\\\sqrt{a^2+b^2}\ne0\end{matrix}\right.\)

                               \(\Leftrightarrow\left\{{}\begin{matrix}a,b\ge0;a,b\le0\\a;b\ne0\end{matrix}\right.\)

                                 \(\rightarrow\left[{}\begin{matrix}a;b>0\\a;b< 0\end{matrix}\right.\)

`b)`\(a=b\)

\(A=\left(1-\dfrac{\sqrt{2ab}}{\sqrt{a^2+b^2}}\right)\left(1+\dfrac{\sqrt{2ab}}{\sqrt{a^2+b^2}}\right)\)

\(A=1^2-\left(\dfrac{\sqrt{2ab}}{\sqrt{a^2+b^2}}\right)^2\)

\(A=1-\dfrac{2ab}{a^2+b^2}\)

\(A=1-\dfrac{2a^2}{2a^2}\)

\(A=1-1=0\)