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Bài 2:
a: \(A=11+\dfrac{3}{13}-2-\dfrac{4}{7}-5-\dfrac{3}{13}\)
\(=4-\dfrac{4}{7}=\dfrac{24}{7}\)
b: \(B=6+\dfrac{4}{9}+3+\dfrac{7}{11}-4-\dfrac{4}{9}\)
\(=5+\dfrac{7}{11}=\dfrac{62}{11}\)
c: \(C=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1+\dfrac{5}{7}=1\)
d: \(D=\dfrac{7}{10}\cdot\dfrac{8}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)
\(=\dfrac{20}{10}\cdot7\cdot\dfrac{8}{3}\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}=2\cdot\dfrac{5}{4}=\dfrac{5}{2}\)
What? Lớp 10? Mí bài nỳ dễ mak! Trên lp cs hc mak k giải đc thì thui lun!
1) \(A=1+2+2^2+2^3+......+2^{2015}\)
\(\Leftrightarrow2A=2+2^2+2^3+......+2^{2016}\)
\(\Leftrightarrow2A-A=\left(2+2^2+2^3+......+2^{2016}\right)-\left(1+2+2^2+2^3+......+2^{2015}\right)\)
\(\Leftrightarrow A=2^{2016}-1\)
Vậy \(A=2^{2016}-1\)
6)Ta có: \(13+23+33+43+.......+103=3025\)
\(\Leftrightarrow2.13+2.23+2.33+2.43+.......+2.103=2.3025\)
\(\Leftrightarrow26+46+66+86+.......+206=6050\)
\(\Leftrightarrow\left(23+3\right)+\left(43+3\right)+\left(63+3\right)+\left(83+3\right)+.......+\left(203+3\right)=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+3.10=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+=6050-30\)
\(\Leftrightarrow23+43+63+83+.......+203+=6020\)
Vậy S=6020
b, B có 19 thừa số
=> \(-B=(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})...(1-\frac{1}{400}) \)
<=>\(-B=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)...(20-1)(20+1)}{4.9.16...400} \)
<=>\(-B=\frac{(1.2.3.4...19)(3.4.5...21)}{(2.3.4.5.6...20)(2.3.4.5...20)} \)
<=>\(-B=\frac{21}{20.2} =\frac{21}{40} \)
<=>\(B=\frac{-21}{40} \)
a: ĐKXĐ: \(\left(2x^2-5x+2\right)\left(x^3+1\right)< >0\)
=>(2x-1)(x-2)(x+1)<>0
hay \(x\notin\left\{\dfrac{1}{2};2;-1\right\}\)
b: ĐKXĐ: x+5<>0
=>x<>-5
c: ĐKXĐ: x4-1<>0
hay \(x\notin\left\{1;-1\right\}\)
d: ĐKXĐ: \(x^4+2x^2-3< >0\)
=>\(x\notin\left\{1;-1\right\}\)
a: \(=\left(\dfrac{1}{15}+\dfrac{14}{15}\right)+\left(\dfrac{9}{10}-2-\dfrac{11}{9}\right)+\dfrac{1}{157}\)
\(=1+\dfrac{1}{157}+\dfrac{81-180-110}{90}\)
\(=\dfrac{158}{157}+\dfrac{-209}{90}\simeq-1.315\)
b: \(=\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{2}{6}\)
=1/3-1/3
=0
c: \(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2015\cdot2017}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
=2016/2017
a: \(=-8\cdot\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-8\cdot\dfrac{1}{2}:\dfrac{27-14}{12}\)
\(=-4\cdot\dfrac{12}{13}=\dfrac{-48}{13}\)
b: \(=\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\)
\(=\dfrac{35}{6}:\dfrac{-31}{30}-\dfrac{11}{31}\)
\(=\dfrac{-35}{6}\cdot\dfrac{30}{31}-\dfrac{11}{31}=-6\)
3.
\(\dfrac{2a^2}{b^2}+2\dfrac{b^2}{c^2}+2\dfrac{c^2}{a^2}\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
áp dụng bất đẳng thức cosi
+ \(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\dfrac{a}{c}\)
......
tương tự với 2 cái sau
Câu 1:
Áp dụng BĐT Cauchy:
\(1+x^3+y^3\geq 3\sqrt[3]{x^3y^3}=3xy\)
\(\Rightarrow \frac{\sqrt{1+x^3+y^3}}{xy}\geq \frac{\sqrt{3xy}}{xy}=\sqrt{\frac{3}{xy}}\)
Hoàn toàn tương tự:
\(\frac{\sqrt{1+y^3+z^3}}{yz}\geq \sqrt{\frac{3}{yz}}; \frac{\sqrt{1+z^3+x^3}}{xz}\geq \sqrt{\frac{3}{xz}}\)
Cộng theo vế các BĐT thu được:
\(\text{VT}\geq \sqrt{\frac{3}{xy}}+\sqrt{\frac{3}{yz}}+\sqrt{\frac{3}{xz}}\geq 3\sqrt[6]{\frac{27}{x^2y^2z^2}}=3\sqrt[6]{27}=3\sqrt{3}\) (Cauchy)
Ta có đpcm
Dấu bằng xảy ra khi $x=y=z=1$
Câu 4:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{2}{x}+\frac{3}{y}\right)(x+y)\geq (\sqrt{2}+\sqrt{3})^2\)
\(\Leftrightarrow 1.(x+y)\geq (\sqrt{2}+\sqrt{3})^2\Rightarrow x+y\geq 5+2\sqrt{6}\)
Vậy \(A_{\min}=5+2\sqrt{6}\)
Dấu bằng xảy ra khi \(x=2+\sqrt{6}; y=3+\sqrt{6}\)
------------------------------
Áp dụng BĐT Cauchy:
\(\frac{ab}{a^2+b^2}+\frac{a^2+b^2}{4ab}\geq 2\sqrt{\frac{ab}{a^2+b^2}.\frac{a^2+b^2}{4ab}}=1\)
\(a^2+b^2\geq 2ab\Rightarrow \frac{3(a^2+b^2)}{4ab}\geq \frac{6ab}{4ab}=\frac{3}{2}\)
Cộng theo vế hai BĐT trên:
\(\Rightarrow B\geq 1+\frac{3}{2}=\frac{5}{2}\) hay \(B_{\min}=\frac{5}{2}\). Dấu bằng xảy ra khi $a=b$
a) \(\left\{{}\begin{matrix}5x+3y=-7\\2x-4y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+3y=-7\\x-2y=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}5x+3y=-7\\x=3+2y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5.\left(3+2y\right)+3y=-7\\x=3+2y\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13y=-22\\x=3+2y\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-22}{13}\\x=3+2.\dfrac{-22}{13}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-22}{13}\\x=\dfrac{-5}{13}\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm là: \(\left\{{}\begin{matrix}y=\dfrac{-22}{13}\\x=\dfrac{-5}{13}\end{matrix}\right.\).
b)\(\left\{{}\begin{matrix}7x+14y=17\\2x+4y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}14x+28y=34\\14x+28y=35\end{matrix}\right.\) (vô nghiệm)
Vậy hệ phương trình vô nghiệm.
2/3+(-2/3)=3/5+(3/-5)=0
i: 2/5-1/10=4/10-1/10=3/10
2/5+(-1/10)=4/10-1/10=3/10
5/6-2/3=5/6-4/6=1/6
5/6+(-2/3)=1/6