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Bạn nhân a+b+c và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)lại với nhau rồi trừ 1 ở mỗi vế, phân tích mẫu ra sẽ đc(a+b)(b+c)(c+a)=0
2) \(\sum\dfrac{x}{x^2-yz+2013}=\sum\dfrac{x^2}{x^3-xyz+2013x}\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\dfrac{1}{x+y+z}\left(đpcm\right)\)
bn dua vao day nay :https://olm.vn/hoi-dap/detail/105816822455.html
Bài 3:
a) \(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)
\(=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}\right):\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a+1}\right)}\right)\)
\(=\dfrac{a-1}{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}-1+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}}:\dfrac{1}{\sqrt{a}-1}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}}\cdot\left(\sqrt{a}-1\right)\)
\(=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
\(=\dfrac{a-1}{\sqrt{a}}\)
b) Thay \(a=3+2\sqrt{2}\) vào biểu thức A:
Ta có: \(\dfrac{3+2\sqrt{2}-1}{\sqrt{3+2\sqrt{2}}}=\dfrac{2+2\sqrt{2}}{\sqrt{\left(1+2\sqrt{2}\right)^2}}=\dfrac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}=2\)
Vậy giá trị biểu thức A tại \(a=3+2\sqrt{2}\)
Bài 1:
Sửa đề: (theo mình là như vậy)
\(x^4-4x^2-12x-9\)
\(=x^4+x^3-x^3-x^2-3x^2-3x-9x-9\)
\(=\left(x^4+x^3\right)-\left(x^3+x^2\right)-\left(3x^2+3x\right)-\left(9x+9\right)\)
\(=x^3.\left(x+1\right)-x^2.\left(x+1\right)-3x.\left(x+1\right)-9.\left(x+1\right)\)
\(=\left(x+1\right).\left(x^3-x^2-3x-9\right)\)
\(=\left(x+1\right).\left(x^3-3x^2+2x-6x+3x-9\right)\)
\(=\left(x+1\right).\left[\left(x^3-3x^2\right)+\left(2x-6x\right)+\left(3x-9\right)\right]\)
\(=\left(x+1\right).\left[x^2.\left(x-3\right)+2x.\left(x-3\right)+3.\left(x-3\right)\right]\)
\(=\left(x+1\right).\left(x-3\right).\left(x^2+2x+3\right)\)
Chúc bạn học tốt!!!
Có: \(\dfrac{a+1}{1+b^2}=\dfrac{\left(1+b^2\right).\left(a+1\right)-b^2\left(a+1\right)}{1+b^2}=a+1-\dfrac{b^2\left(a+1\right)}{1+b^2}\)
Áp dụng bất đẳng thức Cauchy cho 2 số dương 1 và b2 ta được
\(1+b^2\ge2b\Rightarrow-\dfrac{b^2\left(a+1\right)}{1+b^2}\ge-\dfrac{b^2\left(a+1\right)}{2b}=-\dfrac{ab+b}{2}\)
\(\Rightarrow\dfrac{a+1}{1+b^2}\ge a+1-\dfrac{ab+b}{2}\)
CMTT\(\Rightarrow\dfrac{b+1}{1+c^2}\ge b+1-\dfrac{bc+c}{2};\dfrac{c+1}{1+a^2}\ge c+1-\dfrac{ac+a}{2}\)
\(\Rightarrow A\ge\left(a+b+c\right)+3-\dfrac{\left(ab+bc+ac\right)+\left(a+b+c\right)}{2}\)
Ta có \(ab+bc+ca\le\dfrac{1}{3}\left(a+b+c\right)^2\)
\(\Rightarrow ab+ac+bc\le\dfrac{1}{3}.3^2=3\)
\(\Rightarrow A\ge3+3-\dfrac{3+3}{2}=3\)(đpcm)
Chả biết đúng hay sai,làm đại.:v
Dự đoán dấu "=" xảy ra tại a = b = c = 1
Với dự đoán đó,
Xét \(\dfrac{a+1}{1+b^2}=2-\dfrac{a+1}{1+b^2}\ge2-\dfrac{a+1}{2b}\)
Tương tự: \(\dfrac{b+1}{1+c^2}\ge2-\dfrac{b+1}{2c};\dfrac{c+1}{1+a^2}\ge2-\dfrac{c+1}{2a}\)
Cộng theo vế 3BĐT,ta có: \(VT\ge2+2+2-\dfrac{a+1}{2b}+\dfrac{b+1}{2c}+\dfrac{c+1}{2a}\)
\(=6-\dfrac{a+1}{2b}+\dfrac{b+1}{2c}+\dfrac{c+1}{2a}\)
\(\ge6-\dfrac{2b}{2b}+\dfrac{2c}{2c}+\dfrac{2a}{2a}=3^{\left(đpcm\right)}\) (do dự đoán a = b = c = 1 nên \(a+1\le2b\))
Vậy điều ta dự đoán là đúng.
Dấu "=" xảy ra khi a=b=c=1
Đặt \(\left\{{}\begin{matrix}x^{671}=a\\y^{671}=b\end{matrix}\right.\). Bài toán trở thành
Cho \(a+b=0,67\) và \(a^2+b^2=1,34\). Tính \(A=a^3+b^3\)
Giải:
\(a^2+2ab+b^2=0,4489\)
\(\Rightarrow ab=\dfrac{0,4489-1,34}{2}=-0,44555\)
\(A=a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=1,1963185\)
\(4B=\dfrac{4}{\sqrt{5}+1}+\dfrac{4}{\sqrt{6}+\sqrt{2}}+...+\dfrac{4}{\sqrt{2014}+\sqrt{2010}}\)
\(=\dfrac{4\left(\sqrt{5}-1\right)}{5-1}+\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{6-2}+...+\dfrac{4\left(\sqrt{2014}-\sqrt{2010}\right)}{2014-2010}\)
\(=\sqrt{5}-1+\sqrt{6}-\sqrt{2}+...+\sqrt{2014}-\sqrt{2010}\)
\(=-1-\sqrt{2}-\sqrt{3}-\sqrt{4}+\sqrt{2011}+\sqrt{2012}+\sqrt{2013}+\sqrt{2014}\)
\(\Rightarrow B=...\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^3(b+c)}+\frac{a(b+c)}{4}\geq 2\sqrt{\frac{1}{a^3(b+c)}.\frac{a(b+c)}{4}}=2\sqrt{\frac{1}{4a^2}}=\frac{1}{a}=\frac{abc}{a}=bc\)
Tương tự:
\(\frac{1}{b^3(c+a)}+\frac{b(c+a)}{4}\geq \frac{1}{b}=ac\)
\(\frac{1}{c^3(a+b)}+\frac{c(a+b)}{4}\geq \frac{1}{c}=ab\)
Cộng theo vế:
\(\Rightarrow \text{VT}+\frac{ab+bc+ac}{2}\geq ab+bc+ac\)
\(\Rightarrow \text{VT}\geq \frac{ab+bc+ac}{2}\)
Tiếp tục áp dụng AM-GM: \(ab+bc+ac\geq 3\sqrt[3]{a^2b^2c^2}=3\)
\(\Rightarrow \text{VT}\ge \frac{3}{2}\) (đpcm)
Dấu bằng xảy ra khi $a=b=c=1$
Lời giải:
Đặt vế trái là $A$
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}\right)(a+b+b+c+c+c)\geq (1+1+1+1+1+1)^2\)
\(\Leftrightarrow \frac{1}{a}+\frac{2}{b}+\frac{3}{c}\geq \frac{36}{a+2b+3c}\)
Hoàn toàn TT:
\(\frac{1}{b}+\frac{2}{c}+\frac{3}{a}\geq \frac{36}{b+2c+3a}\)
\(\frac{1}{c}+\frac{2}{a}+\frac{3}{b}\geq \frac{36}{c+2a+3b}\)
Cộng theo vế:
\(\Rightarrow 6\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 36A\)
\(\Rightarrow A\leq \frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Theo đkđb: \(ab+bc+ac=abc\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
Do đó: \(A\leq \frac{1}{6}< \frac{3}{16}\) (đpcm)
ĐKXĐ: \(a,b,c\ne0\)
\(\left(a+b+c\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=2013.\dfrac{1}{2013}\)
\(\Leftrightarrow1+1+1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}=1\)
\(\Leftrightarrow\dfrac{a^2c+a^2b+b^2c+ab^2+bc^2+ac^2+2abc}{abc}=0\)
\(\Leftrightarrow a^2c+a^2b+b^2c+ab^2+bc^2+ac^2+2abc=0\)
\(\Leftrightarrow ac\left(a+b\right)+ab\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)
Mà \(a+b+c=2013\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2013\\b=2013\\c=2013\end{matrix}\right.\)(đpcm)