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197(199+98)-199.197=197.199+197.98-199.197
=197.(98+199-199)=197.98=19306
Bai 1:
a)20a30a chia het cho 5=>a=0 hoac 5
Neu a=0
Tong cac chu so cua 20a30a la 2+3=5 (ko chia het cho 3)(loai)
Neu a=5
Tong cac chu so cua 20a30a la 2+5+3+5=15 (chia het cho 3)(chon)
Vay a=5
b) Ta co : 162=2.9^2
576=2^6.9
UCLN(162;576)=2.9=18
Tich cho minh nha !^^
a) -12.(x - 5) + 7(3 - x) = 5
=> -12x + 60 + 21 - 7x = 5
=> -19x + 81 = 5
=> -19x = 5 - 81
=> -19x = -76
=> x = -76 : (-19)
=> x = 4
b) (x + 1) + (x + 2) + (x + 3) + ... + (x + 20) = 250
=> (x + x + x + ... + x) + (1 + 2 + 3 + ... + 20) = 250
=> 20x + 210 = 250
=> 20x = 250 - 210
=> 20x = 40
= > x = 40 : 20
=> x = 2
\(-12\left(x-5\right)+7\left(3-x\right)=5\)
\(\Leftrightarrow-12x+60+21-7x=5\)
\(\Leftrightarrow-19x+81=5\)
\(\Leftrightarrow81-5=19x\)
\(\Leftrightarrow19x=76\)
\(\Leftrightarrow x=4\)
\(\dfrac{1}{2}\) \(\times\) ( \(x\) - \(\dfrac{2}{3}\)) - \(\dfrac{1}{3}\) \(\times\) ( 2\(x\) - 3) = \(x\)
\(\dfrac{1}{2}\) \(\times\) \(\dfrac{3x-2}{3}\) - \(\dfrac{2x-3}{3}\) = \(x\)
\(\dfrac{3x-2}{6}\) - \(\dfrac{4x-6}{6}\) = \(\dfrac{6x}{6}\)
3\(x-2-4x\) + 6 = 6\(x\)
-\(x\) + 4 - 6\(x\) = 0
7\(x\) = 4
\(x\) = \(\dfrac{4}{7}\)
\(3\left(x+2\right)^2-5^2=2.5^2\)
\(\Rightarrow3\left(x+2\right)^2=2.5^2+5^2\)
\(\Rightarrow3\left(x+2\right)^2=5^2\left(2+1\right)\)
\(\Rightarrow3\left(x+2\right)^2=5^2.3\)
\(\Rightarrow\left(x+2\right)^2=5^2\)
\(\Rightarrow\left[{}\begin{matrix}x+2=5\\x+2=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\) \(\Rightarrow x=3\left(x\inℕ\right)\)
3(x + 2)² - 5² = 2.5²
3(x + 2)² - 25 = 50
3(x + 2)² = 50 + 25
3(x + 2)² = 75
(x + 2)² = 75 : 3
(x + 2)² = 25
x + 2 = 5 hoặc x + 2 = -5
*) x + 2 = 5
x = 5 - 2
x = 3 (nhận)
*) x + 2 = -5
x = -5 - 2
x = -7 (loại)
Vậy x = 3
\(\Leftrightarrow-\dfrac{2}{5}\left(4x-3\right)^2=-\dfrac{5}{18}\)
\(\Leftrightarrow\left(4x-3\right)^2=\dfrac{25}{36}\)
\(\Leftrightarrow4x-3\in\left\{\dfrac{5}{6};-\dfrac{5}{6}\right\}\)
hay \(x\in\left\{\dfrac{23}{24};\dfrac{13}{24}\right\}\)
\(2x^4-x^3+2x^2+1=2x^4-2x^3+2x^2+x^3-x^2+x+x^2-x+1\\ \)
\(=2x^2\left(x^2-x+1\right)+x\left(x^2-x+1\right)+\left(x^2-x+1\right)=\left(x^2-x+1\right)\left(2x^2+x+1\right)\)
Vậy a = 2; b = 1; c = 1.
\(\dfrac{20142014}{20152015}\times x+7986=1+3+5+...+199\)
Vì các số ở vế 2 đều cách nhau 2 đơn vị
=> Số số hạng của vế 2 là \(\left(199-1\right)\div2+1=100\) ( số hạng )
=> Tổng của vế 2 là \(\left(199+1\right)\times100\div2=10000\)
Thay vào biểu thức, ta có:
\(\dfrac{20142014}{20152015}\times x+7986=10000\)
\(\dfrac{2014}{2015}\times x=10000-7986=2014\)
\(x=2014\div\dfrac{2014}{2015}\)
\(x=2015\)
\(\dfrac{20142014}{20152015}\)\(x\)+ 7986 = 1 + 3 + 5 + ...+ 197 + 199
\(\dfrac{2014}{2015}\)\(x\) + 7986 = (199 + 3){ (199 -1): 2 + 1}: 2
\(\dfrac{2014}{2015}\)\(x\) + 7986 = 202. 100: 2
\(\dfrac{2014}{2015}x\) = 10000
\(\dfrac{2014}{2015}\)\(x\) = 10000 - 7986
\(\dfrac{2014}{2015}\)\(x\) = 2014
\(x\) = 2014 : \(\dfrac{2014}{2015}\)
\(x\) = 2015