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Gọi 2 số cần tìm là a, b. Ta có: \(\dfrac{a}{b}\)= \(\dfrac{3}{7}\)=k
=> a= 3k; b=7k và a.b= 189
Hay 3k. 7k= 189--> 21. k2= 189=> k2=9=> k=3 hoặc k= -3
Khi k=3=> a=9; b= 21
k=-3=> a=-9; b=-21
Vậy a= 9 hoặc -9; b=21 hoặc -21
Bài 2 thiếu đề thì phải
bài 2 đề đầy đủ, mik biết làm rồi,..thôi dù gì cũng cám ơn bn
1) \(\left(+15\right)+\left(+17\right)=15+17=32\)
2) \(\left(-3\right)+\left(-7\right)=-3-7=-\left(3+7\right)=-10\)
3) \(\left(-25\right)+\left(+4\right)=-25+4=-\left(25-4\right)=-21\)
4) \(\left(-6\right)+\left(-54\right)=-6-54=-\left(6+54\right)=-60\)
5) \(\left(-15\right)+20=20-15=5\)
6) \(\left(-5\right)+8+7+5\)
\(=\left(-5+5\right)+\left(8+7\right)\)
\(=15\)
7) \(\left(-8\right)+\left(-11\right)+\left(-2\right)\)
\(=\left[\left(-8\right)+\left(-2\right)\right]+\left(-11\right)\)
\(=\left(-10\right)+\left(-11\right)\)
\(=-21\)
8) \(15+\left(-5\right)+\left(-14\right)+\left(-16\right)\)
\(=\left[15+\left(-5\right)\right]+\left[\left(-14\right)+\left(-16\right)\right]\)
\(=10+\left(-30\right)\)
\(=-20\)
9) \(\left(-20\right)+\left(-14\right)+3+\left(-86\right)\)
\(=\left[\left(-20\right)+3\right]+\left[\left(-14\right)+\left(-86\right)\right]\)
\(=\left(-17\right)+\left(-100\right)\)
\(=-117\)
10) \(\left(-136\right)+123+\left(-264\right)+\left(-83\right)+240\)
\(=\left[\left(-136\right)+\left(-264\right)\right]+\left[123+\left(-83\right)\right]+240\)
\(=\left(-400\right)+40+240\)
\(=\left(-360\right)+240\)
\(=-120\)
11) \(\left(-596\right)+2001+1999+\left(-404+189\right)\)
\(=\left(-596\right)+2001+1999-404+189\)
\(=\left[\left(-596\right)-404\right]+\left(2001+189\right)+1999\)
\(=\left(-1000\right)+2190+1999\)
\(=1190+1999\)
\(=3189\)
12) \(314+\left(-153\right)+64+121+\left(-247\right)+218\)
\(=\left(314+64+121\right)+\left[\left(-153\right)+\left(-247\right)\right]+218\)
\(=\left(378+121\right)+\left(-400\right)+218\)
\(=499-400+218\)
\(=99+218\)
\(=317\)
\(\text{#}Toru\)
a, = (1-2-3+4)+(5-6-7+8)+....+(2001-2002-2003+2004) = 0+0+...+0 = 0
b, => x-1=0 hoặc x-10=0 hoặc x=0
=> x=1 hoặc x=10 hoặc x=0
c, => 9x=189
=> x=189:9 = 21
k mk nha
tìm x biết:
(3x-1) [- 1/2x+5]=0
1/4+1/3:(2x-1)=-5
[2x+3/5]2 - 9/25=0
-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6
[x+1/2]x [2/3-2x]=0
17/2-|2x-3/4|=-7/4
2/3x-1/2x =5/12
(x+1/5)2+17/25=26/25
[x.44/7+3/7].11/5-3/7=-2
3[3x-1/2]+1/9=0
Toán lớp 6Tìm x
Trả lời Câu hỏi tương tự
Chưa có ai trả lời câu hỏi này,bạn hãy là người đâu tiên giúp nguyenvanhoang giải bài toán này !
a)15/16:3/8x3/4
=15/16:6/16x12/16
=15/16x16/6x12/16
=15/6x12/16
=30/16=15/8
b)5/11x18/29-5/11x8/29+5/11x19/29
=5/11x(18/29-8/29+19/29)
=5/11x1=5/11
Bài 1:
\(A=7^3+7^4+7^5+...+7^{97}+7^{98}.\)
\(A=\left(7^3+7^4\right)+\left(7^5+7^6\right)+...+\left(7^{97}+7^{98}\right).\)
\(A=7^3\left(1+7\right)+7^5\left(1+7\right)+...+7^{97}\left(1+7\right).\)
\(A=7^3.8+7^5.8+...+7^{97}.8.\)
\(A=\left(7^3+7^5+...+7^{97}\right).8⋮8\left(đpcm\right).\)
Vậy.....
Bài 2: Tìm x:
\(3^{47}:\left(189-3x\right)=3^{44}.\)
\(189-3x=3^{47}:3^{44}.\)
\(189-3x=27.\)
\(3x=189-27.\)
\(3x=162.\)
\(x=162:3.\)
\(x=54.\)
Vậy.....
\(27-3.\left(5x+2\right)=6.\)
\(3.\left(5x+2\right)=27-6.\)
\(3.\left(5x+2\right)=21.\)
\(5x+2=21:3.\)
\(5x+2=7.\)
\(5x=7-2.\)
\(5x=5.\)
\(x=5:5.\)
\(x=1.\)
Vậy.....
\(A=7^3+7^4+...+7^{98}\\ \Rightarrow A=\left(7^3+7^4\right)+\left(7^5+7^6\right)+....+\left(7^{97}+7^{98}\right)\\ =7^3\left(1+7\right)+7^5\left(1+7\right)+...+7^{97}\left(1+7\right)\\ =7^3.8+7^5.8+...+7^{97}.8\\ =8\left(7^3+7^5+...+7^{97}\right)⋮8\)
\(189:\left[628-\left(2x-1\right)^2\right]=3^2\cdot7\)
=>\(628-\left(2x-1\right)^2=\dfrac{189}{63}=3\)
=>\(\left(2x-1\right)^2=628-3=625\)
=>\(\left[{}\begin{matrix}2x-1=25\\2x-1=-25\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=26\\2x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=-12\end{matrix}\right.\)
\(189:\left[628-\left(2x-1\right)^2\right]=3^2.7\)
\(\Rightarrow628-\left(2x-1\right)^2=\dfrac{189}{63}\)
\(\Rightarrow628-\left(2x-1\right)^2=3\)
\(\Rightarrow\left(2x-1\right)^2=625\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=25\\2x-1=-25\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=13\\x=-12\end{matrix}\right.\)