\(tinh\)
\(\left(-24\right)+\left(-6\right)=\)
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a) \(\left( { - 3} \right).7 = - \left( {3.7} \right) = - 21\)
b) \(\left( { - 8} \right).\left( { - 6} \right) = 8.6 = 48\)
c) \(\left( { + 12} \right).\left( { - 20} \right) = - \left( {12.20} \right) = - 240\)
d) \(24.\left( { + 50} \right) = 24.50 = 1200\)
Ta có :
\(P=\left(x-4\right)^{\left(x-5\right)^{\left(x-6\right)^{\left(x+6\right)^{\left(x+5\right)}}}}\)
Mà : x=7
\(P=\left(7-4\right)^{\left(7-5\right)^{\left(7-6\right)^{\left(7+6\right)^{\left(7+5\right)}}}}\)
\(P=9\)
Ta có P= (7-4)\(^{\left(7-5\right)^{\left(7-6\right)^{^{\left(7+6^{\left(7+5\right)}\right)}}}}\)
P=9
Cách 1:
a) \(12:6 = 2\)
b) \(24:\left( { - 8} \right)=-(24:8)=-3\)
c) \(\left( { - 36} \right):9=-(36:9)=-4\)
d) \(\left( { - 14} \right):\left( { - 7} \right)=14:7=2\)
Cách 2:
a) Ta có \(12 = 6.2\) nên \(12:6 = 2\).
b) Ta có \(24 = \left( { - 8} \right).\left( { - 3} \right)\)\( \Rightarrow 24:\left( { - 8} \right) = \left( { - 3} \right)\).
c) Ta có \(\left( { - 36} \right) = 9.\left( { - 4} \right)\) nên \(\left( { - 36} \right):9 = \left( { - 4} \right)\).
d) Ta có \(\left( { - 14} \right) = \left( { - 7} \right).2\) nên \(\left( { - 14} \right):\left( { - 7} \right) = 2\)
a) \(\left( { - 5} \right).4 = - \left( {5.4} \right) = - 20\)
b) \(6.\left( { - 7} \right) = - \left( {6.7} \right) = - 42\)
c) \(\left( { - 14} \right).20 = - \left( {14.20} \right) = - 280\)
d) \(51.\left( { - 24} \right) = - \left( {51.24} \right) = - 1224\)
a: \(=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{2}{3}\right)-\dfrac{32}{17}=2-\dfrac{32}{17}=\dfrac{2}{17}\)
b: \(=-8\cdot\dfrac{1}{4}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)=-2:\dfrac{27-14}{12}=\dfrac{-2\cdot12}{13}=-\dfrac{24}{13}\)
c: \(=\dfrac{-5}{3}\left(16+\dfrac{2}{7}+28+\dfrac{2}{7}\right)=\dfrac{-5}{3}\cdot\left(44+\dfrac{4}{7}\right)\)
=-520/7
a) (-24) + 6 + 10 + 24
= [(-24) + 24] + 6 + 10
= 0 + 6 + 10
= 16
b) 15 + 23 + (-25) + (-23)
= [15+ (-25)] + [23 +(-23)]
= -10 + 0
= -10
c) (-3) + (-350) + (-7) + 350
=[-350 + 350] + [-3+(-7)]
= 0 + (-10)
= -10
d) (-9) + (-11) +21 + (-1)
= [ (-9) + (-11) ] + [ 21 + (-1)]
= -20 + 20
= 0
b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)
\(\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)
Xét đa thức bậc 8: \(P\left(x\right)=x^8+\dfrac{x^3-x}{2}\)
Ta có, \(P\left(x\right)-P\left(-x\right)=x^8+\dfrac{x^3-x}{2}-\left(-x\right)^8-\dfrac{\left(-x\right)^3-\left(-x\right)}{2}=x^3-x\)
Thay \(x=1;2;3;4\) đều thỏa mãn
\(\Rightarrow P\left(5\right)-P\left(-5\right)=5^3-5=120\)
- 30
tk mk nha
(-24)+(-6)=-30