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a) 25.6 + 5.5.29 - 45.5 =25.6 - 25.29 + 9.5.5 25.6 - 25.29 +9.(5.5) [có chung 25 nên sẽ đặt 25 ra ngoài rồi trừ cộng trong ngoặc] b) 30.75 + 25.30 - 150 =30.75 + 25.30 - 30.5 [tương tự như trên]
Câu 15:
\(A=2+2^2+2^3+...+2^{100}\)
\(=2\left(1+2\right)+...+2^{99}\left(1+2\right)\)
\(=6\cdot\left(1+...+2^{98}\right)⋮6\)
Đáp án:
Bài 1 :
1) 30.75 + 25.30 – 150
= 30.(75+25) -150
= 30.100 -150
= 3000 -150
= 2850
2) 160 - (4.52 - 3.23)
= 160 – ( 4.25 – 3.8)
= 160 – (100 – 24)
= 160 – 76
= 84
3) [36.4 - 4.(82 - 7.11)2] : 4 - 20220
= [36.4 – 4.(82 – 7.11)2] :4 - 1
= [36.4 – 4.(82 – 77)2] : 4 – 1
= [36.4 – 4.52] : 4 – 1
=[36.4 – 4.25] : 4 – 1
=[4. (36 – 25)] : 4 – 1
= [4. 11] : 4 – 1
=44 : 4 – 1
=11 – 1
= 10
Đây nha bạn!!!
Chúc bạn học tốt!!!
a) 80- (4.52 - 3.23)
= 80- ( 208 -69 )
=80+139 quy tắc đổi dấu trừ tước dấu ngoặc
= 219
e: Ta có: \(2448:\left[119-\left(23-6\right)\right]\)
\(=2448:\left(119-23+6\right)\)
\(=2448:102=24\)
a) 80-(4.52-3.23)=80-100+24=4
b)[36.4-4.(82-7.11)2]:4-20190
={4.[36-(82-7.11)2]}:4-1
=[36-(82-7.11)2]-1
=11-1=10
c)56:54+23.22-12018
=52+25-1
=25+32-1=56
d)303-3.{[655-(18:2+1).43+55]}:100
=303-3.[(655-9-1).43+55]:1
=303-3[655-640+5]
=303-3(20)
=303-60=243
3:
a: \(\dfrac{\left(x-3\right)}{5}=6^2-2^3\cdot4\)
=>\(\dfrac{x-3}{5}=36-8\cdot4=4\)
=>x-3=20
=>x=23
b: \(3^{x+2}+5\cdot2^3=47+\dfrac{18}{4^2-7}\)
=>\(3^{x+2}+5\cdot8=47+\dfrac{18}{16-7}=49\)
=>\(3^{x+2}=9\)
=>x+2=2
=>x=0
c: \(2^{x+1}-2^x=8^2\)
=>\(2^x\cdot2-2^x=2^6\)
=>\(2^x=2^6\)
=>x=6
d: \(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\right)\cdot x^2=99\)
=>\(x^2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=99\)
=>\(x^2\cdot\dfrac{99}{100}=99\)
=>\(x^2=100\)
=>\(\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)
e: \(\left(2x-3\right)^7=\left(2x-3\right)^5\)
=>\(\left(2x-3\right)^5\left[\left(2x-3\right)^2-1\right]=0\)
=>\(\left(2x-3\right)^5\cdot\left(2x-3-1\right)\left(2x-3+1\right)=0\)
=>\(\left(2x-3\right)^5\left(2x-4\right)\left(2x-2\right)=0\)
=>\(\left[{}\begin{matrix}2x-3=0\\2x-4=0\\2x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=1\end{matrix}\right.\)
f: \(\left(x-2\right)^{10}=\left(x-2\right)^8\)
=>\(\left(x-2\right)^8\left[\left(x-2\right)^2-1\right]=0\)
=>\(\left(x-2\right)^8\left(x-2-1\right)\left(x-2+1\right)=0\)
=>\(\left(x-2\right)^8\cdot\left(x-3\right)\left(x-1\right)=0\)
=>\(x\in\left\{2;3;1\right\}\)
a, 80 - 4 . 5 2 - 3 . 2 3
= 80 – (4.25 – 3.8)
= 80 – 76 = 4
b, 5 6 : 5 4 + 2 3 . 2 2 - 1 2018
= 5 2 + 2 5 - 1
= 25 + 32 – 1 = 56
c, [36.4 – 4. 82 - 7 . 11 2 ]:4 – 2019 0
= (144 – 4. 5 2 ):4 – 1
= (144 – 100):4 – 1
= 11 – 1 = 10
d, 303 – 3.{[655 – (18:2+1). 4 3 +5]}: 10 0
= 303 – 3.(655 – 10.64 + 5):1
= 303 – 10 = 293
\(\left[36\times4-4\times\left(82-7\times11\right)^2\right]:4-2022^0\\ =\left[144-4\times\left(82-77\right)^2\right]:4-1\\ =\left[144-4\times5^2\right]:4-1\\ =\left[144-4\times25\right]:4-1\\ =\left[144-100\right]:4-1\\ =44:4-1=11-1=10\)
\(36\times4-4\times\left(82-7\times11\right)^2\div4-2016^0\)
\(=144-4\times\left(82-77\right)^2\div4-1\)
\(=144-4\times5^2\div4-1\)
\(=144-4\times25\div4-1\)
\(=144-100\div4-1\)
\(=144-25-1\)
\(=119-1\)
\(=118\)
a: \(2^3-5^3:5^2+12\cdot2^2\)
\(=8-5+48\)
\(=51\)
b: \(5\cdot\left[\left(85-35:7\right):8+90\right]-5\)
\(=5\cdot\left[10+90\right]-5\)
=495
Answer:
Mình sửa lại đề, sai thì bạn bảo mình nhé!
a) \(25.6+5.5.29-45.5\)
\(=5.5.6+5.5.29-45.5\)
\(=5.30+5.145-45.5\)
\(=5.\left(30+145-45\right)\)
\(=5.130\)
\(=450\)
b) \(30.75+25.30-150\)
\(=30.\left(75+25\right)-150\)
\(=30.100-150\)
\(=3000-150\)
\(=2850\)
c) \(2.[\left(7-3^3:3^2\right):2^2+99]-100\)
\(=2.[\left(7-3\right):4+99]-100\)
\(=2.\left(1+99\right)-100\)
\(=200-100\)
\(=100\)
d) \(160-\left(4.25-3.23\right)\)
\(=160-\left(208-69\right)\)
\(=160-139\)
\(=21\)
e) \(\left(7^{2022}-7^{2021}\right):7^{2021}\)
\(=7^{2021}.\left(7-1\right):7^{2021}\)
\(=7-1\)
\(=6\)
g) \([36.4-4.\left(82-7.11\right).2]:4-2022^0\)
\(=[144-4.2.\left(82-77\right)]:4-1\)
\(=[144-8.5]:4-1\)
\(=104:4-1\)
\(=25\)
Tính toán lại cho mình vs ạ
Cho kĩ nha vì bài này để thi á