A

Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)

\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)

\(\Leftrightarrow x+8+20x-12=0\)

\(\Leftrightarrow x=\dfrac{4}{21}\)

a) Để biểu thức vô nghĩa thì \(\dfrac{3x-2}{5}-\dfrac{x-4}{3}=0\)

\(\Leftrightarrow\dfrac{3x-2}{5}=\dfrac{x-4}{3}\)

\(\Leftrightarrow3\left(3x-2\right)=5\left(x-4\right)\)

\(\Leftrightarrow9x-6=5x-20\)

\(\Leftrightarrow9x-5x=-20+6\)

\(\Leftrightarrow4x=-14\)

\(\Leftrightarrow x=-\dfrac{7}{2}\)

Ta có: 

\(x^2+9x+2x=\left(x+4\right)\left(x+5\right)\)

\(x^2+11x+30=\left(x+6\right)\left(x+5\right)\)

\(x^2+13x+42=\left(x+6\right)\left(x+7\right)\)

ĐK: \(\left\{{}\begin{matrix}x\ne-4\\x\ne-5\\x\ne-6\\x\ne-7\end{matrix}\right.\)

pt \(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{18\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}-\dfrac{18\left(x+4\right)}{18\left(x+4\right)\left(x+7\right)}=\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}\)

\(\Rightarrow18\left(x+7\right)-18\left(x+4\right)=\left(x+4\right)\left(x+7\right)\)

\(\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+13=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=2\end{matrix}\right.\)  (tm)

`20((x-2)/(x+1))^2-5((x+2)/(x-1))^2+48(x^2-4)/(x^2-1)=0(x ne +-1)`

Đặt `(x-2)/(x+1)=a,(x+2)/(x-1)=b`

`pt<=>20a^2-5b^2+48ab=0`

`<=>20a^2+48ab-5b^2=0`

`<=>20a^2-2ab+50ab-5b^2=0`

`<=>2a(a-10b)+5b(10a-b)=0`

`<=>(a-10b)(2a+5b)=0`

Đến đây dễ rồi bạn tự giải tiếp.

ĐKXĐ: x \(\ne\)\(\pm\)1

Ta có: \(20\left(\dfrac{x-2}{x+1}\right)^2-5\left(\dfrac{x+2}{x-1}\right)^2+48\cdot\dfrac{x^2-4}{x^2-1}=0\)

Đặt: \(\dfrac{x-2}{x+1}=a\) ; \(\dfrac{x+2}{x-1}=b\)

=> ab = \(\dfrac{x^2-4}{x^2-1}\)

Do đó, ta có pt mới: 20a² - 5b² + 48ab = 0

<=> 20a² + 50ab - 2ab - 5b² = 0

<=> (10a - b)(2a + 5b) = 0

<=> \(\left[{}\begin{matrix}10a=b\\2a=-5b\end{matrix}\right.\)

TH1: 10a = b => \(10\cdot\dfrac{x-2}{x+1}=\dfrac{x+2}{x-1}\)

<=> 10(x - 2)(x - 1) = (x + 2)(x + 1)

<=> 10x² - 30x + 20 = x² + 3x + 2

<=> 9x² - 33x + 18 = 0

<=> 9x² - 27x - 6x + 18 = 0

<=> (9x - 6)(x - 3) = 0

<=> \(\left[{}\begin{matrix}x=3\\x=\dfrac{2}{3}\end{matrix}\right.\)(tm)

TH2: \(2a=-5b\)=> \(2\cdot\dfrac{x-2}{x+1}=-5\cdot\dfrac{x+2}{x-1}\)

=> (2x - 4)(x - 1) = (-5x - 10)(x + 1)

<=> 2x² - 6x + 4 = -5x² - 15x - 10

<=> 7x² + 9x + 14 = 0

=> pt vn

\(A=\dfrac{1-\sqrt{x}}{\sqrt{x}+2}=\dfrac{3-\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=\dfrac{3}{\sqrt{x}+2}-1\)

Có \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\)\(\Leftrightarrow\dfrac{3}{\sqrt{x}+2}-1\le\dfrac{1}{2}\)\(\Leftrightarrow A\le\dfrac{1}{2}\)

Dấu "=" xảy ra khi x=0 (tm)

Vậy \(A_{max}=\dfrac{1}{2}\)

Bài 2:

Đk: \(x\ge3;y\ge5;z\ge4\)

Pt\(\Leftrightarrow\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}+\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}+\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}=20\)

Áp dụng AM-GM có:

\(\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}\ge2\sqrt{\sqrt{x-3}.\dfrac{4}{\sqrt{x-3}}}=4\)

\(\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}\ge6\)

\(\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}\ge10\)

Cộng vế với vế \(\Rightarrow VT\ge20\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x-3}=\dfrac{4}{\sqrt{x-3}}\\\sqrt{y-5}=\dfrac{9}{\sqrt{y-5}}\\\sqrt{z-4}=\dfrac{25}{\sqrt{z-4}}\end{matrix}\right.\)\(\Leftrightarrow x=7;y=14;z=29\) (tm)

Vậy...

\(\Leftrightarrow2\sqrt{x}+2=\sqrt{x}+3\)

=>\(\sqrt{x}=1\)

hay x=1

\(\dfrac{\sqrt{x+1}}{\sqrt{x+3}}=\dfrac{1}{2}\)

\(\Leftrightarrow2\left(\sqrt{x+1}\right)=\sqrt{x}+3\)

\(\Leftrightarrow2\sqrt{x}+2=\sqrt{x}+3\)

\(\Leftrightarrow2\sqrt{x}+2-\sqrt{x}-3=0\)

\(\Leftrightarrow\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)

\(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{1}{2}\)

=> \(2\left(\sqrt{x}+1\right)=\sqrt{x}+3\)

     \(2\sqrt{x}+2=\sqrt{x}+3\)

     \(2\sqrt{x}-\sqrt{x}=3-2\)

     \(\sqrt{x}=1\)

=> x=1

b: \(\Leftrightarrow\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\)

=>\(\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\)

=>x*(x+20)=400*6=2400

=>x^2+20x-2400=0

=>(x+60)(x-40)=0

=>x=-60 hoặc x=40

c: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)

=>(2x+1)^2-(2x-1)^2=8

=>4x^2+4x+1-4x^2+4x-1=8

=>8x=8

=>x=1(nhận)