\( \left( {\dfrac{x}{{\sqrt x - 2}} - \dfrac{{4x - 4\sqrt x }}{{x - 2\sqrt x }}} \right).\dfrac{6}{{{{\left( {\sqrt x - 2} \right)}^2}}}\\ = \left[ {\dfrac{x}{{\sqrt x - 2}} - \dfrac{{4\sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}} \right].\dfrac{6}{{{{\left( {\sqrt x - 2} \right)}^2}}}\\ = \left( {\dfrac{{x - 4\sqrt x + 4}}{{\sqrt x - 2}}} \right).\dfrac{6}{{{{\left( {\sqrt x - 2} \right)}^2}}}\\ = \dfrac{{{{\left( {\sqrt x - 2} \right)}^2}}}{{\sqrt x - 2}}.\dfrac{6}{{{{\left( {\sqrt x - 2} \right)}^2}}} = \dfrac{6}{{\sqrt x - 2}} \)

mong mọi người giải giúp em vs 

a: ĐKXĐ: \(x\in R\)

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

=>|2x+3|=5

=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

b: ĐKXĐ: \(x\in R\)

\(\sqrt{9\left(x-2\right)^2}=18\)

=>\(\sqrt{9}\cdot\sqrt{\left(x-2\right)^2}=18\)

=>\(3\cdot\left|x-2\right|=18\)

=>\(\left|x-2\right|=6\)

=>\(\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: x>=2

\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)

=>\(3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)

=>\(4\sqrt{x-2}=40\)

=>\(\sqrt{x-2}=10\)

=>x-2=100

=>x=102(nhận)

d: ĐKXĐ: \(x\in R\)

\(\sqrt{4\left(x-3\right)^2}=8\)

=>\(\sqrt{\left(2x-6\right)^2}=8\)

=>|2x-6|=8

=>\(\left[{}\begin{matrix}2x-6=8\\2x-6=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=14\\2x=-2\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

e: ĐKXĐ: \(x\in R\)

\(\sqrt{4x^2+12x+9}=5\)

=>\(\sqrt{\left(2x\right)^2+2\cdot2x\cdot3+3^2}=5\)

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

=>|2x+3|=5

=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

f: ĐKXĐ:x>=6/5

\(\sqrt{5x-6}-3=0\)

=>\(\sqrt{5x-6}=3\)

=>\(5x-6=3^2=9\)

=>5x=6+9=15

=>x=15/5=3(nhận)

Tưởng bn lớp 5 ạ?? Sao lại đăng câu hỏi lp 9 ạ??:)

minh lop 5 dang chi minh muon nick cua minh

Ta có : x(2x + 3)² - 4x² + 9 = 0 

<=> x(2x + 3)² - (4x² - 9) = 0 

<=> x(2x + 3)² - (2x - 3)(2x + 3) = 0 

<=> (2x + 3)[x(2x + 3) - 2x + 3] = 0 

<=> (2x + 3)(2x² + 3x - 2x + 3) = 0 

<=> (2x + 3)(2x² + x + 3) = 0 

<=> 2x + 3 = 0 (vì 2x² + x + 3 > 0 với mọi x)

<=> 2x = -3

<=> x = -3/2

ĐKXĐ: \(0\le x\le5\).

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{5-x}=b\end{matrix}\right.\left(a,b\ge0\right)\).

PT đã cho tương đương với: \(\left(8-ab\right)\left(a-b\right)=2\left(a-b\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\ab=6\end{matrix}\right.\).

+) \(a=b\Leftrightarrow\sqrt{x}=\sqrt{5-x}\Leftrightarrow x=2,5\left(TMĐK\right)\).

+) \(ab=6\Leftrightarrow\sqrt{x\left(5-x\right)}=6\Leftrightarrow x^2-5x+6=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(TMĐK\right)\\x=3\left(TMĐK\right)\end{matrix}\right.\).

Vậy...

ĐK: \(0\le x\le5\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\\\sqrt{5-x}=b\end{matrix}\right.\left(a,b\ge0\right)\)

\(pt\Leftrightarrow\left(8-ab\right)\left(a-b\right)=2\left(a^2-b^2\right)\)

\(\Leftrightarrow\left(a-b\right)\left(8-ab-2a-2b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\ab+2a+2b=8\end{matrix}\right.\)

TH1: \(a=b\Leftrightarrow\sqrt{x}=\sqrt{5-x}\Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)

TH2: \(ab+2a+2b=8\)

\(\Leftrightarrow\sqrt{5x-x^2}+2\sqrt{5-x}+2\sqrt{x}=8\)

\(\Leftrightarrow\left(\sqrt{5-x}+\sqrt{x}-3\right)\left(\sqrt{5-x}+\sqrt{x}+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{5-x}+\sqrt{x}=-7\left(l\right)\\\sqrt{5-x}+\sqrt{x}=3\end{matrix}\right.\)

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

\(\Leftrightarrow5+2\sqrt{5x-x^2}=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

Vậy ...

cho mk hỏi một chút là đây đích thực có phải lớp 1 ko ak?