Bài 1: Tính
a) 324 + [ 112 – ( 112 + 324 ) – 230 ]
b) -567 – (-113) + (-69) – ( 113 – 567 )
c) 2011 + { 743 – [ 2011 – (-257) ] }
d) 3 . (-5)2 ( mũ 2) + 2 . (-5) – 20
Bài 2: Tìm x, biết
a) 3 . |x-1| = 27
b) -7 + 2x = -37 – (-26)
c) 23 – ( x – 23 ) = 34
d) 3 . |x – 1| + 5 = 17
( giúp e với ạ)
a) 324 + [ 112 – ( 112 + 324 ) – 230 ]
= 324 + 112 – 112 – 324 – 230
= -230
b) -567 – (-113) + (-69) – ( 113 – 567 )
= -567 + 113 – 69 – 113 + 567
= -69
c) 2011 + { 743 – [ 2011 – (-257) ] }
= 2011 + 743 – 2011 – 257
= 486
d) 3 . (-5)2 ( mũ 2) + 2 . (-5) – 20
= 75 – 10 – 20
= 45
Bài 2: Tìm x, biết
a) 3 . |x-1| = 27
⇔ |x-1| = 9
⇔\(\left[ \begin{array}{l}x-1=9\\x-1=-9\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=10\\x=-8\end{array} \right.\)
Vậy x ∈ { 10; -8 }
b) -7 + 2x = -37 – (-26)
⇔ 2x = -37 + 26 + 7
⇔ x = -2
Vậy x = -2
c) 23 – ( x – 23 ) = 34
⇔ x – 23 = -11
⇔ x = 12
Vậy x = 12
d) 3 . |x – 1| + 5 = 17
⇔ 3|x-1| = 12
⇔ |x-1| = 4
⇔ \(\left[ \begin{array}{l}x-1=4\\x-1=-4\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=5\\x=-3\end{array} \right.\)
Vậy x ∈ {5; -3}
Chúc bạn học tốt!!!
Bài 1:
$\begin{array}{l}a) 324 + [ 112 – ( 112 + 324 ) – 230 ]\\=324+[112-436-230]\\=324+(-554)=-230\\b) -567 – (-113) + (-69) – ( 113 – 567 )\\=-523-(-454)=-69\\c) 2011 + { 743 – [ 2011 – (-257) ] }\\=2011+{743-2267}\\=2011+(-1524)=487\\d) 3.(-5)^{2}+2.(-5)-20\\=75+(-10)-20=45\end{array}$
Bài 2:
$\begin{array}{l}a) 3 . |x-1| = 27\\\Leftrightarrow \left | x-1 \right |=9\\\Leftrightarrow \left[ \begin{array}{l}x-1=9\\x-1=-9\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=10\\x=-8\end{array} \right.\end{array}$
Vậy $x\in \left \{ 10;-8 \right \}$
$\begin{array}{l}b) -7 + 2x = -37 – (-26)\\\Leftrightarrow -7+2x=-11\\\Leftrightarrow 2x=-4\\\Leftrightarrow x=-2\end{array}$
Vậy $x=-2$
$\begin{array}{l}c) 23 – ( x – 23 ) = 34\\\Leftrightarrow x-23=-11\\\Leftrightarrow x=12\end{array}$
Vậy $x=12$
$\begin{array}{l}d) 3 . |x – 1| + 5 = 17\\\Leftrightarrow 3\left | x-1 \right |=12\\\Leftrightarrow \left | x-1 \right |=4\\\Leftrightarrow \left[ \begin{array}{l}x-1=4\\x-1=-4\end{array} \right.\\\Leftrightarrow \left[ \begin{array}{l}x=5\\x=-3\end{array} \right.\end{array}$
Vậy $x\in \left \{ 5;-3 \right \}$