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TD
1
26 tháng 7 2023
B=x5-15x4+16x3-29x2+13x
B= 145-15.144+16.143-29.142+13.14
B=14.144-15.144+16.143-29.142+13.14
B=(14-15).144+16.143-29.142+13.14
B= (-1).144+16.143-29.142+13.14
B= (-1).144+16.142.14-29.142+13.14
B=(-1).144+224.142-29.142+13.14
B= (-1).144+(224-29).142+13.14
B=(-1).144+195.142+13.14
B=[(-1).143].14+195.14.14+13.14
B= (-2744).14+2730.14+13.14
B= 14.[(-2744)+2730+13]
B= 14.(-1)
B= -14
4T
1
NH
5 tháng 3 2023
1)
`7x^2 -49x=0`
`<=>x(7x-49)=0`
\(< =>\left[{}\begin{matrix}x=0\\7x-49=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)
2)
`8x^2 -16x=0`
`<=>x(8x-16)=0`
\(< =>\left[{}\begin{matrix}x=0\\8x-16=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
3)
`2x^3 +40x=0`
`<=>x(2x^2 +40)=0`
`<=>x=0` hoặc`2x^2 +40=0`
`<=>x=0` hoặc `2x^2 =-40` (vô lí vì `2x^2` luôn lớn hơn hoặc bằng 0)
`<=>x=0`
4)
`-x^3 +16x=0`
`<=>x^3 -16x=0`
`<=>x(x^2 -16)=0`
\(< =>\left[{}\begin{matrix}x=0\\x^2-16=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x^2=16\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
EJ
14 tháng 8 2017
1,\(2:\left(\dfrac{1}{2}-\dfrac{2}{3}\right)^3\) = \(2:\left(\dfrac{-1}{216}\right)\)=\(-432\)
2, \(a,2^6.5^6.5^4.4^4=10^6.20^4\)
\(b,\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}\)
NT
1
29 tháng 7 2017
\(B=1+4^2+4^4+4^6+...+4^{100}\)
\(4B=4+4^3+4^5+...+4^{101}\)
\(4B+B=\left(1+4^2+4^4+...+4^{100}\right)+\left(4+4^3+4^5+...+4^{101}\right)\)
\(5B=1+4+4^2+4^3+...+4^{100}+4^{101}\)
\(=1+4+4.4+4^2.4+...+4^{100}.4\)
\(=1+4\left(1+4+4^2+...+4^{100}\right)\)
\(=1+4\left(4^{101}-1\right)\)
\(=1+4^{102}-4\)
\(=4^{102}-3\)
\(B=\frac{4^{102}-3}{5}\)
T i c k cho mình nha ^^
4 tháng 8 2017
\(8^{30}+8^{31}+8^{32}\)
\(=8^{30}.1+8^{30}.8+8^{30}.8^2\)
\(=8^{30}.1+8^{30}.8+8^{30}.64\)
\(=8^{30}\left(1+8+64\right)\)
\(=8^{30}.73\)
\(=\left(2^3\right)^{30}.73\)
\(=2^{90}.73\)
\(=2^{89}.146⋮146\rightarrowđpcm\)
\(4^{25}+4^{26}+4^{27}+4^{28}+4^{29}+4^{30}\)
\(=4^{25}.1+4^{25}.4+4^{25}.4^2+4^{25}.4^3+4^{25}.4^4+4^{25}.4^5\)
\(=4^{25}.1+4^{25}.4+4^{25}.16+4^{25}.64+4^{25}.256+4^{25}.1024\)
\(=4^{25}\left(1+4+16+64+256+1024\right)\)
\(=4^{25}.1365\)
\(=4^{25}.195.7⋮7\rightarrowđpcm\)
4 tháng 8 2019
a) \(\frac{1}{4}+\frac{3}{4}x=\frac{3}{4}\Leftrightarrow\frac{3}{4}x=\frac{1}{2}\Leftrightarrow x=\frac{1}{2}\times\frac{4}{3}\Leftrightarrow x=\frac{2}{3}\)
b)\(1\frac{3}{4}x+1\frac{1}{2}=-\frac{4}{5}\Leftrightarrow\frac{7}{4}x+\frac{3}{2}=-\frac{4}{5}\Leftrightarrow\frac{7}{4}x=-\frac{23}{10}\)
\(\Leftrightarrow x=-\frac{23}{10}\times\frac{4}{7}\Leftrightarrow x=-\frac{46}{35}\)
c)\(\frac{3}{4}x+\frac{2}{5}x=1,2\Leftrightarrow x\left(\frac{3}{4}+\frac{2}{5}\right)=1,2\Leftrightarrow\frac{23}{20}x=1,2\)
\(\Leftrightarrow x=1,2\times\frac{20}{23}\Leftrightarrow x=\frac{24}{23}\)
d)\(\frac{3}{7}+\frac{1}{7}:x=\frac{3}{14}\Leftrightarrow\frac{1}{7x}=\frac{3}{14}-\frac{3}{7}\Leftrightarrow\frac{1}{7x}=-\frac{3}{14}\Leftrightarrow14=-3\times7x\)
\(\Leftrightarrow-21x=14\Leftrightarrow x=-\frac{2}{3}\)
e) \(-\frac{3}{4}-\left|\frac{4}{5}-x\right|=-1\Leftrightarrow\left|\frac{4}{5}-x\right|=-\frac{3}{4}+1\Leftrightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{11}{20}\\x=\frac{21}{20}\end{matrix}\right.\)
4 tháng 8 2019
a, \(\frac{1}{4}+\frac{3}{4}x=\frac{3}{4}\\ \Rightarrow\frac{3}{4}x=\frac{1}{2}\\ \Rightarrow x=\frac{2}{3}\)
Vậy \(x=\frac{2}{3}\)
b, \(1\frac{3}{4}x+1\frac{1}{2}=\frac{-4}{5}\\ \frac{7}{4}x+\frac{3}{2}=\frac{-4}{5}\\ \Rightarrow\frac{7}{4}x=\frac{-23}{10}\\ \Rightarrow x=\frac{-46}{35}\)
Vậy \(x=\frac{-46}{35}\)
c, \(\frac{3}{4}x+\frac{2}{5}x=1,2\\ x\left(\frac{3}{4}+\frac{2}{5}\right)=\frac{6}{5}\\ x\cdot\frac{23}{20}=\frac{6}{5}\\ \Rightarrow x=\frac{24}{23}\)
Vậy \(x=\frac{24}{23}\)
d, \(\frac{3}{7}+\frac{1}{7}:x=\frac{3}{14}\\ \Rightarrow\frac{1}{7}:x=\frac{-3}{14}\\ \Rightarrow x=\frac{-2}{3}\)
Vậy \(x=\frac{-2}{3}\)
e, \(\frac{-3}{4}-\left|\frac{4}{5}-x\right|=-1\\ \Rightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\\ \Rightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=\frac{-1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{11}{20}\\x=\frac{21}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{11}{20};\frac{21}{20}\right\}\)
G
1
PA
31 tháng 5 2016
1.
a.
\(\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{7}\right)\)
\(=\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\)
\(=\frac{35-21-15}{105}\)
\(=-\frac{1}{105}\)
b.
\(\frac{3}{5}-\left(\frac{3}{4}-\frac{1}{2}\right)\)
\(=\frac{3}{5}-\frac{3}{4}+\frac{1}{2}\)
\(=\frac{12-15+10}{20}\)
\(=\frac{7}{20}\)
c.
\(\frac{4}{7}-\left(\frac{2}{5}+\frac{1}{3}\right)\)
\(=\frac{4}{7}-\frac{2}{5}-\frac{1}{3}\)
\(=\frac{60-42-35}{105}\)
\(=-\frac{17}{105}\)
2.
a.
\(S=-\frac{1}{1\times2}-\frac{1}{2\times3}-\frac{1}{3\times4}-...-\frac{1}{\left(n-1\right)\times n}\)
\(S=-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{\left(n-1\right)\times n}\right)\)
\(S=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(S=-\left(1-\frac{1}{n}\right)\)
\(S=-1+\frac{1}{n}\)
b.
\(S=-\frac{4}{1\times5}-\frac{4}{5\times9}-\frac{4}{9\times13}-...-\frac{4}{\left(n-4\right)\times n}\)
\(S=-\left(\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{\left(n-4\right)\times n}\right)\)
\(S=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)
\(S=-\left(1-\frac{1}{n}\right)\)
\(S=-1+\frac{1}{n}\)
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DP
1
4 - 16 x 3
= 4 - 48
= -42
bằng -44 nha bạn