Đáp án:
Bai 1
a. $66$
b. $\frac{19}{121}$
Bai 2
a. \(\left[ \begin{array}{l}x=-0,5\\x=-2,5\end{array} \right.\)
b. $x = 3$
Giải thích các bước giải:
Bài 1
a. $-4\frac{5}{7}×33\frac{1}{3} + 4\frac{5}{7}×47\frac{1}{3}$
= $\frac{-33}{7}×\frac{100}{3} + \frac{33}{7}×\frac{142}{3}$
= $\frac{33}{7}×( \frac{-100}{3} + \frac{142}{3} )$
= $\frac{33}{7}×\frac{42}{3}$
= $\frac{33}{7}×14$
= $66$
b. $(\frac{3}{5} + 5\frac{2}{3} ) : \frac{121}{19} + ( -5\frac{2}{3} + \frac{2}{5} ) : \frac{121}{19}$
= $( \frac{3}{5} + 5\frac{2}{3} - 5\frac{2}{3} + \frac{2}{5} ) : \frac{121}{19}$
= $[ ( \frac{3}{5} + \frac{2}{5} ) + ( 5\frac{2}{3} - 5\frac{2}{3} ) ] : \frac{121}{19}$
= $( 1 + 0 ) : \frac{121}{19}$
= $\frac{19}{121}$
Bai 2
a. $3,5 + | x + \frac{3}{2} | = -1,5×( -\sqrt[]{9} )$
⇔ $3,5 + | x + 1,5 | = -1,5×( -3 )$
⇔ $3,5 + | x + 1,5 | = 4,5$
⇔ $| x + 1,5 | = 1$
⇔ \(\left[ \begin{array}{l}x+1,5=1\\x+1,5=-1\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-0,5\\x=-2,5\end{array} \right.\)
b. $3^{x+2} + 3^{x} = 270$
⇔ $3^{x}×3^{2} + 3^{x} = 270$
⇔ $3^{x}×9 + 3^{x} = 270$
⇔ $3^{x}×10 = 270$
⇔ $3^{x} = 27$
⇔ $3^{x} = 3^{3}$
⇔ $x = 3$