bài 1:
a) $\frac{1}{4}$ + $\frac{3}{4}$. 19$\frac{1}{3}$+$\frac{3}{4}$. 39$\frac{1}{3}$
= $\frac{1}{4}$ + $\frac{3}{4}$.(19$\frac{1}{3}$ + 39$\frac{1}{3}$)
= $\frac{1}{4}$ +$\frac{3}{4}$.58$\frac{1}{3}$
= $\frac{1}{4}$+$\frac{175}{4}$
= 44.
b) [√$\frac{4}{9}$+($\frac{-1}{2}$)²] : 0,75 +1$\frac{1}{3}$.|1-$\frac{11}{12}$|
= ($\frac{2}{3}$+$\frac{1}{4}$) : 0,75 + $\frac{4}{3}$.$\frac{1}{12}$
= $\frac{11}{12}$ : 0,75+ $\frac{1}{9}$
= $\frac{11}{9}$ + $\frac{1}{9}$
= $\frac{4}{3}$.
bài 2:
a) $\frac{3}{4}$.x + $\frac{1}{2}$ =5
$\frac{3}{4}$x = $\frac{9}{2}$
x = 6
Vậy x = 6.
b) 1$\frac{1}{4}$ - |x+$\frac{5}{6}$| = $\frac{-5}{7}$.$\frac{21}{6}$
$\frac{5}{4}$ - |x+$\frac{5}{6}$| =$\frac{-5}{2}$
|x+$\frac{5}{6}$| = $\frac{-5}{4}$ (là khẳng định sai)
Vậy ko có giá trị x thỏa mãn.
c) (x²+√16)(|x|-$\frac{1}{3}$)
(x²+4)(|x|-$\frac{1}{3}$) = 0
\(\left[ \begin{array}{l}x²+4=0\\|x|-\frac{1}{3}=0\end{array} \right.\)
\(\left[ \begin{array}{l}x²=-4(là khẳng định sai)\\|x|=\frac{1}{3}\end{array} \right.\)
\(\left[ \begin{array}{l}\frac{1}{3}\\x=\frac{-1}{3}\end{array} \right.\)
Vậy x=$\frac{1}{3}$ hoặc x=$\frac{-1}{3}$
bài 3:
a) f(0)= 2.0² + 3 =3
f($\frac{1}{2}$) = 2.$\frac{1}{2}$² +3 =4
b) f(x) =2x² +3 =11
2x²=8
x²=4
\(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\)
@xin ctlhn nha :3
