Đáp án:
51. a. x = 4
b. x = 8
c. x = { 0; 1; 2; 3; 4 }
52. a. x = { $\frac{-1}{35}$ ; $\frac{-13}{35}$ }
b. x = $\frac{2}{3}$
c. x = $\frac{3}{4}$
53. a. $\left \{ {{x=0} \atop {y=\frac{1}{10}}} \right.$
b. (x; y) = (10; $\frac{1}{2}$ ); ( 10; $\frac{-1}{2}$ )
Giải thích các bước giải:
51. a. 8 < $2^{x}$ ≤ $2^{9}$ × $2^{-5}$
⇔ 2³ < $2^{x}$ ≤ $2^{9+(-5)}$
⇔ 2³ < $2^{x}$ ≤ $2^{4}$
⇔ 3 < x ≤ 4 ⇔ x = 4
b. 27 < 81³ : $3^{x}$ < 243
⇔ 3³ < ($3^{4}$ )³ : $3^{x}$ < $3^{5}$
⇔ 3³ < $3^{12}$ : $3^{x}$ < $3^{5}$
⇔ 3³ < $3^{12-x}$ < $3^{5}$
⇔ 3 < 12 - x < 5 ⇔ 12 - x = 4 ⇔ x = 8
c. $(\frac{2}{5})^{x}$ > $(\frac{5}{2})^{-3}$ × $(\frac{-2}{5})^{2}$
⇔ $(\frac{2}{5})^{x}$ > $(\frac{2}{5})^{3}$ × $(\frac{2}{5})^{2}$
⇔ $(\frac{2}{5})^{x}$ > $(\frac{2}{5})^{(3+2)}$
⇔ $(\frac{2}{5})^{x}$ > $(\frac{2}{5})^{5}$
⇔ x < 5 ( vì $\frac{2}{5}$ < 1 )
Mà x ∈ N nên x = { 0; 1; 2; 3; 4 }
52. a. (5x + 1)² = $\frac{36}{49}$
⇔ (5x + 1)² = $(\frac{±6}{7})^{2}$
⇔ \(\left[ \begin{array}{l}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{array} \right.\)
b. ( x - $\frac{2}{9}$ )³ = $(\frac{2}{3})^{6}$
⇔ ( x - $\frac{2}{9}$ )³ = ( $\frac{4}{9}$ )³
⇔ x - $\frac{2}{9}$ = $\frac{4}{9}$ ⇔ x = $\frac{2}{3}$
c. $(8x-1)^{2n+1}$ = $5^{2n+1}$ ( n ∈ N )
⇔ 8x - 1 = 5 ⇔ x = $\frac{3}{4}$
53. a. x² + $( y- \frac{1}{10} )^{4}$ = 0
Nhận xét vế trái luôn ≥ 0 ∀ x; y
Dấu "=" xảy ra ⇔ $\left \{ {{x=0} \atop {y=\frac{1}{10}}} \right.$
b. $( \frac{1}{2}x - 5 )^{20}$ + $(y² - \frac{1}{4} )^{10}$ = 0
Nhận xét vế trái luôn ≥ 0 ∀ x; y
Dấu "=" xảy ra ⇔ $\left \{ {{x=10} \atop {y²=\frac{1}{4}}} \right.$
⇔ (x; y) = (10; $\frac{1}{2}$ ); ( 10; $\frac{-1}{2}$ )
