a) 2x(x – 1/7) = 0 => TH1: 2x = 0 <=> x = 0 $\\$ => TH2: x – 1/7 = 0 <=> x = 1/7 KL: S = {0; 1/7} $\\$ b) (x -1)$^{x + 2}$ = (x – 1)$^{x + 6}$ <=> x + 2 = x + 6 <=> x – x = 6 – 2 <=> 0 = 4 (vô lý) KL: S $\in$ $\emptyset$ #Ling Trả lời
a)2x(x-1/7)=0=> \(\left[ \begin{array}{l}2x=0\\x-\dfrac{1}{7}=0\end{array} \right.\) => \(\left[ \begin{array}{l}x=0\\x=\dfrac{1}{7}\end{array} \right.\) Vậy x∈{0;1/7}– Áp dụng : A.B=0=> \(\left[ \begin{array}{l}A=0\\B=0\end{array} \right.\) b)(x-1)^(x+2) =(x-1)^(x+6)=>x+2=x+6=>x-x=6-2=>0=4\text{ ( vô lí )Vậy x∈∅– Áp dụng : a^m =a^n =>m=nAA m;n Trả lời
=> \(\left[ \begin{array}{l}2x=0\\x-\dfrac{1}{7}=0\end{array} \right.\)
=> \(\left[ \begin{array}{l}x=0\\x=\dfrac{1}{7}\end{array} \right.\)
Vậy x∈{0;1/7}
– Áp dụng : A.B=0=> \(\left[ \begin{array}{l}A=0\\B=0\end{array} \right.\)
b)(x-1)^(x+2) =(x-1)^(x+6)
=>x+2=x+6
=>x-x=6-2
=>0=4\text{ ( vô lí )
Vậy x∈∅
– Áp dụng : a^m =a^n =>m=nAA m;n