a,ĐK: x≥4

Ta có: \(2\sqrt{x-4}-\dfrac{1}{3}\sqrt{9x-36}=4-\sqrt{x-4}\)

      \(\Leftrightarrow2\sqrt{x-4}-\sqrt{x-4}=4-\sqrt{x-4}\)

      \(\Leftrightarrow2\sqrt{x-4}=4\)

      \(\Leftrightarrow\sqrt{x-4}=2\Leftrightarrow x-4=4\Leftrightarrow x=8\left(tm\right)\)

b, ĐK: x≥2

Ta có: \(3\sqrt{x-2}-\sqrt{x^2-4}=0\)

      \(\Leftrightarrow3\sqrt{x-2}-\sqrt{\left(x-2\right)\left(x+2\right)}=0\)

      \(\Leftrightarrow\sqrt{x-2}\left(3-\sqrt{x+2}\right)=0\)

      \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\3-\sqrt{x+2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x+2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=7\end{matrix}\right.\)

Lời giải:

a. ĐKXĐ: $x\in\mathbb{R}$

PT $\Leftrightarrow \sqrt{(x-2)^2}=5$

$\Leftrightarrow |x-2|=5$

$\Leftrightarrow x-2=5$ hoặc $x-2=-5$

$\Leftrightarrow x=7$ hoặc $x=-3$ (đều tm)

b. ĐKXĐ: $x\geq -1$

PT $\Leftrightarrow \sqrt{16}.\sqrt{x+1}-3\sqrt{x+1}+\sqrt{4}.\sqrt{x+1}=16-\sqrt{x+1}$

$\Leftrightarrow 4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}$

$\Leftrightarrow 4\sqrt{x+1}=16$

$\Leftrightarrow \sqrt{x+1}=4$

$\Leftrightarrow x+1=16$

$\Leftrightarrow x=15$ (tm)

a. \(\sqrt{2x+5}=\sqrt{1-x}\)

<=>  2x + 5 = 1 - x

<=> 2x + x = 1 - 5

<=> 3x = -4

<=> x = \(\dfrac{-4}{3}\)

Vậy ...............

b. \(\sqrt{x^2-x}=\sqrt{3-x}\)

<=> x² - x = 3 - x

<=> x² - x + x = 3

<=> x² = 3

<=> x = \(\sqrt{3}\)

Vậy ..................

c. \(\sqrt{2x^2-3}=\sqrt{4x-3}\)

<=> 2x² - 3 = 4x - 3

<=> 2x² - 4x = -3 + 3

<=> 2x² - 4x = 0

<=> x(x - 4) = 0

\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy .................

a,\(ĐK:-\dfrac{5}{2}\le x\le1\)

Ta có: \(\left(1\right)\Leftrightarrow2x+5=1-x\)

                \(\Leftrightarrow3x=-4\Leftrightarrow x=-\dfrac{4}{3}\left(tm\right)\)

b,\(ĐK:1\le x\le3\)

Ta có: \(\left(1\right)\Leftrightarrow x^2-x=3-x\)

                \(\Leftrightarrow x^2=3\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\left(tm\right)\\x=-\sqrt{3}\left(loại\right)\end{matrix}\right.\)

c,\(ĐK:\left\{{}\begin{matrix}x\ge\sqrt{\dfrac{3}{2}}\\x\le-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\)

Ta có: \(\left(1\right)\Leftrightarrow2x^2-3=4x-3\)

                \(\Leftrightarrow2x^2-4x=0\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\left(tm\right)\end{matrix}\right.\)

sorry bn mik quên ĐKXĐ và bn thêm x = \(-\sqrt{3}\) vào câu b giùm mik nha

giúp với mọi người

Câu 1: ĐKXĐ: ...

\(\Leftrightarrow4x\left(3x-1\right)+x-1=4x\sqrt{3x+1}\)

\(\Leftrightarrow12x^2-3x-1-4x\sqrt{3x+1}=0\)

\(\Leftrightarrow16x^2-\left(4x^2+4x\sqrt{3x+1}+3x+1\right)=0\)

\(\Leftrightarrow16x^2-\left(2x+\sqrt{3x+1}\right)^2=0\)

\(\Leftrightarrow\left(2x-\sqrt{3x+1}\right)\left(6x+\sqrt{3x+1}\right)=0\)

\(\Leftrightarrow...\)

Câu 2:

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x^2-4\right)=y^3+2y\\x^2-4=-3y^2\end{matrix}\right.\)

\(\Leftrightarrow x\left(-3y^2\right)=y^3+2y\)

\(\Leftrightarrow y\left(y^2+3xy+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=0\Rightarrow...\\y^2+3xy+2=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow3xy=-y^2-2\Rightarrow x=\frac{-y^2-2}{3y}\)

\(\Rightarrow\left(\frac{y^2+2}{3y}\right)^2-1=3\left(1-y^2\right)\)

\(\Leftrightarrow\left(\frac{y^2-3y+2}{3y}\right)\left(\frac{y^2+3y+2}{3y}\right)=3\left(1-y^2\right)\)

\(\Leftrightarrow\frac{\left(y-1\right)\left(y-2\right)\left(y+1\right)\left(y+2\right)}{9y^2}=3\left(1-y^2\right)\)

\(\Leftrightarrow\frac{\left(y^2-1\right)\left(y^2-4\right)}{9y^2}=3\left(1-y^2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\\frac{y^2-4}{9y^2}=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y^2-1=0\\28y^2=4\end{matrix}\right.\)

\(3x-1+\frac{x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{4x\left(3x-1\right)+x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{12x^2-4x+x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{12x^2-3x-1}{4x}=\sqrt{3x+1}\)

\(\Leftrightarrow\frac{\left(12x^2-3x-1\right)^2}{16x^2}=3x+1\)

\(\Leftrightarrow\left(12x^2-3x-1\right)^2=16x^2\left(3x+1\right)\)

\(\Leftrightarrow144x^4-120x^3-31x^2+6x+1=0\)

\(\Leftrightarrow144x^4-144x^3+24x^3-24x^2-7x^2+7x-x+1=0\)

\(\Leftrightarrow144x^3\left(x-1\right)+24x^2\left(x-1\right)+7x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(144x^3+24x^2+7x-1\right)=0\)

Tìm được mỗi nghiệm thôi à :v