b,
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. 2xyz+4xyz-\(\frac{1}{2}\) xyz =(2+4-\(\frac{1}{2}\) )(xxxyyyzzz) =3.5x3 y3z3 b)\(\frac{1}{9}.\frac{2}{145}-4\frac{1}{3}.\frac{2}{145}+\frac{2}{145}\) \(=\frac{2}{145}.\left(\frac{1}{9}-\frac{13}{3}+1\right)\) \(=\frac{2}{145}.\left(-\frac{29}{9}\right)\) \(=\frac{-2}{45}\) a: \(=\left(\dfrac{5}{15}-\dfrac{12}{9}\right)+\left(\dfrac{14}{15}+\dfrac{11}{25}\right)+\dfrac{2}{7}\) \(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{70+33}{75}+\dfrac{2}{7}\) \(=-1+\dfrac{2}{7}+\dfrac{103}{75}=\dfrac{-5}{7}+\dfrac{103}{75}=\dfrac{346}{525}\) b: \(4\cdot\left(-\dfrac{1}{2}\right)^3+\dfrac{1}{2}\) \(=4\cdot\dfrac{-1}{8}+\dfrac{1}{2}=\dfrac{-1}{2}+\dfrac{1}{2}=0\) c: \(\dfrac{10^3+5\cdot10^2+5^3}{6^3+3\cdot6^2+3^3}=\dfrac{5^3\cdot8+5\cdot5^2\cdot2^2+5^3}{3^3\cdot2^3+3\cdot2^2\cdot3^2+3^3}\) \(=\dfrac{5^3\left(8+4+1\right)}{3^3\left(8+4+1\right)}=\dfrac{125}{27}\) e: \(\dfrac{2^8\cdot9^2}{6^4\cdot8^2}=\dfrac{2^8\cdot3^4}{3^4\cdot2^4\cdot2^6}=\dfrac{1}{4}\) 1.a)\(2.x-\dfrac{5}{4}=\dfrac{20}{15}\) \(\Leftrightarrow2.x=\dfrac{20}{15}+\dfrac{5}{4}=\dfrac{4}{3}+\dfrac{5}{4}=\dfrac{16+15}{12}=\dfrac{31}{12}\) \(\Leftrightarrow x=\dfrac{31}{12}:2=\dfrac{31}{12}.\dfrac{1}{2}=\dfrac{31}{24}\) b)\(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{8}\right)\) \(\Leftrightarrow\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\) \(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\) \(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}=-\dfrac{5}{6}\) 2.Theo đề bài, ta có: \(\dfrac{a}{2}=\dfrac{b}{3}\) và \(a+b=-15\) Áp dụng tính chất của dãy tỉ số bằng nhau, ta có: \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{a+b}{2+3}=\dfrac{-15}{5}=-3\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=-3\Rightarrow a=-6\\\dfrac{b}{3}=-3\Rightarrow b=-9\end{matrix}\right.\) 3.Ta xét từng trường hợp: -TH1:\(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\)\(\Rightarrow x\in\left\{0;1\right\}\) -TH2:\(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\)\(\Rightarrow x\in\varnothing\) Vậy \(x\in\left\{0;1\right\}\) 4.\(B=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^9=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^9=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^{18}=\left(\dfrac{3}{7}\right)^3=\dfrac{27}{343}\) \(0,3+\dfrac{-13}{2}+\dfrac{6}{5}\) \(=\dfrac{3}{10}+\dfrac{-13}{2}+\dfrac{6}{5}\) \(=-5\) 0,3+\(\dfrac{-13}{2}+\dfrac{6}{5}\) = \(\dfrac{3}{10}+\dfrac{-65}{10}+\dfrac{12}{10}\) = \(\dfrac{-50}{10}\)=-5 \(a,\dfrac{2^4.6^7}{9^3.4^6}=\dfrac{2^4.\left(2.3\right)^7}{\left(3^2\right)^3.\left(2^2\right)^6}=\dfrac{2^4.2^7.3^7}{3^6.2^{12}}=\dfrac{2^{11}.3^7}{3^6.2^{12}}=\dfrac{3}{2}\) \(b,\dfrac{3^{18}.24^4}{9^4.81^5}=\dfrac{3^{18}.\left(3.2^3\right)^4}{\left(3^2\right)^4.\left(3^4\right)^5}=\dfrac{3^{18}.3^4.2^{12}}{3^8.3^{20}}=\dfrac{3^{22}.2^{12}}{3^{28}}=\dfrac{2^{12}}{3^8}\) a: =>||12x-1/2|-2|=-2/3x3/4=-6/12=-1/2(loại) b: =>2/3-1/3x-1/2+2/3x=2x+2/3 =>-5/3x=1/2 =>x=-1/2:5/3=-1/2x3/5=-3/10 c: =>|3/2x+1/4|=2+3/4=11/4 =>3/2x+1/4=11/4 hoặc 3/2x+1/4=-11/4 =>3/2x=5/2 hoặc 3/2x=-3 =>x=3/5 hoặc x=-3:3/2=-2 \(\dfrac{x-2}{4}=\dfrac{y+1}{5}=\dfrac{z+3}{7}\) \(\Rightarrow\dfrac{2\left(x-2\right)}{8}=\dfrac{y+1}{5}=\dfrac{2\left(z+3\right)}{14}\) \(\Rightarrow\dfrac{2x-4}{8}=\dfrac{y+1}{5}=\dfrac{2z+6}{14}\) Dựa vào tính chất dãy tỉ số bằng nhau ta có: \(=\dfrac{2x-4+y+1-2z-6}{8+5-14}\) \(=\dfrac{2x+y-2z-9}{-1}\) \(=\dfrac{7-9}{-1}=2\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-2}{4}=2\Rightarrow x-2=8\Rightarrow x=10\\\dfrac{y+1}{5}=2\Rightarrow y+1=10\Rightarrow y=9\\\dfrac{z+3}{7}=2\Rightarrow z+3=14\Rightarrow z=11\end{matrix}\right.\)
K
Khách
TN
Tram Nguyen
1 tháng 4 2018
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