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1,a) 695- [200+ (11- 12)]
= 695- [200+ (11- 1)]
= 695- [200+ 10]
= 695- 210
= 485
b) (519: 517+ 3): 7
= (52+ 3): 7
= (25+ 3): 7
= 28: 7
= 4
c) 129- 5[29- (6- 12)]
= 129- 5[29- (6- 1)]
= 129- 5[29- 5]
= 129- 5. 24
= 129- 120
= 9
3,a) 2x- 49= 5. 32
2x- 49= 5. 9
2x- 49= 45
2x = 45+ 49
2x = 94
x = 94: 2
x = 47
c) 2x- 15= 17
2x = 17+ 15
2x = 32
2x = 25
=> x = 5
Câu 3b bạn tự làm nhé, xin loiosxn vì không giúp được cả bài.
CHÚC BẠN HỌC GIỎI !!!
MÌNH TÌM RA CÁCH LÀM CÂU 3b RỒI !!!
5x+ 2x= 45+ 20: 15
5x+ 2x= 45+ \(\frac{4}{3}\)
5x+ 2x= \(\frac{139}{3}\)
(5+ 2)x=\(\frac{139}{3}\)
7x =\(\frac{139}{3}\)
x =\(\frac{139}{3}\): 7
x =\(\frac{139}{21}\)
CHÚC BẠN HỌC GIỎI !!!
1)
a) \(\left(-175\right)-123+3+175+\left(-3\right)\)
\(=\left[\left(-175\right)+175\right]+\left[3+\left(-3\right)\right]-123\)
\(=0+0-123\)
\(=-123\)
b) \(17.85+15.17-150\)
\(=17.\left(85+15\right)-150\)
\(=17.100-150\)
\(=1700-150\)
\(=1550\)
c) \(176:\left(4.5^2-3.2^2\right)\)
\(=176:\left(4.25-3.4\right)\)
\(=176:\left(100-12\right)\)
\(=176:88\)
\(=2\)
2)
a) \(8-x=-12\)
\(\Rightarrow x=8-\left(-12\right)\)
\(\Rightarrow x=8+12\)
\(\Rightarrow x=20\)
Vậy \(x=20\)
b) \(20+2^3.x=5^2.4\)
\(\Rightarrow20+8.x=25.4\)
\(\Rightarrow20+8.x=100\)
\(\Rightarrow8.x=100-20\)
\(\Rightarrow8.x=80\)
\(\Rightarrow x=80:8\)
\(\Rightarrow x=10\)
Vậy \(x=10\)
c) \(2\left|x-1\right|+5=15\)
\(\Rightarrow2\left|x-1\right|=15-5\)
\(\Rightarrow2\left|x-1\right|=10\)
\(\Rightarrow\left|x-1\right|=10:2\)
\(\Rightarrow\left|x-1\right|=5\)
Xét trường hợp 1: \(x-1=5\)
\(\Rightarrow x=5+1\)
\(\Rightarrow x=6\)
Xét trường hợp 2: \(x-1=-5\)
\(\Rightarrow x=\left(-5\right)+1\)
\(\Rightarrow x=-5+1\)
\(\Rightarrow x=-\left(5-1\right)\)
\(\Rightarrow x=-4\)
Vậy \(x=6\) hoặc \(x=-4\)
Bài 1:
\(a)\left(-175\right)-123+3+175+\left(-3\right)\)
\(=\left[\left(-175\right)+175\right]+\left[\left(-3+3\right)\right]-123\)
\(=0+0-123\)
\(=-123\)
\(b)17.85+15.17-150\)
\(=17.\left(85+15\right)-150\)
\(=17.100-150=1700-150=1550\)
\(c)176:\left(4.5^2-3.2^2\right)\)
\(=176:\left(4.25-3.4\right)\)
\(=176:\left(100-12\right)=176:88=2\)
Bài 2:
\(a)8-x=-12\)
\(\Rightarrow x=8-\left(-12\right)\)
\(\Rightarrow x=20\)
\(b)20+2^3.x=5^2.4\)
\(\Rightarrow20+8.x=25.4\)
\(\Rightarrow20+8.x=100\)
\(\Rightarrow8.x=100-20\)
\(\Rightarrow8.x=80\)
\(\Rightarrow x=80:8=10\)
\(c)2\left|x-1\right|+5=15\)
\(\Rightarrow2\left|x-1\right|=15-5\)
\(\Rightarrow2\left|x-1\right|=10\)
\(\Rightarrow\left|x-1\right|=10:2\)
\(\Rightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
a, \(2^x-15=17\)
\(\Rightarrow2^x=17+15\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
b, \(\left(7x-11\right)^3=2^5.5^2+200\)
\(\Rightarrow\left(7x-11\right)^3=32.25+200\)
\(\Rightarrow\left(7x-11\right)^3=1000\)
\(\Rightarrow\left(7x-11\right)^3=10^3\)
\(\Rightarrow7x-11=10\)
\(\Rightarrow7x=10+11\)
\(\Rightarrow7x=21\)
\(\Rightarrow x=21:7\)
\(\Rightarrow x=3\)
c, \(x^{10}=1^x\)
\(\Rightarrow x\in\left\{1;0\right\}\)
\(2^x-15=17\)
\(\Rightarrow2^x=17+15\)
\(\Rightarrow2^x=32=2^4\)
\(\Rightarrow x=4\)
\(\left(7x-11\right)^3=2^5.5^2+200\)
Phần này mk ko bt làm đâu
\(x^{10}=1^x\)
\(\Rightarrow\)\(x^{10}=1\)
\(\Rightarrow x=1\)
a) \(2^x=32\)=>\(2^x=2^5\)=>\(x=5\)
b) \(64\times4^x=4^5\)
\(4^3.4^x=4^5\)
\(4^{3+x}=4^5\)
=>\(3+x=5\)
=> x=2
c)\(2^x-15=17\)
\(2^x=17+15\)
\(2^x=32\)
=>\(2^x=2^5\)
=>\(x=5\)
a, 2 mũ x = 32
2 mũ x = 2 mũ 5
x = 5
b, 64 x 4 mũ x = 4 mũ 5
64 x 4 mũ x = 1024
4 mũ x = 1024 : 64
4 mũ x = 16
4 mũ x = 4 mũ 2
x = 2
c, 2 mũ x - 15 = 17
2 mũ x = 17 + 15
2 mũ x = 32
2 mũ x = 2 mũ 5
x = 5
2 mũ 4 - 4 mũ 2=16-16=0
( 2 mũ 17 + 17 mũ 2 ) . ( 9 mũ 15 - 3 mũ 15 ).0=0
đ/s: 0
1) \(2^x-15=17\)
\(\Leftrightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
2) \(\left(7x-11\right)^3=25\cdot5^2+200\)
\(\Leftrightarrow\left(7x-11\right)^3=825\)
\(\Leftrightarrow7x-11=\sqrt[3]{825}\)
\(\Leftrightarrow7x=11+\sqrt[3]{825}\)
\(\Rightarrow x=\frac{11+\sqrt[3]{825}}{7}\)
3) \(\left(x+1\right)^{100}-3\left(x+1\right)^{99}=0\)
\(\Leftrightarrow\left(x+1\right)^{99}\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^{99}=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
4) \(4x+5\left(x+3\right)=105\)
\(\Leftrightarrow9x+15=105\)
\(\Leftrightarrow9x=90\)
\(\Rightarrow x=10\)
5) \(5\cdot\left(x-2\right)+10\left(x+3\right)=170\)
\(\Leftrightarrow5\left[x-2+2\left(x+3\right)\right]=170\)
\(\Leftrightarrow3x+4=34\)
\(\Leftrightarrow3x=30\)
\(\Rightarrow x=10\)
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