Đáp án + Giải thích các bước giải:
a) $(2x + 3)^{2}$ = $\dfrac{9}{121}$
=> $(2x + 3)^{2}$ = $(\dfrac{3}{11})^{2}$ hoặc $(\dfrac{-3}{11})^{2}$
=> 2x + 3 = $\dfrac{3}{11}$ hoặc $\dfrac{-3}{11}$
$TH_{1}$:
2x + 3 = $\dfrac{3}{11}$
=> 2x = $\dfrac{3}{11}$ - 3
=> 2x = $\dfrac{-30}{11}$
=> x = $\dfrac{-30}{11}$ : 2
=> x = $\dfrac{-30}{11}$ . $\dfrac{1}{2}$
=> x = $\dfrac{-15}{11}$
$TH_{2}$:
2x + 3 = $\dfrac{-3}{11}$
=> 2x = $\dfrac{-3}{11}$ - 3
=> 2x = $\dfrac{-36}{11}$
=> x = $\dfrac{-36}{11}$ : 2
=> x = $\dfrac{-36}{11}$ . $\dfrac{1}{2}$
=> x = $\dfrac{-18}{11}$
b) $(3x - 1)^{3}$ = $\dfrac{-8}{27}$
=> $(3x - 1)^{3}$ = $(\dfrac{-2}{3})^{3}$
=> 3x - 1 = $\dfrac{-2}{3}$
=> 3x = $\dfrac{-2}{3}$ + 1
=> 3x = $\dfrac{1}{3}$
=> x = $\dfrac{1}{3}$ : 3
=> x = $\dfrac{1}{9}$
Học Tốt UwU<3
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