1)

ĐKXĐ: x>4

Ta có: \(\dfrac{\sqrt{x+5}}{\sqrt{x-4}}=\dfrac{\sqrt{x-2}}{\sqrt{x+3}}\)

\(\Leftrightarrow x^2+8x+15=x^2-6x+8\)

\(\Leftrightarrow8x+6x=8-15\)

\(\Leftrightarrow14x=-7\)

hay \(x=-\dfrac{1}{2}\)(loại)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)

1:

a: =>(|x|+4)(|x|-1)=0

=>|x|-1=0

=>x=1; x=-1

b: =>x^2-4>=0

=>x>=2 hoặc x<=-2

d: =>|2x+5|=2x-5

=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0

=>x=0(loại)

Giải cách phương trình sau

2.(x+3/5)=5-(13/5+x)

=> 2x + 6/5 = 5 - 13/5 - x

=> 2x + x = 12/5 - 6/5

=> 3x = 6/5

=> x = 2/5

a)

\(\frac{4}{x-5}+\frac{9}{x+5}=\frac{1}{2}\\ \Leftrightarrow\frac{4x+20}{2\cdot\left(x^2-25\right)}+\frac{9x-40}{2\cdot\left(x^2-25\right)}-\frac{x^2-25}{2\cdot\left(x^2-25\right)}=0\\ \Leftrightarrow\frac{4x+20+9x-40-x^2+25}{2\cdot\left(x^2-25\right)}=0\\ \Leftrightarrow\frac{13x+5-x^2}{2\cdot\left(x^2-25\right)}=0\\ \Rightarrow13x+5-x^2=0\\ \Rightarrow x=\left[{}\begin{matrix}-\frac{-13+3\sqrt{21}}{2}\\-\frac{-13-3\sqrt{21}}{2}\end{matrix}\right.\)

b)

\(x^4-4x^2+3=0\\ \Leftrightarrow\left(x^2-3\right)\cdot\left(x^2-1\right)=0\\ \Rightarrow x=\left[{}\begin{matrix}\sqrt{3}\\-\sqrt{3}\\1\\-1\end{matrix}\right.\)

Bài làm

a) \(\frac{4}{x-5}+\frac{9}{x+5}=\frac{1}{2}\) ĐKXĐ: x khác 5, x khác -5

\(\Rightarrow2.4.\left(x+5\right)+9.2\left(x-5\right)=2\left(x+5\right)\left(x-5\right)\)

\(\Leftrightarrow8x+40+18x-90=2x^2-50\)

\(\Leftrightarrow-2x^2+26x=0\)

\(\Leftrightarrow2x\left(-x+13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\-x+13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=13\end{matrix}\right.\)( tmđkxđ )

Vậy S = { 0; 13 }

b) x⁴ - 4x² + 3 = 0

<=> x⁴ - x² - 3x² + 3 = 0

<=> x²( x² - 1 ) - 3( x² - 1 ) = 0

<=> ( x² - 3 )( x² - 1 ) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x^2-3=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{3}\\x=\pm1\end{matrix}\right.\)

Vậy S = { 3; -3; 1; -1 }

cộng cả 2 vế với -1

x=105

Ta có :\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)

<=> \(\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)=\left(\frac{x-100}{5}-1\right)+\left(\frac{x-101}{4}-1\right)+\left(\frac{x-102}{3}-1\right)\)

<=> \(\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)

<=> \(\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}\right)=\left(x-105\right)\left(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}\right)\)

<=> \(\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)

<=> x - 105 = 0 (Vì \(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\))

<=> x = 105

Vậy nghiệm phương trình là x = 105

#muon roi ma sao con

\(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)

\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}-\frac{x-105}{5}-\frac{x-101}{4}-\frac{x-102}{3}=0\)

( cả 2 vế trừ đi 3 và từng phân thức trừ đi 1 ) 

\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\ne0\right)=0\Leftrightarrow x=105\)

Vậy tập nghiệm của pt là S = { 105 } 

a ) \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\)

\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\)

\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x^2-3x^2\right)+\left(6x+3x\right)+\left(8-1\right)=17\)

\(\Leftrightarrow9x+7=17\)

\(\Leftrightarrow9x=10\)

\(\Leftrightarrow x=\dfrac{10}{9}\)

Vậy nghiệm của p/t là : \(\dfrac{10}{9}\)

b ) \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)

\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3+8\right)=3\)

\(\Leftrightarrow x^3-25x-x^3-8=3\)

\(\Leftrightarrow-25x-8=3\)

\(\Leftrightarrow-25x=11\)

\(\Leftrightarrow x=-\dfrac{11}{25}\)

Vậy nghiệm của p/t là : \(-\dfrac{11}{25}\)