Thực hiện phép tính theo cách hợp lý:
a)25(75 – 49) + 75.|25 – 49|
b)23(64 – 51) -51(-23 – 64) -26.(-64)
2)Tìm số nguyên x , biết:
a) – 7 – 2x là số nguyên âm lớn nhất
b) 125 – 3(x + 18) = 77
c) (x+3)(2 –x) = 0
d) (-100): (2x -7)2 = – 4
e) (1-3x)3 = – 8
f) -9 – (21 – 3x) = -215 – (84 – 215)
g) 5(3-x) – 2(7-x) = -14
h) 3|2x -1| + 5 = 14
i) (x+2)(x-3) <0 k) (x2 +1)(25-x2) = 0
Đáp án:
$\begin{array}{l}
a)25.\left( {75 – 49} \right) + 75\left| {25 – 49} \right|\\
= 25.75 – 25.49 + 75.\left( {49 – 25} \right)\\
= 25.75 – 25.49 + 75.49 – 75.25\\
= 49.\left( {75 – 25} \right)\\
= 49.50\\
= 2450\\
b)23.\left( {64 – 51} \right) – 51.\left( { – 23 – 64} \right) – 26.\left( { – 64} \right)\\
= 23.64 – 23.51 + 51.23 + 51.64 + 26.64\\
= \left( {23.64 + 26.64} \right) + 51.\left( { – 23 + 23 + 64} \right)\\
= 64.\left( {23 + 26 + 51} \right)\\
= 64.100\\
= 6400\\
2)a) – 7 – 2x = – 1\\
\Rightarrow 2x = – 6\\
\Rightarrow x = – 3\\
Vậy\,x = – 3\\
b)125 – 3.\left( {x + 18} \right) = 77\\
\Rightarrow 3\left( {x + 18} \right) = 125 – 77\\
\Rightarrow 3.\left( {x + 18} \right) = 48\\
\Rightarrow x + 18 = 16\\
\Rightarrow x = – 2\\
Vậy\,x = – 2\\
c)\left( {x + 3} \right)\left( {2 – x} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = – 3\\
x = 2
\end{array} \right.\\
Vậy\,x = – 3;x = 2\\
d)\left( { – 100} \right):{\left( {2x – 7} \right)^2} = – 4\\
\Rightarrow {\left( {2x – 7} \right)^2} = 25\\
\Rightarrow \left[ \begin{array}{l}
2x – 7 = 5\\
2x – 7 = – 5
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
2x = 12\\
2x = 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 6\\
x = 1
\end{array} \right.\\
Vậy\,x = 6;x = 1\\
e){\left( {1 – 3x} \right)^3} = – 8\\
\Rightarrow 1 – 3x = – 2\\
\Rightarrow 3x = 3\\
\Rightarrow x = 1\\
Vậy\,x = 1\\
f) – 9 – \left( {21 – 3x} \right) = – 215 – \left( {84 – 215} \right)\\
\Rightarrow – 9 – 21 + 3x = – 215 – 84 + 215\\
\Rightarrow 3x – 30 = – 84\\
\Rightarrow 3x = – 54\\
\Rightarrow x = – 18\\
Vậy\,x = – 18\\
g)5\left( {3 – x} \right) – 2\left( {7 – x} \right) = – 14\\
\Rightarrow 15 – 5x – 14 + 2x = – 14\\
\Rightarrow 3x = 15\\
\Rightarrow x = 5\\
Vậy\,x = 5\\
h)3\left| {3x – 1} \right| + 5 = 14\\
\Rightarrow \left| {3x – 1} \right| = 3\\
\Rightarrow \left[ \begin{array}{l}
3x – 1 = 3\\
3x – 1 = – 3
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \frac{4}{3}\\
x = \frac{2}{3}
\end{array} \right.\\
i)\left( {x + 2} \right)\left( {x – 3} \right) < 0\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 2 > 0\\
x – 3 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 < 0\\
x – 3 > 0
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > – 2\\
x < 3
\end{array} \right.\\
\left\{ \begin{array}{l}
x < – 2\\
x > 3
\end{array} \right.\left( {ktm} \right)
\end{array} \right.\\
Vậy\, – 2 < x < 3
\end{array}$
$\begin{array}{l}
k)\left( {{x^2} + 1} \right)\left( {25 – {x^2}} \right) = 0\\
\Rightarrow 25 – {x^2} = 0\left( {do:{x^2} + 1 > 0} \right)\\
\Rightarrow {x^2} = 25\\
\Rightarrow x = \pm 5
\end{array}$
$\begin{array}{l} a)25.\left( {75 – 49} \right) + 75\left| {25 – 49} \right|\\ = 25.75 – 25.49 + 75.\left( {49 – 25} \right)\\ = 25.75 – 25.49 + 75.49 – 75.25\\ = 49.\left( {75 – 25} \right)\\ = 49.50\\ = 2450\\ b)23.\left( {64 – 51} \right) – 51.\left( { – 23 – 64} \right) – 26.\left( { – 64} \right)\\ = 23.64 – 23.51 + 51.23 + 51.64 + 26.64\\ = \left( {23.64 + 26.64} \right) + 51.\left( { – 23 + 23 + 64} \right)\\ = 64.\left( {23 + 26 + 51} \right)\\ = 64.100\\ = 6400\\ 2)a) – 7 – 2x = – 1\\ \Rightarrow 2x = – 6\\ \Rightarrow x = – 3\\ Vậy\,x = – 3\\ b)125 – 3.\left( {x + 18} \right) = 77\\ \Rightarrow 3\left( {x + 18} \right) = 125 – 77\\ \Rightarrow 3.\left( {x + 18} \right) = 48\\ \Rightarrow x + 18 = 16\\ \Rightarrow x = – 2\\ Vậy\,x = – 2\\ c)\left( {x + 3} \right)\left( {2 – x} \right) = 0\\ \Rightarrow \left[ \begin{array}{l} x = – 3\\ x = 2 \end{array} \right.\\ Vậy\,x = – 3;x = 2\\ d)\left( { – 100} \right):{\left( {2x – 7} \right)^2} = – 4\\ \Rightarrow {\left( {2x – 7} \right)^2} = 25\\ \Rightarrow \left[ \begin{array}{l} 2x – 7 = 5\\ 2x – 7 = – 5 \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} 2x = 12\\ 2x = 2 \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} x = 6\\ x = 1 \end{array} \right.\\ Vậy\,x = 6;x = 1\\ e){\left( {1 – 3x} \right)^3} = – 8\\ \Rightarrow 1 – 3x = – 2\\ \Rightarrow 3x = 3\\ \Rightarrow x = 1\\ Vậy\,x = 1\\ f) – 9 – \left( {21 – 3x} \right) = – 215 – \left( {84 – 215} \right)\\ \Rightarrow – 9 – 21 + 3x = – 215 – 84 + 215\\ \Rightarrow 3x – 30 = – 84\\ \Rightarrow 3x = – 54\\ \Rightarrow x = – 18\\ Vậy\,x = – 18\\ g)5\left( {3 – x} \right) – 2\left( {7 – x} \right) = – 14\\ \Rightarrow 15 – 5x – 14 + 2x = – 14\\ \Rightarrow 3x = 15\\ \Rightarrow x = 5\\ Vậy\,x = 5\\ h)3\left| {3x – 1} \right| + 5 = 14\\ \Rightarrow \left| {3x – 1} \right| = 3\\ \Rightarrow \left[ \begin{array}{l} 3x – 1 = 3\\ 3x – 1 = – 3 \end{array} \right.\\ \Rightarrow \left[ \begin{array}{l} x = \frac{4}{3}\\ x = \frac{2}{3} \end{array} \right.\\ i)\left( {x + 2} \right)\left( {x – 3} \right) < 0\\ \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x + 2 > 0\\ x – 3 < 0 \end{array} \right.\\ \left\{ \begin{array}{l} x + 2 < 0\\ x – 3 > 0 \end{array} \right. \end{array} \right. \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x > – 2\\ x < 3 \end{array} \right.\\ \left\{ \begin{array}{l} x < – 2\\ x > 3 \end{array} \right.\left( {ktm} \right) \end{array} \right.\\ Vậy\, – 2 < x < 3 \end{array}$