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1: \(a^2-4b^2=\left(a-2b\right)\left(a+2b\right)\)
2: \(16a^2-b^4=\left(4a-b^2\right)\left(4a+b^2\right)\)
3: \(4a^4-81b^4=\left(2a^2-9b^2\right)\left(2a^2+9b^2\right)\)
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\(D=\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+...+\left(1-\frac{1}{2015.2016}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2015.2016}\right)\)
\(=2015-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
\(=2015-\left(1-\frac{1}{2016}\right)\)
\(=2015-\frac{2015}{2016}\)
TO LẮM
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13+23+33+...+1003
=1+2+1.2.3+3+2.3.4+100+99.100.101
=(1+2+3+...+100)+(1.2.3+2.3.4+...+99.100.101)
=5050+101989800
=101994850
NHỚ T.I.C.K và KB với mk nha
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Tính:
a) \(\dfrac{5^{15}.7^{14}}{5^{16}.7^{16}}\)=\(\dfrac{1}{5.7^2}\)=\(\dfrac{1}{5.49}\)=\(\dfrac{1}{245}\)
b)\(\dfrac{3^{12}.7^{13}}{3^{10}.7^{11}}\)=\(\dfrac{3^2.7^2}{1}\)=\(\dfrac{9.49}{1}\)=\(\dfrac{441}{1}\)
c)\(\dfrac{8^{10}.3^{15}}{9^8.2^{83}}\)=\(\dfrac{\left(2^3\right)^{10}.3^{15}}{\left(3^2\right)^8.2^{83}}\)=\(\dfrac{2^{30}.3^{15}}{3^{16}.2^{83}}\)=\(\dfrac{1}{3.2^{53}}\)= bài c) hình như bị lỗi rồi
não mình úng nước rồi, không biết làm d), e) và f)
Tìm x:
a) x - \(\dfrac{1}{2}\)=\(\dfrac{3}{4}\)
x =\(\dfrac{3}{4}\)+\(\dfrac{1}{2}\)
x = \(\dfrac{3}{4}+\dfrac{2}{4}\)
x = \(\dfrac{5}{4}\)
b)\(\dfrac{1}{7}\) - x = \(-\dfrac{5}{6}\)
- x = \(-\dfrac{5}{6}\)\(-\dfrac{1}{7}\)
- x = \(-\dfrac{35}{42}-\dfrac{6}{42}\)
- x = \(-\dfrac{41}{42}\)
x = \(\dfrac{41}{42}\)
c) - x + \(\dfrac{1}{2}\) = \(-\dfrac{10}{11}\)
-x = \(-\dfrac{10}{11}-\dfrac{1}{2}\)
- x = \(-\dfrac{20}{22}-\dfrac{11}{22}\)
- x = \(-\dfrac{31}{22}\)
x = \(\dfrac{31}{22}\)
d) - x \(-\dfrac{3}{4}\) = \(-\dfrac{7}{5}\)
- x = \(-\dfrac{7}{5}+\dfrac{3}{4}\)
- x = \(-\dfrac{28}{20}+\dfrac{15}{20}\)
- x = \(-\dfrac{13}{20}\)
x = \(\dfrac{13}{20}\)
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cái này giờ mình phải đi học nên chưa giải hộ bạn được
đợi tối mình giải cho
Bài 15:
a, Không phải. Vì 2+3<6.
b, Không phải. Vì 2+4=6.
c,Phải. Vì 3+4=6.
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c. \(\left|\dfrac{8}{4}-\left|x-\dfrac{1}{4}\right|\right|-\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{8}{4}-x+\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{8}{4}+x-\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{9}{4}-x\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{7}{4}+x\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\dfrac{9}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\\x=\dfrac{9}{4}-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}\dfrac{7}{4}+x-\dfrac{1}{2}=\dfrac{3}{4}\\-\dfrac{7}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\\x=-3\end{matrix}\right.\)
Ở nơi x=9/4-1/2 là x-9/4-1/2 nha
a. -1,5 + 2x = 2,5
<=> 2x = 2,5 + 1,5
<=> 2x = 4
<=> x = 2
b. \(\dfrac{3}{2}\left(x+5\right)-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{3}{2}x+\dfrac{15}{2}-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{9x}{6}+\dfrac{45}{6}-\dfrac{3}{6}=\dfrac{8}{6}\)
<=> 9x + 45 - 3 = 8
<=> 9x = 8 + 3 - 45
<=> 9x = -34
<=> x = \(\dfrac{-34}{9}\)