50[( 2 . 32 2 - 8 ): 2 +30 ]-215
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![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
Tổng số tiền An dùng mua đồ:
125 000 + 85 000 + 60 000 + 65 000 = 335 000 (đồng)
Số tiền An còn lại sau khi mua đồ:
350 000 - 335 000 = 15 000 (đồng)
Đ.số: 15 000 đồng
Bài 2:
a, IV: Bốn(4); XXVII: Hai mươi bảy (27), XXX: ba mươi (30), M:một nghìn (1000)
b, 7: VII; 15: XV; 29: XXIX
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a) (2x-32):16=3
2x-32=3.16
2x-32=48
2x=48+32
2x=80
x=80:2
x=40
b) (7x -8:2):5=8
(7x-4):5=8
7x-4=8.5
7x-4=40
7x=40+4
a) (2x - 32) : 16 = 3 \(=\left(2x-32\right)=3.16=48\)
= \(\left(2x-32\right)=48\)
= \(2x=48+32=80\)
= \(2x=80\Rightarrow x=80:2=40\)
b) ( 7x - 8 :2) : 5 = 8
= \(\left(7x-8:2\right)=8.5=40\)
= \(\left(7x-8:2\right)=40\)
= \(\left(7x-8\right)=40.2=80\)
= \(\left(7x-8\right)=80\)
= \(7x=80+8=88\)
=>x = 88 : 7 = ......
c) 215 + ( x - 21) : 2 = 235
\(=\left(x-21\right):2=235-215=20\)
= \(\left(x-21\right):2=20\)
= \(\left(x-21\right)=20.2=40\)
= \(\left(x-21\right)=40\)
= \(\Rightarrow x=40+21=61\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2,
a) \(315-\left(135-x\right)=215\)
\(\Rightarrow135-x=315-215\)
\(\Rightarrow135-x=100\)
\(\Rightarrow x=135-100\)
\(\Rightarrow x=35\)
b) \(x-320:32=25\cdot16\)
\(\Rightarrow x-10=5^2\cdot4^2\)
\(\Rightarrow x-10=20^2\)
\(\Rightarrow x-10=400\)
\(\Rightarrow x=410\)
c) \(3\cdot x-2018:2=23\)
\(=3\cdot x-1009=23\)
\(\Rightarrow3\cdot x=1032\)
\(\Rightarrow x=1032:3\)
\(\Rightarrow x=344\)
d) \(280-9\cdot x-x=80\)
\(\Rightarrow280-x\cdot\left(9+1\right)=80\)
\(\Rightarrow280-10\cdot x=80\)
\(\Rightarrow10\cdot x=280-80\)
\(\Rightarrow10\cdot x=200\)
\(\Rightarrow x=20\)
e) \(38\cdot x-12\cdot x-x\cdot16=40\)
\(\Rightarrow x\cdot\left(38-12-16\right)=40\)
\(\Rightarrow x\cdot10=40\)
\(\Rightarrow x=40:10\)
\(\Rightarrow x=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(20-\left[30-\left(5-1\right)^2\right]\)
\(=20-\left[30-4^2\right]\)
\(=20-14=6\)
b: \(71+\dfrac{50}{5+3\left(57-6\cdot7\right)}\)
\(=71+\dfrac{50}{5+3\cdot\left(57-42\right)}\)
\(=71+\dfrac{50}{5+3\cdot15}=71+\dfrac{50}{50}=72\)
c: \(4\cdot\left\{270:\left[50-\left(2^5+45:5\right)\right]\right\}\)
\(=4\cdot\left\{270:\left[50-32-9\right]\right\}\)
\(=4\cdot\left\{\dfrac{270}{50-41}\right\}=4\cdot\dfrac{270}{9}=4\cdot30=120\)
d: \(411-\left[\dfrac{\left(107+3\right)}{5}-2^2\right]\)
\(=411-\left[\dfrac{110}{5}-4\right]\)
=410-22+4
=410-18
=392
e: \(450-5\left[3^2\left(7^5:7^3-41\right)-12\right]+18\)
\(=450-5\left[9\cdot\left(7^2-41\right)-12\right]+18\)
\(=450-5\cdot\left[9\cdot8-12\right]+18\)
=468-5*60
=468-300
=168
f:
\(102-150:\left[18-2\cdot\left(10-8\right)^2\right]+1018^0\)
\(=102-150:\left[18-2\cdot4\right]+1\)
\(=103-\dfrac{150}{18-8}=103-15=88\)
![](https://rs.olm.vn/images/avt/0.png?1311)
16) M = (-32 + 215) - (-30 + 215)
= -32 + 215 + 30 - 215
= (215 - 215) + (30 - 32)
= 0 - 2 = -2
17) N = -5 + (-37 - 45 + 151) - (-37 + 151)
= -5 - 37 - 45 + 151 + 37 - 151
= (151 - 151) - (5+45) + (37 - 37)
= 0 - 50 + 0 = -50
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}\)
\(=3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}\)
\(=-10\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{32}+\sqrt{50}-2\sqrt{8}+\dfrac{1}{3}\sqrt{18}\)
\(=\sqrt{4^2\cdot2}+\sqrt{5^2\cdot2}-2\cdot2\sqrt{2}+\dfrac{1}{3}\cdot\sqrt{3^2\cdot2}\)
\(=4\sqrt{2}+5\sqrt{2}-4\sqrt{2}+\dfrac{1}{3}\cdot3\sqrt{2}\)
\(=\left(4\sqrt{2}-4\sqrt{2}\right)+5\sqrt{2}+\sqrt{2}\)
\(=5\sqrt{2}+\sqrt{2}\)
\(=6\sqrt{2}\)
\(=50\cdot\left[32^2-8+1\right]-215=50\cdot1017-215=50635\)
\(50\left[\left(2.32^2-8\right):2+3^0\right]-215\)
\(50\left[\left(2.1024-8\right):2+1\right]-215\)
\(50\left[\left(2048-8\right):2+1\right]-215\)
\(50\left[2040:2+1\right]-215\)
\(50\left[1020+1\right]-215\)
\(50.1021-215\)
\(51050-215\)
\(=50835\)