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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: \(D=\dfrac{7}{10}\cdot\dfrac{8}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)
\(=\dfrac{20}{10}\cdot7\cdot\dfrac{8}{3}\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}=2\cdot\dfrac{5}{4}=\dfrac{5}{2}\)
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) A = {\(\dfrac{1}{n\left(n+1\right)}\)| \(n\in\mathbb{N},1\le n\le5\)}
b) B = {\(\dfrac{1}{n^2-1}\)|\(n\in\mathbb{N},2\le n\le6\)\(\)}
câu 1: \(VT=\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
Bài 3:
Ta có: \(a^2+b^2+c^2=3\ge ab+bc+ca\) ( tự cm bđt nha )
Áp dụng bất đẳng thức Schwarz ta có:
\(\dfrac{a^3}{b+c}+\dfrac{b^3}{c+a}+\dfrac{c^3}{a+b}=\dfrac{a^4}{ab+bc}+\dfrac{b^4}{bc+ab}+\dfrac{c^4}{ac+bc}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
Dấu " = " khi a = b = c = 1
Bài 4:
Ta có: \(\dfrac{a^3}{a^2+b^2}=\dfrac{a\left(a^2+b^2\right)-ab^2}{a^2+b^2}=a-\dfrac{ab^2}{a^2+b^2}\ge a-\dfrac{ab^2}{2ab}=a-\dfrac{b}{2}\)
( BĐT AM - GM )
Tương tự \(\Rightarrow\dfrac{b^3}{c^2+a^2}\ge b-\dfrac{c}{2}\)
\(\dfrac{c^3}{c^2+a^2}\ge c-\dfrac{a}{2}\)
\(\Rightarrow VT\ge\left(a+b+c\right)-\dfrac{1}{2}\left(a+b+c\right)=\dfrac{a+b+c}{2}\)
Dấu " = " khi a = b = c
Tiếp sức cho Tú đệ
Bài 1: \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(\ge\left(a+b\right)\left(2ab-ab\right)=ab\left(a+b\right)\)
\(\Rightarrow\dfrac{a^3+b^3}{ab}\ge\dfrac{ab\left(a+b\right)}{ab}=a+b\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT\ge VP."="\Leftrightarrow a=b=c\)
Bài 2: Holder:
\(\left(\dfrac{a^4}{bc^2}+\dfrac{b^4}{ca^2}+\dfrac{c^4}{ab^2}\right)\left(\dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c}\right)\left(c+a+b\right)\ge\left(a+b+c\right)^3\)
Cần chứng minh \(\dfrac{bc}{a}+\dfrac{ca}{b}+\dfrac{ab}{c}\ge a+b+c\)
AM-GM: \(\dfrac{bc}{a}+\dfrac{ca}{b}\ge2\sqrt{\dfrac{bc}{a}\cdot\dfrac{ca}{b}}=2c\)
Tương tự rồi cộng theo vế:
\("=" \Leftrightarrow a=b=c\)
Câu 2:
\(A=2\cdot\dfrac{1}{2}+3\cdot\dfrac{1}{2}+1=1+1+1=3\)
Bài 3:
\(cos^2a=1-\left(\dfrac{12}{13}\right)^2=\dfrac{25}{169}\)
mà cosa>0
nên cosa=5/13
=>tan a=12/5; cot a=5/12
Câu 4: \(sin^2a=1-\dfrac{1}{4}=\dfrac{3}{4}\)
mà sina <0
nên sin a=-căn 3/2
=>tan a=-căn 3
\(A=-\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\cdot\left(-\sqrt{3}\right)=-\sqrt{3}\)
a: \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{2}{x}-\dfrac{8}{y}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y}=11\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\\dfrac{1}{x}=-3+\dfrac{4}{y}=-3+4=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{36}{x-3}-\dfrac{15}{y+2}=189\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{44}{x-3}=176\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=\dfrac{1}{4}\\\dfrac{15}{y+2}=-13-\dfrac{8}{x-3}=-13-32=-45\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=-\dfrac{1}{3}-2=-\dfrac{7}{3}\end{matrix}\right.\)
Từng sau em hạn chế đăng nhiều bài cùng một lúc như thế này nhé.
Bài 1:
Ta có: \(a+\frac{4}{(a-b)(b+1)^2}=(a-b)+\frac{b+1}{2}+\frac{b+1}{2}+\frac{4}{(a-b)(b+1)^2}-1\)
Áp dụng BĐT AM-GM cho các số không âm ta có:
\((a-b)+\frac{b+1}{2}+\frac{b+1}{2}+\frac{4}{(a-b)(b+1)^2}\geq 4\sqrt[4]{\frac{4(a-b)(b+1)^2}{4(a-b)(b+1)^2}}=4\)
\(\Rightarrow a+\frac{4}{(a-b)(b+1)^2}=(a-b)+\frac{b+1}{2}+\frac{b+1}{2}+\frac{4}{(a-b)(b+1)^2}-1\geq 4-1\)
\(\Leftrightarrow a+\frac{4}{(a-b)(b+1)^2}\geq 3\)
Ta có đpcm.
Dấu bằng xảy ra khi \(a-b=\frac{b+1}{2}=\frac{4}{(a-b)(b+1)^2}\)
\(\Leftrightarrow a=2; b=1\)
Bài 2:
Đặt \(\left(\frac{a}{b}, \frac{b}{c}, \frac{c}{a}\right)\mapsto (x,y,z)\Rightarrow xyz=1\)
BĐT cần chứng minh tương đương với:
\(x^2+y^2+z^2\geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\Leftrightarrow x^2+y^2+z^2\geq \frac{xy+yz+xz}{xyz}=xy+yz+xz(*)\)
Áp dụng BĐT AM-GM:
\(x^2+y^2\geq 2\sqrt{x^2y^2}=2xy\)
\(y^2+z^2\geq 2\sqrt{y^2z^2}=2yz\)
\(z^2+x^2\geq 2\sqrt{z^2x^2}=2zx\)
Cộng theo vế: \(\Rightarrow 2(x^2+y^2+z^2)\geq 2(xy+yz+xz)\)
\(\Leftrightarrow x^2+y^2+z^2\geq xy+yz+xz\)
Do đó (*) đúng, ta có đpcm.
Dấu bằng xảy ra khi \(x=y=z=1\Leftrightarrow a=b=c\)
Bài 3:
Ta có: \(\text{VT}=(\frac{b}{\sqrt{a}}+\frac{c}{\sqrt{b}}+\frac{a}{\sqrt{c}})+(\frac{c}{\sqrt{a}}+\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}})\)
Áp dụng BĐT Bunhiacopxky:
\((\frac{b}{\sqrt{a}}+\frac{c}{\sqrt{b}}+\frac{a}{\sqrt{c}})(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq (\sqrt{b}+\sqrt{c}+\sqrt{a})^2\)
\(\frac{b}{\sqrt{a}}+\frac{c}{\sqrt{b}}+\frac{a}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}(1)\)
Áp dụng BĐT AM-GM:
\(\frac{c}{\sqrt{a}}+\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}\geq 3\sqrt[3]{\frac{abc}{\sqrt{abc}}}=3(2)\) do $abc=1$
Từ \((1); (2)\Rightarrow \text{VT}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=1\)
a: \(=\left(\dfrac{-48}{12}+\dfrac{-8}{12}+\dfrac{21}{12}\right)\cdot\dfrac{-12}{13}\)
\(=\dfrac{-35}{12}\cdot\dfrac{-12}{13}=\dfrac{35}{13}\)
b: \(=\dfrac{-3}{6}+\dfrac{5}{6}-\dfrac{312}{100}+\dfrac{51}{10}\)
\(=\dfrac{1}{3}-\dfrac{312}{100}+\dfrac{51}{10}=\dfrac{347}{150}\)
c: \(=\left(\dfrac{48}{300}+\dfrac{175}{300}-\dfrac{135}{100}\right)\cdot\dfrac{5}{2}+\dfrac{1}{4}\)
\(=\dfrac{88}{300}\cdot\dfrac{5}{2}+\dfrac{1}{4}=\dfrac{59}{60}\)