ĐKXĐ:...

\(x^2+\frac{36}{x^2}-4\left(x-\frac{6}{x}\right)-17=0\)

Đặt \(x-\frac{6}{x}=a\Rightarrow a^2=x^2+\frac{36}{x^2}-12\Rightarrow x^2+\frac{36}{x^2}=a^2+12\)

\(a^2+12-4a-17=0\)

\(\Leftrightarrow a^2-4a-5=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-\frac{6}{x}=-1\\x-\frac{6}{x}=5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+x-6=0\\x^2-5x-6=0\end{matrix}\right.\)

hiu hiu

help me !!!!

ko biết ok

sử dụng BDT cosi là ra bn à

\(x^2-12+\frac{36}{x^2}-4x+\frac{24}{x}=5\)

\(\Leftrightarrow x^2+\frac{36}{x^2}-4x+\frac{24}{x}=5+12\)

\(\Leftrightarrow x^2+\frac{36}{x^2}-4x+\frac{24}{x}=17\)

\(\Leftrightarrow x^2.x^2+\frac{36}{x^2}.x^2-4x.x^2+\frac{24}{x}.x^2=17x^2\)

\(\Leftrightarrow x^4+36-4x^3+24x=17x^2\)

\(\Leftrightarrow x^4+36-4x^3+24x=17x^2-17x^2\)

\(\Leftrightarrow x^4+36-4x^3+24x=0\)

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

\(\Rightarrow x\in\left\{-1;2;-3;4\right\}\)

\(ĐKXĐ:x\ne1;5;9\)

\(pt\Leftrightarrow\frac{2x-1}{\left(x-1\right)\left(x-5\right)}+\frac{\left(x-2\right)}{\left(x-1\right)\left(x-9\right)}=\frac{3x-12}{\left(x-9\right)\left(x+5\right)}\)

\(\Rightarrow\left(2x-1\right)\left(x-9\right)+\left(x-2\right)\left(x-9\right)=\left(3x-12\right)\left(x-1\right)\)

\(=>2x^2-x-18x+9+x^2-2x+5x-10=3x^2-12-3x+12\)

\(=>3x^2-16x-1=3x^2-15x+12\)

=>x=-13

a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

la`thu'hai nga`y 19 nhe

ĐK \(x\ne\left\{1;2;3;4\right\}\)

Ta có \(\frac{x^2-2x+2}{x-1}+\frac{x^2-8x+20}{x-4}=\frac{x^2-4x+6}{x-2}+\frac{x^2-6x+12}{x-3}\)

\(\Leftrightarrow\frac{\left(x-1\right)^2+1}{x-1}+\frac{\left(x-4\right)^2+4}{x-4}=\frac{\left(x-2\right)^2+2}{x-2}+\frac{\left(x-3\right)^2+3}{x-3}\)

\(\Leftrightarrow x-1+\frac{1}{x-1}+x-4+\frac{4}{x-4}=x-2+\frac{2}{x-2}+x-3+\frac{3}{x-3}\)

\(\Leftrightarrow\frac{1}{x-1}+\frac{4}{x-4}=\frac{2}{x-2}+\frac{3}{x-3}\)

\(\Leftrightarrow\frac{5x-8}{x^2-5x+4}=\frac{5x-12}{x^2-5x+6}\)\(\Leftrightarrow\left(5x-8\right)\left(x^2-5x+6\right)=\left(5x-12\right)\left(x^2-5x+4\right)\)

\(\Leftrightarrow5x^3-25x^2+30x-8x^2+40x-48=5x^3-25x^2+20x-12x^2+60x-48\)

\(\Leftrightarrow4x^2-10x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{2}\end{cases}\left(tm\right)}\)

Vậy x=0 hoặc x=5/2

\(ĐKXĐ:x\ne-3;x\ne2;x\ne-1;x\ne\frac{1}{2}\)

Xét\(VT=\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}\)

\(=\frac{5\left(x+1\right)}{\left(x+3\right)\left(x-2\right)\left(x+1\right)}-\frac{2\left(x-2\right)}{\left(x+1\right)\left(x+3\right)\left(x-2\right)}\)

\(=\frac{5x+5-2x+4}{\left(x+3\right)\left(x-2\right)\left(x+1\right)}\)

\(=\frac{3x+9}{\left(x+3\right)\left(x-2\right)\left(x+1\right)}=\frac{3}{\left(x-2\right)\left(x+1\right)}\)

\(pt\Leftrightarrow\frac{3}{\left(x-2\right)\left(x+1\right)}=\frac{3}{4x-2}\)

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

\(\Leftrightarrow x^2-x-2=4x-2\)

\(\Leftrightarrow x^2-5x=0\)

\(\Leftrightarrow x\left(x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)(tm)

Vậy tập nghiệm của phương trình là {0;5}

ĐKXĐ: \(x\ne-3,2,-1\)

\(\frac{5}{x^2+x-6}-\frac{2}{x^2+4x+3}=\frac{3}{4x-2}\)

\(\Leftrightarrow\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=\frac{3}{2\left(x-2\right)}\)

\(\Leftrightarrow10\left(x+1\right)\left(2x-1\right)-4\left(x-2\right)\left(2x-1\right)=3\left(x-2\right)\left(x+3\right)\left(x+1\right)\)

\(\Leftrightarrow12x^2+30x-18=3x^2+6x^2-15x-18\)

\(\Leftrightarrow12x^2+30x=3x^3+6x^2-15\)

\(\Leftrightarrow12x^2+30x-3x^3-6x^2+15x=0\)

\(\Leftrightarrow6x^2+45x-3x^2=0\)

\(\Leftrightarrow3x\left(2x+15-x^2\right)=0\)

\(\Leftrightarrow-x\left(x^2-2x-15\right)=0\)

\(\Leftrightarrow-x\left(x-5\right)\left(x+3\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}-x=0\\x-5=0\\x+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\left(tm\right)\\x=5\left(tm\right)\\x=-3\left(ktm\right)\end{cases}}\)

Vậy: tập nghiệm của phương trình là: S = {0, 5}