Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{\text{4}5-x}{1963}+\frac{\text{4}0-x}{1968}+\frac{35-x}{1973}+\frac{30-x}{1978}+\text{4}=0\)
tham khảo nhé
https://olm.vn/hoi-dap/detail/103171879928.html
\(\frac{45-x}{1963}+\frac{40-x}{1968}+\frac{35-x}{1973}+\frac{30-x}{1978}+4=0\)
\(\Leftrightarrow\left(\frac{45-x}{1963}+1\right)+\left(\frac{40-x}{1968}+1\right)+\left(\frac{35-x}{1973}+1\right)+\left(\frac{30-x}{1978}+1\right)=0\)
\(\Leftrightarrow\frac{2008-x}{1963}+\frac{2008-x}{1968}+\frac{2008-x}{1973}+\frac{2008-x}{1973}=0\)
\(\Leftrightarrow\left(2008-x\right)\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0\)
\(\Leftrightarrow x=2008\)
Các câu dễ tự làm :v
\(\dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1981}=-4\) (sau khi đã sửa đề)
\(\Rightarrow\left(\dfrac{45-x}{1968}+1\right)+\left(\dfrac{40-x}{1973}+1\right)+\left(\dfrac{35-x}{1978}+1\right)+\left(\dfrac{30-x}{1981}+1\right)=0\)\(\Rightarrow\dfrac{2013-x}{1968}+\dfrac{2013-x}{1973}+\dfrac{2013-x}{1978}+\dfrac{2013-x}{1981}=0\)
\(\Rightarrow\left(2013-x\right)\left(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1981}\right)=0\)
\(\Rightarrow2013-x=0\Rightarrow x=2013\)
\(1+5+9+13+17+.....+x=5050\)
Số các số hạng là:
\(\dfrac{x-1}{4}+1=\dfrac{1}{4}x+\dfrac{3}{4}\)
Như vậy có :
\(\left(\dfrac{1}{4}x+\dfrac{3}{4}\right):2\) số hạng
Theo đề bài ta có:
\(\left(\dfrac{1}{4}x+\dfrac{3}{4}\right):2\left(x+1\right)=5050\)
\(\Rightarrow\left(\dfrac{1}{4}x+\dfrac{3}{4}\right)\left(x+1\right)=10100\)
\(\Rightarrow\dfrac{1}{4}x^2+\dfrac{1}{4}x+\dfrac{3}{4}x+\dfrac{3}{4}=10100\)
\(\Rightarrow\dfrac{1}{4}x^2+x+\dfrac{3}{4}=10100\)
Kiệt sức.đến đây ko nghĩ nổi nx
a,
\(5^x+5^{x+2}=650\\ 5^x\left(1+5^2\right)=650\\ 5^x\cdot26=650\\ 5^x=25\\ 5^x=5^2\\ \Rightarrow x=2\)
Vậy \(x=2\)
b,
\(\left(x+2\right)^2=81\\ \Rightarrow\left[{}\begin{matrix}x+2=9\\x+2=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-11\end{matrix}\right.\)
Vậy \(x=7\) hoặc \(x=-11\)
d,
\(\dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1983}=-4\\ \dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1983}+4=0\\ \dfrac{45-x}{1968}+1+\dfrac{40-x}{1973}+1+\dfrac{35-x}{1978}+1+\dfrac{30-x}{1983}+1=0\\ \dfrac{2013-x}{1968}+\dfrac{2013-x}{1973}+\dfrac{2013-x}{1978}+\dfrac{2013-x}{1983}=0\\ \left(2013-x\right)\left(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1983}\right)=0\)
Vì \(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1983}\ne0\) nên
\(2013-x=0\\ x=2013\)
Vậy \(x=2013\)
e,
\(\dfrac{1}{2016}:2015x=\dfrac{-1}{2015}\\ 2015x=\dfrac{-2015}{2016}\\ x=\dfrac{-1}{2016}\)
Vậy \(x=\dfrac{-1}{2016}\)
a. 7804 - x . 6 + 3829 = 5315
=> 11633 - x . 6 = 5315
=> x . 6 = 11633 - 5315
=> x . 6 = 6318
=> x = 6318 : 6
=> x = 1053
b. x . 23 + x . 35 - 1968 = 10270
=> x . 58 = 10270 + 1968
=> x . 58 = 12238
=> x = 12238 : 58
=> x = 211
c. 3.(x-2) - 2.(x+1) = -30
=> 3.x - 6 - 2.x - 2 = -30
=> x - 8 = -30
=> x = -30 + 8
=> x = -22
d. 42 + (3x + 7) : 2 = 25 + 34
=> 42 + (3x + 7) : 2 = 32 + 81
=> (3x + 7) : 2 = 32 + 81 - 42
=> (3x + 7) : 2 = 71
=> 3x + 7 = 71 . 2
=> 3x + 7 = 142
=> 3x = 142 - 7
=> 3x = 135
=> x = 135:3
=> x = 45
\(5x+40:2=45\)
\(5x+20=45\)
\(5x=45-20\)
\(5x=25\)
\(x=25:5\)
\(x=5\)
_______
\(x+30=40-2.5^2\)
\(x+30=40-2.25\)
\(x+30=40-50=-10\)
\(x=\left(-10\right)-30\)
\(x=-40\)
1)tính
a)\(-17+\left(-19\right)+25\)
\(=-17-19+25\)
\(=-\left(17+19\right)+25\)
\(=-36+25\)
\(=-\left(36-25\right)\)
\(=-11\)
b)\(-26+\left(-25\right)-18\)
\(=-26-25-18\)
\(=-\left(26+25\right)-18\)
\(=-51-18\)
\(=-\left(51+18\right)\)
\(=-69\)
c)\(35-\left(-19\right)-19-8\)
\(=35+19-19-8\)
\(=35+\left(19-19\right)-8\)
\(=35+0-8\)
\(=35-8\)
\(=27\)
d)\(1963-\left(-180\right)-1990\)
\(=1963+180-1990\)
\(=2143-1990\)
\(=153\)
2)Tìm x thuộc z biết
a)\(x-\left(-18\right)=25+\left(-17\right)\)
\(\Leftrightarrow x+18=25-17\)
\(\Leftrightarrow x+18=8\)
\(\Leftrightarrow x=8-18\)
\(\Leftrightarrow x=-10\)
Vậy x=-10
b)\(x-28=-27-30\)
\(\Leftrightarrow x-28=-\left(27+30\right)\)
\(\Leftrightarrow x-28=-57\)
\(\Leftrightarrow x=-57+28\)
\(\Leftrightarrow x=-\left(57-28\right)\)
\(\Leftrightarrow x=-29\)
Vậy x=-29
c)\(35-\left(-x\right)=27-18-\left(-8\right)\)
\(\Leftrightarrow35+x=27-18+8\)
\(\Leftrightarrow35+x=9+8\)
\(\Leftrightarrow35+x=17\)
\(\Leftrightarrow x=17-35\)
\(\Leftrightarrow x=-18\)
Vậy x=-18
d)\(40+\left(-x\right)=35-\left(-27\right)\)
\(\Leftrightarrow40-x=35+27\)
\(\Leftrightarrow40-x=62\)
\(\Leftrightarrow x=40-62\)
\(\Leftrightarrow x=-22\)
Vậy x=-22
\(a,45-\left(30+x\right)=x-\left(37-8\right)\)
\(\Leftrightarrow45-30-x=x-37+8\)
\(\Leftrightarrow-x-x=-37+8-45+30\)
\(\Leftrightarrow-2x=-44\)
\(\Leftrightarrow x=22\)
\(b,\left(x-12\right)-15=\left(30-8\right)-\left(40+x\right)\)
\(\Leftrightarrow x-12-15=30-8-40-x\)
\(\Leftrightarrow x+x=30-8-40+12+15\)
\(\Leftrightarrow2x=9\)
\(\Leftrightarrow x=4,5\)
271-[(-43)+271+(-13)] = 56
40.(45-135)-40.(45+65)=-8000
32x+41=35-(-70) => x=2
tìm x thuộc z,y thuộc z biết:(x-1)(xy-5)=5 Chịu
a/ -x + 8 = -17-30
-x + 8 = -47
-x = -47 - 8
-x = -55
x = 55
Vậy x = 55
b/ 35 – x = 37
x = 35 - 37
x = -2
Vậy x = -2
c/ (-19 – x) + 23 = -20
-19 – x = -20 - 23
-19 – x = -43
x = -19 - (-43)
x = 24
Vậy x = 24
d/ (x – 45) : 2 = -17
x – 45 = -17 * 2
x – 45 = -34
x = -34 + 45
x = 11
Vậy x = 11
a/ -x + 8 = -17-30
-x + 8 = -47
-x =-47 - 8
-x = -55
x = 55
Vậy x = 55
Phần a mình bị lỗi nên sửa lại nha
Vì 612 chia hết cho a và 680 chia hết cho a nên a ∈ ƯC(612,680)
Ta có : 612 = 2 2 . 3 2 . 17 ; 680 = 2 3 . 5 . 17 => ƯCLN(612,680) = 2 2 . 17 = 68
Mà Ư(68) = {1;2;4;17;34;68}
=> ƯC(612,680) = {1;2;4;17;34;68}
=> a ∈ {1;2;4;17;34;68}
Vì a lớn hơn 30 nên a ∈ {34;68}
\(\frac{45-x}{1963}+\frac{40-x}{1968}+\frac{35-x}{1973}+\frac{30-x}{1978}=-4\)
\(\left(\frac{45-x}{1963}+1\right)+\left(\frac{40-x}{1968}+1\right)+\left(\frac{35-x}{1973}+1\right)+\left(\frac{30-x}{1978}+1\right)=0\)
\(\frac{2008-x}{1963}+\frac{2008-x}{1968}+\frac{2008-x}{1973}+\frac{2008-x}{1978}=0\)
\(\left(2008-x\right)\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0\)
=> 2008 - x = 0 ( vì 1/ 1963 + ... khác 0 )
=> x = 2008