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\(a,4\frac{5}{9}:\frac{\left(-5\right)}{7}+\frac{4}{9}:\frac{-5}{7}\)
\(=\frac{41}{9}.\frac{-7}{5}+\frac{4}{9}.\frac{-7}{5}\)
\(=\frac{-7}{5}.\left(\frac{41}{9}+\frac{4}{9}\right)\)
\(=-\frac{7}{9}.5\)
\(=-7\)
a)Bn Kaito Kid làm rùi!
B)Không viết lại đề
\(=\frac{11}{7}\cdot\left(-\frac{3}{5}+\frac{4}{9}-\frac{2}{5}+\frac{5}{9}\right)=\frac{11}{7}\cdot0=0\)
c)Không viết lại đề
\(A=\left(2+4+...+100\right)\left(\frac{3}{5}\cdot\frac{10}{7}-\frac{6}{7}\right):\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(2+4+6+...+100\right)\cdot0\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)=0\)
\(=\frac{7}{6}\cdot\left(\frac{3}{26}-\frac{3}{13}+\frac{1}{10}-\frac{8}{5}\right)=\frac{7}{6}\left(\frac{-3}{26}+\frac{-17}{10}\right)=\frac{7}{6}\cdot\frac{236}{130}=\frac{413}{195}\)
D)
\(\dfrac{7}{8}\): \(\left(\dfrac{2}{9}-\dfrac{1}{18}\right)\)+\(\dfrac{7}{8}\): \(\left(\dfrac{1}{36}-\dfrac{5}{12}\right)\)
= \(\dfrac{7}{8}\): \(\dfrac{1}{6}\)+\(\dfrac{7}{8}\): \(\left(\dfrac{-14}{36}\right)\)
= \(\dfrac{7}{8}\): \(\left\{\dfrac{1}{6}+\left(\dfrac{-14}{36}\right)\right\}\)
= \(\dfrac{7}{8}\): \(\left(\dfrac{-2}{9}\right)\)
= \(\dfrac{7}{8}\). \(\left(\dfrac{-9}{2}\right)\)
= \(\dfrac{-63}{16}\)
Giải:
\(\dfrac{7}{8}:\left(\dfrac{2}{9}-\dfrac{1}{18}\right)+\dfrac{7}{8}:\left(\dfrac{1}{36}-\dfrac{5}{12}\right)\)
\(=\dfrac{7}{8}:\left(\dfrac{4}{18}-\dfrac{1}{18}\right)+\dfrac{7}{8}:\left(\dfrac{1}{36}-\dfrac{15}{36}\right)\)
\(=\dfrac{7}{8}:\dfrac{3}{18}+\dfrac{7}{8}:\left(-\dfrac{7}{18}\right)\)
\(=\dfrac{7}{8}.6+\dfrac{7}{8}.\dfrac{18}{-7}\)
\(=\dfrac{7}{8}\left(6+\dfrac{18}{-7}\right)\)
\(=\dfrac{7}{8}.\dfrac{24}{7}\)
\(=3\)
Vậy giá trị của biểu thức trên là 3.
Chúc bạn học tốt!!!
\(5^{2016}-5^{2015}-....-5-1\)
\(=5^{2016}-\left(5^{2015}+5^{2014}+....+5+1\right)\)
\(=5^{2016}-\left(5^{2016}-1\right)\)
\(=5^{2016}-5^{2016}+1\) \(=0+1=1\)
<br class="Apple-interchange-newline"><div id="inner-editor"></div>5M=5(52016−52015−52014−...−5−1)
=>5M=52017−52016−52015−...−52−5
=>5M−M=(52017−52016−52015−...−52−5)−(52016−52015−52014−...−5−1)
=>
\(\frac{8^5.\left(-5\right)^8+\left(-2\right)^5.10^9}{2^{16}.5^7+20^8}\)
\(=\frac{\left(2^3\right)^5.5^8+\left(-2\right)^5.\left(2.5\right)^9}{2^{16}.5^7+\left(2^2.5\right)^8}\)
\(=\frac{2^{15}.5^8+\left(-2\right)^5.2^9.5^9}{2^{16}.5^7+2^{16}.5^8}\)
\(=\frac{2^{15}.5^8-2^{14}.5^9}{2^{16}.5^7\left(1+5\right)}\)
\(=\frac{2^{14}.5^8\left(2-5\right)}{2^{16}.5^7.\left(1+5\right)}\)
\(=\frac{2^{14}.5^8.\left(-3\right)}{2^{16}.5^7.6}\)
\(=\frac{-5}{8}\)
Bài này bạn xét 2 trường hợp:
TH1: \(x-\frac{8}{7}\ge0
\Rightarrow
x\ge\frac{8}{7}\)
Khi đó:
\(\frac{4}{7}< x-\frac{8}{7}< \frac{5}{7}\)
\(\Leftrightarrow\frac{4}{7}+\frac{8}{7}< x-\frac{8}{7}+\frac{8}{7}< \frac{5}{7}+\frac{8}{7}\) (Cộng 8/7 vào mỗi vế)
\(\Leftrightarrow\frac{12}{7}< x< \frac{13}{7}\) (thỏa mãn điều kiện x > 8/7)
TH2: \(x-\frac{8}{7}\le0
\Rightarrow
x\le\frac{8}{7}\)
Khi đó:
\(\frac{4}{7}< \frac{8}{7}-x< \frac{5}{7}
\)
\(\frac{4}{7}-\frac{8}{7}< -x< \frac{5}{7}-\frac{8}{7}\)
\(-\frac{4}{7}< -x< -\frac{3}{7}\)
\(\frac{3}{7}< x< \frac{4}{7}\) (thỏa mãn x < 8/7) (*bất đẳng thức đổi chiều*)
Vậy: ......
Theo mik nghĩ thì bài này nên dành cho h/s lớp 8, vì lớp 7 chưa học bất đẳng thức đổi chiều...
\(\frac{4}{7}< \left|x-\frac{8}{7}\right|< \frac{5}{7}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{4}{7}< x-\frac{8}{7}< \frac{5}{7}\\\frac{-4}{7}>\frac{8}{7}>\frac{-5}{7}\end{cases}}\)
\(TH1:\)\(\orbr{\frac{4}{7}< x-\frac{8}{7}< \frac{5}{7}\Leftrightarrow\frac{12}{7}< x< \frac{13}{7}}\)
\(TH2:\)\(\orbr{\frac{-4}{7}>x-\frac{8}{7}>\frac{-5}{7}\Leftrightarrow\frac{4}{7}>x>\frac{3}{7}}\)