a) x(2x - 1) - (x - 2)(2x + 3) = 5

2x² - x - 2x² - 3x + 4x + 6 = 5

0x = -1 (vô lý)

Vậy không tìm được x

b) (x - 3)² - 25 = 0

(x - 3)² - 5² = 0

(x - 3 - 5)(x - 3 + 5) = 0

(x - 8)(x + 2) = 0

\(\Rightarrow\) x - 8 = 0 hoặc x + 2 = 0

*) x - 8 = 0

x = 0 + 8

x = 8

*) x + 2 = 0

x = 0 - 2

x = -2

Vậy x = 8; x = -2

c) (x - 1)(2 - x) + (x + 3)² = 4 - 2x

2x - x² - 2 + x + x² + 6x + 9 = 4 - 2x

9x + 7 = 4 - 2x

9x + 2x = 4 - 7

11x = -3

x = \(\dfrac{-3}{11}\)

Vậy x = \(\dfrac{-3}{11}\)

Nhập biểu thức ở dấu Σ cho dễ nhìn em ạ

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

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-2x}{x-9}:\dfrac{\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-x-3\sqrt{x}}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{-\sqrt{x}+5}\)

\(=\dfrac{-\sqrt{x}\left(\sqrt{x}+3\right)\cdot\sqrt{x}\left(\sqrt{x}-3\right)}{\left(x-9\right)\left(-\sqrt{x}+5\right)}=\dfrac{-x}{-\sqrt{x}+5}\)

\(A=\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+5}+\dfrac{\sqrt{x}}{\sqrt{x}-5}+\dfrac{x+9}{25-x}\right):\dfrac{1-\sqrt{x}}{5+\sqrt{x}}\)

\(=\dfrac{x-7\sqrt{x}+10+x+5\sqrt{x}-x-9}{\left(x-25\right)}\cdot\dfrac{\sqrt{x}+5}{1-\sqrt{x}}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}=\dfrac{1-\sqrt{x}}{\sqrt{x}-5}\)

\(B=\left(\dfrac{1}{x-4}-\dfrac{1}{x-4\sqrt{x}+4}\right):\dfrac{\sqrt[2]{x}}{2\sqrt{x}-x}\)

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

\(=\dfrac{-4}{x-4}\)

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Tìm X

a) \(2x+\dfrac{3}{24}=3x-\dfrac{1}{32}\)

\(\Leftrightarrow\left(2x+\dfrac{3}{24}\right)-\left(3x-\dfrac{1}{32}\right)=0\)

\(\Leftrightarrow2x+\dfrac{3}{24}-3x+\dfrac{1}{32}=0\)

\(\Leftrightarrow\left(\dfrac{3}{24}+\dfrac{1}{32}\right)+\left(2x-3x\right)=0\)

\(\Leftrightarrow\dfrac{5}{32}-x=0\)

\(\Leftrightarrow x=\dfrac{5}{32}\)

a) \(\left(x+3\right)^2-\left(x-2\right)^3=\left(x+5\right)\left(x^2-5x+25\right)-108\)

\(\Leftrightarrow x^2+6x+9-x^2+4x-4=x^3-5x^2+25x+5x^2-25x+125-108\)

\(\Leftrightarrow x^3-10x+12=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+6\right)=0\)

\(\Leftrightarrow x=2\)( do \(x^2+2x+6=\left(x+1\right)^2+4\ge4>0\))

\(a,4+3x=25-4x\\ \Leftrightarrow7x=21\\ \Leftrightarrow x=3\\ b,\left(x-1\right)^2+\left(x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-1+x+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(2x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

c, ĐKXĐ:\(x\ne-1,x\ne2\)

\(\dfrac{1}{x+1}+\dfrac{3}{x-2}=\dfrac{9}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}+\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{9}{\left(x+1\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x-2+3x+3-9}{\left(x+1\right)\left(x-2\right)}=0\\ \Rightarrow4x-8=0\\ \Leftrightarrow x=2\left(ktm\right)\)