\(\frac72\) -\(x\) = \(\frac23\) +\(\frac14\)
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\(P=\dfrac{\dfrac{8}{12}-\dfrac{3}{12}+\dfrac{5}{11}}{\dfrac{5}{12}+\dfrac{12}{12}-\dfrac{7}{11}}=\dfrac{\dfrac{5}{12}+\dfrac{5}{11}}{\dfrac{17}{12}-\dfrac{7}{11}}=\dfrac{115}{132}:\dfrac{103}{132}=\dfrac{115}{103}\)

a) \(x+2x+3x+...+100x=-213\)
\(\Rightarrow x.\left(1+2+3+...+100\right)=-213\)
\(\Rightarrow x.5050=-213\Rightarrow x=\frac{-213}{5050}\)
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-\frac{25}{6}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{-47}{12}\)
\(\Rightarrow\frac{1}{2}x=\frac{-43}{12}\Rightarrow x=\frac{-43}{6}\)
d) \(\frac{x+1}{3}=\frac{x-2}{4}\Rightarrow4\left(x+1\right)=3\left(x-2\right)\Rightarrow4x+4=3x-6\)
\(\Rightarrow4x-3x=-6-4\Rightarrow x=-10\)
c) \(3\left(x-2\right)+2\left(x-1\right)=10\)
\(\Rightarrow3x-6+2x-2=10\)
\(\Rightarrow5x=18\Rightarrow x=\frac{18}{5}\)
a) \(x+2x+3x+4x+...+100x=-213\)
\(x.\left(1+2+3+4+...+100\right)=-213\)
\(x.5050=-213\)
\(x=-\frac{213}{5050}\)
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)
\(\frac{1}{2}x-\frac{1}{3}=-\frac{47}{12}\)
\(\frac{1}{2}x=-\frac{43}{12}\)
\(x=\frac{-43}{6}\)

a)P(x) = x^5 + 7x^4 - 9x^3 - 2x^2 - 1/4x
Q(x) = x^5 + 5x ^ 4 - 2x ^ 3 + 4x^2 - 1/4
b) P(x)+Q(x)
= (x^5 – 2x^2 + 7x^4 – 9x^3 – ¼ x ) + (5x^4 – x^5 + 4x^2 – 2x^3 – 1/4)
= x^5 – 2x^2 + 7x^4 – 9x^3 – ¼ x + 5x^4 – x^5 + 4x^2 – 2x^3 – 1/4
= (x^5 - x^5 ) + ( 7x^4 + 5x^4) + (-2x^3-9x^3) + ( -2x^2 +4x^2) + 1/4x+1/4
= 0 + 12x^4 + -11x^3 + 2x^2 + 1/4x + 1/4
= 12x^4 - 11x^3 + 2x^2 + 1/4x + 1/4
P(x) – Q(x)
= (x^5 – 2x^2 + 7x^4 – 9x^3 – ¼ x ) - (5x^4 – x^5 + 4x^2 – 2x^3 – 1/4)
= x^5 – 2x^2 + 7x^4 – 9x^3 – ¼ x - 5x^4 + x^5 - 4x^2 + 2x^3 + 1/4
=(x^5 + x^5 ) + ( 7x^4 - 5x^4) + (2x^3 - 9x^3) + ( -2x^2 - 4x^2) + 1/4x+1/4
= 2x^5 + 2x^4 + -7x^3 + -6x^2 + 1/4x + 1/4
=2x^5 + 2x^4 - 7x^3 - 6x^2 + 1/4x + 1/4

\(Q=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2018}\right)\)
\(Q=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2017}{2018}\)
\(Q=\dfrac{1.2.3...2017}{2.3.4...2018}\)
\(Q=\dfrac{1}{2018}\)
Vậy \(Q=\dfrac{1}{2018}\)

\(Q=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right).........\left(1-\dfrac{1}{2018}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right).........\left(\dfrac{2018}{2018}-\dfrac{1}{2018}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.......\dfrac{2017}{2018}\)
\(=\dfrac{1}{2018}\)
Tính giá trị biểu thức
P=(x-a) (x-b) (x-c)... (x-y) (x-z) tại x=2018 câu khó hơn nhiều

a) \(\left(m-\dfrac{1}{4}\right)^3=\left(m^3-3m^2.\dfrac{1}{4}+3m\left(\dfrac{1}{4}\right)^2-\left(\dfrac{1}{4}\right)^3\right)\\ =\left(m^3-\dfrac{3}{4}m^2+\dfrac{3}{16}m-\dfrac{1}{64}\right)\)
b)\(\left(\dfrac{2}{3}-n\right)^3=\left(\dfrac{2}{3}\right)^3-3\left(\dfrac{2}{3}\right)^2n+3.\dfrac{2}{3}n^2-n^3\\ =\dfrac{8}{27}-\dfrac{4}{3}n+2n^2-n^3\)
c)\(m^3-125=m^3-5^3=\left(m-5\right)\left(m^2+5m+25\right)\)
d)\(m^3+\dfrac{1}{64}=m^3+\left(\dfrac{1}{4}\right)^3=\left(m+\dfrac{1}{4}\right)\left(m^2-\dfrac{1}{4}m+\dfrac{1}{16}\right)\)
7/2 -x =11/12
x=7/2 - 11/12
x= 31/12
`7/2 -x =2/3 +1/4`
`7/2-x=8/12+3/12`
`7/2-x=11/12`
`x=7/2-11/12`
`x=42/12-11/12`
`x=31/12`
Vậy `x=31/12`