Question

[leanhduy123]

Giải phương trình sau " 

a, \(\sqrt{2x-1}=\sqrt{x^2+2x-5}.\)

b, \(\sqrt{x\left(x^3-3x+1\right)}=\sqrt{x\left(x^3-x\right)}\)

c, \(\sqrt{4x+1}-\sqrt{3x+4}=\sqrt{x-2}\)

d, \(\sqrt{x-1}-\sqrt{5x-1}=\sqrt{3x-2}\)

e, \(\sqrt{x+2}-\sqrt{2x-3}=\sqrt{3x-5}\)

f, \(\sqrt{x\left(x-1\right)+\sqrt{x\left(2x-1\right)}=x}\)

g, \(\sqrt{x+1}+\sqrt{x-1}=2\)

h, \(\sqrt{2x-3}-\sqrt{4x+3}=-3\)

Mn giúp với cần gấp bài toan nâng co giải dc thì tick nhiều

Answer 1

[•๖ۣۜMã ๖ۣۜMã ( ๖ۣۜTεαм ๖ۣۜTαм ๖ۣۜGĭ...]

a, \(\sqrt{x^2+2x-5}\)= \(\sqrt{2x-1}\)( x \(\ge\frac{1}{2}\))

\(\Leftrightarrow x^2+2x-5=2x-1\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=-2\left(ktm\right)\end{cases}}\)

#mã mã#

Answer 2

[•๖ۣۜMã ๖ۣۜMã ( ๖ۣۜTεαм ๖ۣۜTαм ๖ۣۜGĭ...]

b, \(\sqrt{x\left(x^3-3x+1\right)}\)\(=\sqrt{x\left(x^3-x\right)}\)\(\left(x\ge1\right)\)

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

\(\Leftrightarrow\)x( x³ - 3x + 1 ) - x ( x³ - 1 ) = 0

\(\Leftrightarrow\)x ( x³ - 3x + 1 - x³ + 1 ) = 0

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2-3x=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=\frac{2}{3}\left(ktm\right)\end{cases}}\)

vậy pt vô nghiệm

#mã mã#