Đáp án:
Giải thích các bước giải:
e)$\frac{x-3}{2}$-x-2 = $\frac{2xx+1}{5}$ -$\frac{8x-11}{10}$
⇒$\frac{5(x-3)}{10}$ -$\frac{10(x-2))}{10}$=$\frac{2(2x+1)}{10}$- $\frac{8x-11}{10}$
⇔5(x-3)-10(x-2)=2(2x+1)-8x-11
⇔5x-15-10x+20=4x+2-8x-11
⇔5x-10x-4x+8x=2-11+15-20
⇔-x=-14⇔x=14
Vậy S=[14}
F) 2(x-1)³=x²(2x-6)
⇔2(x-1)³-x²(2x-6)=0
⇔2(x³-3x²+3x-1)-2x³-6x²=0
⇔2x³-6x²+6x-2-2x³-6x²=0
⇔6x-2=0
⇔6x=2
⇔x=1/3
vậy S=1/3
g)$\frac{x+101}{2001}$ +$\frac{x+99}{2003}$=$\frac{x+100}{2002}$+ $\frac{x+98}{2004}$
⇔$\frac{x+101}{2001}$ +$\frac{x+99}{2003}$-$\frac{x+100}{2002}$- $\frac{x+98}{2004}$ =0
⇔$\frac{x+101}{2001}$+1+ $\frac{x+99}{2003}$+1- $\frac{x+100}{2002}$+1- $\frac{x+98}{2004}$+1=0
⇔$\frac{x+101+2001}{2001}$+ $\frac{x+99+2003}{2003}$- $\frac{x+100+2002}{2002}$- $\frac{x+98+2004}{2004}$=0
⇔$\frac{x+2102}{2001}$+ $\frac{x+2102}{2003}$- $\frac{x+2102}{2002}$- $\frac{x+2102}{2004}$=0
⇔(x+2102)($\frac{1}{2001}$+ $\frac{1}{2003}$ -$\frac{1}{2002}$ -$\frac{1}{2004}$ )=0
⇔x+2102=0( vì$\frac{1}{2001}$+ $\frac{1}{2003}$ -$\frac{1}{2002}$ -$\frac{1}{2004}$ khác 0)
⇔x+2102=0
⇔x=-2102
Vậy S=2102
h)$\frac{10-x}{100}$+ $\frac{20-x}{110}$ +$\frac{30-x}{120}$ =3
⇔$\frac{10-x}{100}$+ $\frac{20-x}{110}$ +$\frac{30-x}{120}$ -3=0
⇔$\frac{10-x}{100}$-1+ $\frac{20-x}{110}$-1 +$\frac{30-x}{120}$-1=0
⇔$\frac{10-x-100}{100}$+ $\frac{20-x-110}{110}$ +$\frac{30-x-120}{120}$=0
⇔$\frac{-x-90}{100}$+ $\frac{-x-90}{110}$ +$\frac{-x-90}{120}$=0
⇔(-x-90)($\frac{1}{100}$+ $\frac{1}{110}$ +$\frac{1}{120}$)=0
⇔(-x-90)=0(vì$\frac{1}{100}$+ $\frac{1}{110}$ +$\frac{1}{120}$ khác 0)
⇔-x-90=0⇔-x=90⇔x=-90 Vậy S=90
( chúc bạn học tốt)
