Toán Lớp 7: a) |3x-2|=|2x-3|
b) (2√x+3)^2+1/4=5/2
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Toán Lớp 7: a) |3x-2|=|2x-3|
b) (2√x+3)^2+1/4=5/2
Comments ( 2 )
<=> \(\left[ \begin{array}{l}3x-2=+(2x-3)\\3x-2=-(2x-3)\end{array} \right.\)
<=> \(\left[ \begin{array}{l}3x-2=2x-3\\3x-2=-2x+3\end{array} \right.\)
<=> \(\left[ \begin{array}{l}3x-2x=2-3\\3x-(-2x)=2+3\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=-1\\x=-1\end{array} \right.\)
Vậy x=1 hoặc x=-1
– Áp dụng : |x|=|a|=> \(\left[ \begin{array}{l}x=a\\x=-a\end{array} \right.\)
b) $(2\sqrt{x}+3)²+\dfrac{1}{4}=\dfrac{5}{2}$
<=> $(2\sqrt{x}+3)²=\dfrac{5}{2}-\dfrac{1}{4}$
<=> $(2\sqrt{x}+3)²=\dfrac{9}{4}$
<=> $(2\sqrt{x}+3)²=$ (+-3/2)^2
– Áp dụng : a^m =a^n =>a=b
<=> $\left[ \begin{array}{l}2\sqrt{x}+3=\dfrac{3}{2}\\2\sqrt{x}+3=-\dfrac{3}{2}\end{array} \right.$
<=> $\left[ \begin{array}{l}2\sqrt{x}=-\dfrac{3}{2}\\2\sqrt{x}=\dfrac{-9}{2}\end{array} \right.$
<=> $\left[ \begin{array}{l}\sqrt{x}=\dfrac{-3}{4}\\\sqrt{x}=\dfrac{-9}{4}\end{array} \right.$
Mà $\sqrt{x}≥0$
=>x∈∅
Vậy x∈∅