Question

[hiha 0803]

a) (2x-1)(x^2-4)=0

b) x+3/x+1+x-2/x=2

c) 2x+2/5+3/10<3x-2/4

Answer 1

[๖²⁴ʱ๖ۣۜTɦủү❄吻༉]

a, \(\left(2x-1\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\pm2\end{cases}}\)

b, \(\frac{x+3}{x+1}+\frac{x-2}{x}=2\)ĐKXĐ : \(x\ne0;-1\)

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

Khử mẫu ta đc : \(x^2+3x+x^2-x-2=2x^2+2x\)

\(\Leftrightarrow2x^2+2x-2-2x^2-2x=0\)

\(\Leftrightarrow-2\ne0\)

Vậy pt vô nghiệm 

c, \(\frac{2x+2}{5}+\frac{3}{10}< \frac{3x-2}{4}\)

\(\Leftrightarrow\frac{8x+8}{20}+\frac{6}{20}< \frac{15x-10}{20}\)

\(\Leftrightarrow8x+8+6< 15x-10\)

\(\Leftrightarrow8x+14< 15x-10\)

\(\Leftrightarrow24< 7x\Leftrightarrow\frac{24}{7}< x\)

Answer 2

[l҉o҉n҉g҉ d҉z҉]

( 2x - 1 )( x² - 4 ) = 0

<=> 2x - 1 = 0 hoặc x² - 4 = 0

* 2x - 1 = 0 <=> 2x = 1 => x = 1/2

* x² - 4 = 0 <=> x² = 4 => x = \(\pm\sqrt{4}=\pm2\)

Vậy nghiệm của phương trình là S = { 1/2 ; \(\pm2\)}

\(\frac{x+3}{x+1}+\frac{x-2}{x}=2\)( đkxđ : \(x\ne0;x\ne-1\))

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

<=> \(x^2+3x+x^2-x-2=2x^2+2x\)

<=> \(x^2+3x+x^2-x-2x^2-2x=2\)

<=> \(0x=2\)( vô lí )

Vậy phương trình vô nghiệm 

\(\frac{2x+2}{5}+\frac{3}{10}< \frac{3x-2}{4}\)

<=> \(\frac{4\left(2x+2\right)}{20}+\frac{6}{20}< \frac{5\left(3x-2\right)}{20}\)

<=> \(8x+8+6< 15x-10\)

<=> \(8x-15x< -10-8-6\)

<=> \(-7x< -24\)

<=> \(-7x\div\left(-7\right)>-24\div\left(-7\right)\)

<=> \(x>\frac{24}{7}\)

Vậy nghiệm của bất phương trình là \(x>\frac{24}{7}\)