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1.
a.
\(\left(\dfrac{-4}{5}+\dfrac{2}{3}\right)\cdot\dfrac{7}{11}+\left(\dfrac{-1}{5}+\dfrac{1}{3}\right)\cdot\dfrac{7}{11}\\ =\dfrac{7}{11}\cdot\left(\dfrac{-4}{5}+\dfrac{2}{3}+\dfrac{-1}{5}+\dfrac{1}{3}\right) \\ =\dfrac{7}{11}\cdot\left[\left(\dfrac{-4}{5}+\dfrac{-1}{5}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\right]\\ =\dfrac{7}{11}\cdot\left[\left(-1\right)+1\right]\\ =\dfrac{7}{11}\cdot0\\ =0\)
b.
\(\left(-3^2\right)\cdot\left(\dfrac{3}{4}-0,25\right)-\left|-2\right|\\ =\left(-9\right)\cdot0,5-2\\ =-4,5-2\\ =-6,5\)
2.
\(y=f\left(x\right)=\left(m+1\right)x\\ \Rightarrow4=f\left(2\right)=\left(m+1\right)\cdot2\\ \Rightarrow m+1=2\\ \Leftrightarrow m=1\)
Tự
3.
a.
\(\left|x-\dfrac{2}{5}\right|=\dfrac{3}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=\dfrac{3}{4}\\x-\dfrac{2}{5}=\dfrac{-3}{4}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{20}\\x=\dfrac{-7}{20}\end{matrix}\right.\)
b.
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2y}{6}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2y}{6}=\dfrac{x+2y-z}{5+6-4}=\dfrac{14}{7}=2\\ \Rightarrow\left\{{}\begin{matrix}x=10\\y=6\\z=8\end{matrix}\right.\)
1)\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2017}{2018}\)
\(B=\dfrac{1}{2018}\)
2)a)\(x^2-2x-15=0\)
\(\Leftrightarrow x^2-2x+1-16=0\)
\(\Leftrightarrow\left(x-1\right)^2-16=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
3)\(\dfrac{a}{b}=\dfrac{d}{c}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a}{b}\cdot\dfrac{d}{c}=\dfrac{ad}{bc}\)
Lại có:\(\dfrac{a^2}{b^2}=\dfrac{d^2}{c^2}=\dfrac{a^2+d^2}{b^2+c^2}\)
\(\Rightarrow\dfrac{a^2+d^2}{b^2+c^2}=\dfrac{ad}{bc}\)
4)Ta có:\(g\left(x\right)=-x^{101}+x^{100}-x^{99}+...+x^2-x+1\)
\(g\left(x\right)=-x^{101}+\left(x^{100}-x^{99}+...+x^2-x+1\right)\)
\(g\left(x\right)=-x^{101}+f\left(x\right)\)
\(\Rightarrow f\left(x\right)-g\left(x\right)=f\left(x\right)+x^{101}-f\left(x\right)=x^{101}\)
Tại x=0 thì f(x)-g(x)=0
Tại x=1 thì f(x)-g(x)=1
f, \(\dfrac{2^9.4^{10}}{8^8}=\dfrac{2^9.\left(2^2\right)^{10}}{\left(2^3\right)^8}=\dfrac{2^9.2^{20}}{2^{24}}=\dfrac{2^{29}}{2^{24}}=2^5=32\)
a: \(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{14}{25}+\dfrac{11}{25}+\dfrac{2}{7}=\dfrac{2}{7}\)
b: \(=\dfrac{3}{7}-\dfrac{5}{2}-\dfrac{3}{5}+\dfrac{4}{7}+\dfrac{3}{2}-\dfrac{2}{5}=1-1-1=-1\)
c: \(=\dfrac{4}{25}+\dfrac{7}{5}\cdot\dfrac{5}{2}-2=\dfrac{4}{25}+\dfrac{7}{2}-2=\dfrac{83}{50}\)
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
Bài 1:
\(B=16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)-28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)\)
\(B=16\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)-28\dfrac{2}{7}.\left(-\dfrac{5}{3}\right)\)
\(B=\left(-\dfrac{5}{3}\right).\left(16\dfrac{2}{7}-28\dfrac{2}{7}\right)\)
\(B=\left(-\dfrac{5}{3}\right).\left(-2\right)\)
\(B=\dfrac{10}{3}=3\dfrac{1}{3}\)
a.\(12,5.\left(-\dfrac{5}{7}\right)+1,5.\left(-\dfrac{5}{7}\right)\)
\(=\left(-\dfrac{5}{7}\right).\left(12,5+1,5\right)\)
\(=-10\)
b,\(\left(-\dfrac{2}{5}-\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{5}+\dfrac{3}{7}\right):\dfrac{4}{5}\)
\(=\left(-\dfrac{2}{5}-\dfrac{3}{7}-\dfrac{1}{5}+\dfrac{3}{7}\right):\dfrac{4}{5}\)
\(=-\dfrac{3}{5}:\dfrac{4}{5}\)
\(=-\dfrac{3}{4}\)
c,\(12.\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\)
\(=12.\dfrac{4}{9}+\dfrac{4}{3}\)
\(=\dfrac{16}{3}+\dfrac{4}{3}\)
\(=\dfrac{20}{3}\)
d,\(1:\left(\dfrac{2}{3}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{1}{1}:\dfrac{1}{144}\)
\(=144\)
e,\(15.\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)
\(=15.\dfrac{4}{9}-\dfrac{7}{3}\)
\(=\dfrac{20}{3}-\dfrac{7}{3}\)
\(=\dfrac{13}{3}\)
a) = ( 12,5 +1,5 ). \(\left(-\dfrac{5}{7}\right)\)
= 14 . \(\left(-\dfrac{5}{7}\right)\)
= -10
b) = (\(-\dfrac{2}{5}+-\dfrac{1}{5}\)) + \(\left(\dfrac{3}{7}-\dfrac{3}{7}\right)\): \(\dfrac{4}{5}\)
= \(\left(-\dfrac{3}{5}+0\right)\): \(\dfrac{4}{5}\)
= \(\dfrac{3}{4}\)
c) = \(\left(12.-\dfrac{2}{9}\right)\) + \(\dfrac{4}{3}\)
= \(\dfrac{8}{3}\) + \(\dfrac{4}{3}\)
= \(-\dfrac{4}{3}\)
d) = 1: \(\dfrac{23}{48}\)
=\(\dfrac{48}{23}\)
e) =\(\left(15.-\dfrac{2}{9}\right)-\dfrac{7}{3}\)
= \(\left(-\dfrac{10}{3}\right)-\dfrac{7}{3}\)
=\(-\dfrac{17}{3}\)
f) = 10 485.76
a, \(\dfrac{20^5.5^{10}}{100^5}=\dfrac{20^5.5^{10}}{\left(20.5\right)^5}=\dfrac{20^5.5^{10}}{20^5.5^5}=5^5\)
b,\(\dfrac{\left(0,9\right)^5}{\left(0,3\right)^6}=\dfrac{\left(0,3.3\right)^5}{\left(0,3\right)^6}=\dfrac{\left(0,3\right)^5.3^5}{\left(0,3\right)^6}=\dfrac{3^5}{\left(0,3\right)}\)
Bài 1:
a: \(=\dfrac{15-32}{40}\cdot10+\dfrac{1}{4}\)
\(=\dfrac{-17}{4}+\dfrac{1}{4}=-\dfrac{16}{4}=-4\)
b: \(=\left(\dfrac{9}{6}-\dfrac{5}{6}\right)^2+\dfrac{5}{2}+\dfrac{2}{3}\)
\(=\dfrac{4}{9}+\dfrac{5}{2}+\dfrac{2}{3}\)
\(=\dfrac{8}{18}+\dfrac{45}{18}+\dfrac{12}{18}=\dfrac{65}{18}\)