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\(-\dfrac{3}{4}.31\dfrac{11}{23}-0,75.8\dfrac{12}{23}\)
\(=-\dfrac{3}{4}.31\dfrac{11}{23}-\dfrac{3}{4}.8\dfrac{12}{23}\)
\(=-\dfrac{3}{4}.\left(31+\dfrac{11}{23}+8+\dfrac{12}{12}\right)\)
\(=-\dfrac{3}{4}.\left(31+8+1\right)\)
\(=-\dfrac{3}{4}.40\)
\(=-3.10\)
\(=-30\)
- \(\dfrac{3}{4}\).31\(\dfrac{11}{23}\) - 0,75.8\(\dfrac{12}{23}\)
= - \(\dfrac{3}{4}\).\(\dfrac{724}{23}\) - \(\dfrac{3}{4}\)\(\dfrac{196}{23}\)
= - \(\dfrac{3}{4.23}.\left(724+196\right)\)
= - \(\dfrac{3}{92}\) . 920
= - 30
a, 1 - 7x = 3x - 4
=> -7x - 3x = - 4 - 1
=> - 10x = - 5
=> x = 1/2
vậy_
b, đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3A-A=1-\frac{1}{3^{99}}\)
\(A=\frac{1-\frac{1}{3^{99}}}{2}\)
mk chỉ bt lm mấy phần hui à!
d)\(\frac{5}{17}+\frac{-4}{7}-\frac{20}{31}+\frac{12}{17}-\frac{11}{31}\)\(=\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}-\frac{11}{31}\right)+\frac{-4}{7}\)
\(=\frac{17}{17}+\frac{-31}{31}+\frac{-4}{7}\)\(=1+\left(-1\right)+\frac{-4}{7}\)\(=0+\frac{-4}{7}\)\(=-\frac{4}{7}\)
e)\(\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{20}{7}-\frac{13}{3}+\frac{13}{23}}\)
a) Ta có: \(C=-\left|x+2\right|\le0\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left|x+2\right|=0\Rightarrow x=-2\)
Vậy Max(C) = 0 khi x = -2
b) Ta có: \(D=1-\left|2x-3\right|\le1\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left|2x-3\right|=0\Rightarrow x=\frac{3}{2}\)
Vậy Max(D) = 1 khi x = 3/2
d) \(D=-\left|x+\frac{5}{2}\right|\le0\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left|x+\frac{5}{2}\right|=0\Rightarrow x=-\frac{5}{2}\)
Vậy Max(D) = 0 khi x = -5/2
e) \(P=4-\left|5x-3\right|-\left|3y+12\right|\le4\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left|5x-3\right|=0\\\left|3y+12\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{3}{5}\\y=-4\end{cases}}\)
Vậy Max(P) = 4 khi \(\hept{\begin{cases}x=\frac{3}{5}\\y=-4\end{cases}}\)
\(-\dfrac{3}{4}.31\dfrac{11}{23}-0,75.8\dfrac{11}{23}\)
\(=\dfrac{3}{4}.\left(-31\dfrac{11}{23}\right)-\dfrac{3}{4}.8\dfrac{11}{23}\)
\(=\dfrac{3}{4}.\left(-31\dfrac{11}{23}-8\dfrac{11}{23}\right)\)
\(=\dfrac{3}{4}.\left(-39\right)\)
\(=-\dfrac{117}{4}\)
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
