(-20).29+(-2).(-99)+(-2).(-30)=?
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(-2) . 29 + (-2) . (-99) + (-2) . (-30)
= (-2) . [29 + (-99) + (-30)]
= (-2) . [ (-70)+(-30)]………………
= (-2).(-100)=200………………….
(-2) . 29 + (-2) . (-99) + (-2) . (-30)
= (-2) . [ 29 + (-99) + (-30)]
=(-2) . (-100)
=200
Câu 11:
(\(\dfrac{11}{4}\). \(\dfrac{-5}{9}\) - \(\dfrac{4}{9}\).\(\dfrac{11}{4}\)).\(\dfrac{8}{33}\)
= \(\dfrac{11}{4}\).(\(\dfrac{-5}{9}\) - \(\dfrac{4}{9}\)). \(\dfrac{8}{33}\)
= \(\dfrac{11}{4}\).(-1).\(\dfrac{8}{33}\)
= - \(\dfrac{2}{3}\)
b, ( cái (99) là -99 hay +99)
\(c,=\left[\left(-25\right).\left(-4\right)\right].\left[9.\left(-7\right)\right]\\ =100.\left(-63\right)=-6300\\ d,=37.\left(-84\right)+37.\left(-16\right)\\ =37.\left[\left(-84\right)+\left(-16\right)\right]\\ =37.\left(-100\right)=-3700\)
b) (-2).29+(-2).(99)+(-2).(-30)
=1.(-2).29+(-2).(99)+(-2)
=-2. (1+99+30)
=-2.130
=-260
c)(-25).9.(-4).(-7)
=[-25.(-4)].[9.(-7)]
=100.(-63)
=-6300
d) (-37).84+37.(-16)
=(-37+37).[84+(-16)]
=0.[84+(-16)]
=0
\(\left( { - 2} \right).29 + \left( { - 2} \right).\left( { - 99} \right)\)\( + \left( { - 2} \right).\left( { - 30} \right)\)\( = \left( { - 2} \right)\left( {29 - 99 - 30} \right)\)\( = \left( { - 2} \right).\left( { - 100} \right) = 200\)
(-2).99 + (-2).(-99) + (-2).(-30)
= 99.(-2 + 2) + 60
= 99.0 + 60
= 0 + 60
= 60
a.
\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow x\left(x+1\right).\left(x-1\right)\left(x+2\right)-24=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)
Đặt \(a=x^2+x-1\) , ta có pt:
\(\left(a+1\right)\left(a-1\right)-24=0\)
\(\Leftrightarrow a^2-1-24=0\)
\(\Leftrightarrow a^2-25=0\)
\(\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=5\\a=-5\end{matrix}\right.\)
*Với a = 5 ta được:
\(x^2+x-1=5\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2+3x-2x-6=0\)
\(\Leftrightarrow\left(x^2+3x\right)-\left(2x+6\right)=0\)
\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
*Với a = -5 ta được:
\(x^2+x-1=-5\)
\(\Leftrightarrow x^2+x+4=0\)
\(\Leftrightarrow x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{15}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\) ( loại)
Vậy pt có tập nghiệm là: \(s=\left\{-3;2\right\}\)
c)(ĐKXĐ: x khác 30;29)
\(\Leftrightarrow\dfrac{x-29}{30}-1+\dfrac{x-30}{29}-1=\dfrac{29}{x-30}-1+\dfrac{30}{x-29}-1\)
\(\Leftrightarrow\dfrac{x-59}{30}+\dfrac{x-59}{29}=\dfrac{x-59}{30-x}+\dfrac{x-59}{29-x}\)
\(\Leftrightarrow x=59\)(tm) or \(\dfrac{1}{30}+\dfrac{1}{29}-\dfrac{1}{30-x}-\dfrac{1}{29-x}=0\)
\(\Leftrightarrow\dfrac{-x}{30\left(30-x\right)}+\dfrac{-x}{29\left(29-x\right)}=0\)
\(\Leftrightarrow x=0\)(tm) or \(\dfrac{1}{30\left(30-x\right)}+\dfrac{1}{29\left(29-x\right)}=0\)
\(\Leftrightarrow1741-59x=0\)
\(\Leftrightarrow x=\dfrac{1741}{59}\left(tm\right)\)
Vậy S={0;\(\dfrac{1741}{59}\);59}
\(a)A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}\)
\(A=2\left(\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{15.16}\right)\)
\(A=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(A=2\left(\dfrac{4}{16}-\dfrac{1}{16}\right)\)
\(A=\dfrac{2.3}{16}\)
\(=\dfrac{3}{8}\)
\(b)B=\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+\dfrac{29}{30}+\dfrac{41}{42}+\dfrac{55}{56}\)
\(B=\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{12}\right)+...+\left(1-\dfrac{1}{56}\right)\)
\(B=6-\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{7.8}\right)\)
\(B=6-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)\)
\(B=6-\left(\dfrac{4}{8}-\dfrac{1}{8}\right)\)
\(B=\dfrac{48}{8}-\dfrac{3}{8}\)
\(B=\dfrac{45}{8}\)
Lời giải:
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}\)
\(=5-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\right)\)
\(=5-\left(\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+\frac{5-4}{4\times 5}+\frac{6-5}{5\times 6}\right)\)
\(=5-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)
\(=5-\left(1-\frac{1}{6}\right)=5-\frac{5}{6}=\frac{25}{6}\)
(-20).29+(-2).(-99)+(-2).(-30)
=(-2).10.29+(-2).(-99)+(-2).(-30)
=(-2).(290-99-30)
=(-2).161
=-322
HT