a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

b: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

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

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

c: Thay \(x=4-2\sqrt{3}\) vào P, ta được:

\(P=\dfrac{-3}{\sqrt{3}-1+3}=\dfrac{-3}{2+\sqrt{3}}=-6+3\sqrt{3}\)

a: Để P nguyên thì \(-3⋮\sqrt{x}+3\)

\(\Leftrightarrow\sqrt{x}+3=3\)

hay x=0

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

Đặt \(2x^2=a;\sqrt{2-x}=b\left(a,b\ge0\right)\)

Phương trình trở thành: \(a+b=ab\Leftrightarrow a-ab+b=0\)

Tới đây bí :v

a) ĐK:\(x\ge0;x\ne9\)

\(P=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{2\sqrt{x}-2-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)

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

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

b)\(P=-\dfrac{3}{\sqrt{x}+3}\) 

Có \(\sqrt{x}+3\ge3;\forall x\ge0\)

\(\Leftrightarrow-\dfrac{3}{\sqrt{x}+3}\ge-\dfrac{1}{3}\)

\(P_{min}=-\dfrac{1}{3}\Leftrightarrow x=0\)

a) Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

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

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

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

a) ĐKXĐ: x <= 2/3

Pt --> 2 - 3x = 4

<=> 3x = -2

<=> x = -2/3 (thỏa)

b) ĐKXĐ: x >= 2

Pt --> x^2 + 4x + 4 = x^2 - 4x + 4

<=> 8x = 0<=> x = 0(loại)

a: Ta có: \(\sqrt{2-3x}=2\)

\(\Leftrightarrow2-3x=4\)

\(\Leftrightarrow3x=-2\)

hay \(x=-\dfrac{2}{3}\)

b: Ta có: \(\sqrt{x^2+4x+4}=x-2\)

\(\Leftrightarrow\left|x+2\right|=x-2\)

\(\Leftrightarrow x+2=2-x\left(x< -2\right)\)

\(\Leftrightarrow x=0\left(loại\right)\)

1) Ta có: \(\sqrt{21-x}+1=x\)

\(\Leftrightarrow21-x=\left(x-1\right)^2\)

\(\Leftrightarrow x^2-2x+1-21+x=0\)

\(\Leftrightarrow x^2-3x-20=0\)

\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-20\right)=9+80=89\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{3+\sqrt{89}}{2}\\x_2=\dfrac{3-\sqrt{89}}{2}\end{matrix}\right.\)

1)\(\sqrt{21-x}+1=x\)

\(\Leftrightarrow21-x=\left(x-1\right)^2\)

\(\Leftrightarrow21-x=x^2-2x+1\)

\(\Leftrightarrow x^2-x-20=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)

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

\(\Leftrightarrow8-x=\left(x-2\right)^2\)

\(\Leftrightarrow8-x=x^2-4x+4\)

\(\Leftrightarrow x^2-3x-4=0\Leftrightarrow\left(x-4\right)\left(x+1\right)=0\)

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

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

\(\Leftrightarrow3x+1=\left(3x-1\right)^2\)

\(\Leftrightarrow3x+1=9x^2-6x+1\)

\(\Leftrightarrow9x^2-9x=0\Leftrightarrow9x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

ai giúp vớ cần gấp

ĐK: \(\frac{2}{3}\le x\le\frac{3}{2}\)

(Vế phải và vế trái đều không âm nên có thể bình phương 2 vế theo một phương trình tương đương)

pt <=> \(x^2\left(3x-2\right)+\left(3-2x\right)+2\sqrt{x^2\left(3x-2\right)\left(3-2x\right)}=x^3+x^2+x+1\)

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

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

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

<=> \(-x^2\left(3-2x\right)-\left(3x-2\right)+2\sqrt{\left(3x-2\right).x^2\left(3-2x\right)}=0\)

<=> \(x^2\left(3-2x\right)+\left(3x-2\right)-2\sqrt{\left(3x-2\right).x^2\left(3-2x\right)}=0\)

<=> \(\left(\sqrt{x^2\left(3-2x\right)}-\sqrt{3x-2}\right)^2=0\)

<=> \(\sqrt{x^2\left(3-2x\right)}-\sqrt{3x-2}=0\)

<=> \(\sqrt{x^2\left(3-2x\right)}=\sqrt{3x-2}\)

<=> \(x^2\left(3-2x\right)=3x-2\)

<=> \(-2x^3+3x^2-3x+2=0\)

<=> \(\left(x-1\right)\left(-2x^2+x-2\right)=0\)

<=> x=1  (tm) 

ĐKXĐ: \(\frac{2}{3}\le x\le\frac{3}{2};x\in R\)

Pt cho tương đương: \(x\sqrt{3x-2}+\sqrt{3-2x}=\sqrt{\left(x+1\right)\left(x^2+1\right)}\)

Đặt \(\sqrt{3x-2}=a;\sqrt{3-2x}=b\left(a,b\ge0\right)\). Khi đó, ta được phương trình:

\(ax+b=\sqrt{\left(a^2+b^2\right)\left(x^2+1\right)}\Leftrightarrow a^2x^2+2abx+b^2=a^2x^2+b^2x^2+a^2+b^2\)

\(\Leftrightarrow2abx-b^2x^2-a^2=0\Leftrightarrow a^2-2abx+b^2x^2=0\)

\(\Leftrightarrow\left(a-bx\right)^2=0\Leftrightarrow a=bx\) hay \(\sqrt{3x-2}=x\sqrt{3-2x}\Leftrightarrow3x-2=3x^2-2x^3\)

\(\Leftrightarrow2x^3-3x^2+3x-2=0\Leftrightarrow2\left(x-1\right)\left(x^2+x+1\right)-3x\left(x-1\right)=9\)

\(\Leftrightarrow\left(x-1\right)\left(2x^2-x+2\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=1\left(tm\right)\\2x^2-x+2=0\left(vn\right)\end{cases}}\)

Vậy PT cho có nghiệm duy nhất x=1.

Cái chỗ " 2(x-1)(x²+x+1) - 3x(x-1) = 9" bn sửa 9 thành 0 nhé, tại mik gõ vội :(

oki mơn bạn

1) \(\sqrt[3]{x+1}=5\)

\(\Rightarrow x+1=125\)

\(\Rightarrow x=124\)

2) \(\sqrt[3]{1-3x^3}=-2\)

\(\Rightarrow1-3x^3=-8\)

\(\Rightarrow3x^3=9\)

\(\Rightarrow x=\sqrt[3]{3}\)

a, \(\sqrt{2x^2-3}=\sqrt{4x-3}\) (x \(\ge\) \(\sqrt{\dfrac{3}{2}}\))

Vì hai vế ko âm, bp 2 vế ta được:

2x² - 3 = 4x - 3

\(\Leftrightarrow\) 2x² = 4x

\(\Leftrightarrow\) x² = 2x

\(\Leftrightarrow\) x² - 2x = 0

\(\Leftrightarrow\) x(x - 2) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(KTM\right)\\x=2\left(TM\right)\end{matrix}\right.\)

Vậy S = {2}

b, \(\sqrt{2x-1}=\sqrt{x-1}\) (x \(\ge\) 1)

Vì hai vế ko âm, bp 2 vế ta được:

2x - 1 = x - 1

\(\Leftrightarrow\) x = 0 (KTM)

Vậy x = \(\varnothing\)

c, \(\sqrt{x^2-x-6}=\sqrt{x-3}\) (x \(\ge\) 3)

Vì hai vế ko âm, bp 2 vế ta được:

x² - x - 6 = x - 3

\(\Leftrightarrow\) x² - 2x - 3 = 0

\(\Leftrightarrow\) x² - 3x + x - 3 = 0

\(\Leftrightarrow\) x(x - 3) + (x - 3) = 0

\(\Leftrightarrow\) (x - 3)(x + 1) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(TM\right)\\x=-1\left(KTM\right)\end{matrix}\right.\)

Vậy S = {3}

d, \(\sqrt{x^2-x}=\sqrt{3x-5}\) (x \(\ge\) \(\dfrac{5}{3}\))

Vì hai vế ko âm, bp 2 vế ta được:

x² - x = 3x - 5

\(\Leftrightarrow\) x² - 4x + 5 = 0

\(\Leftrightarrow\) x² - 4x + 4 + 1 = 0

\(\Leftrightarrow\) (x - 2)² + 1 = 0

Vì (x - 2)² \(\ge\) 0 với mọi x \(\ge\) \(\dfrac{5}{3}\) \(\Rightarrow\) (x - 2)² + 1 > 0 với mọi x \(\ge\) \(\dfrac{5}{3}\)

\(\Rightarrow\) Pt vô nghiệm

Vậy S = \(\varnothing\)

Chúc bn học tốt!

Nguyễn Lê Phước Thịnh nhờ anh xíu ạ