Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có BĐT:
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
\(\Leftrightarrow6\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)+2016\le6\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2016\)
\(\Leftrightarrow7.\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\le6\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2016\)
\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\le2016\)
Xét \(P=\frac{1}{\sqrt{3\left(2x^2+y^2\right)}}+\frac{1}{\sqrt{3\left(2y^2+z^2\right)}}+\frac{1}{\sqrt{3\left(2z^2+x^2\right)}}\)
\(P^2=\left(\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2x^2+y^2}}+\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2y^2+z^2}}+\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2z^2+x^2}}\right)^2\)
Áp dụng BĐT Bunhiacopxki ta có:
\(P^2\le\left(\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2\right)\left(\left(\frac{1}{\sqrt{2x^2+y^2}}\right)^2+\left(\frac{1}{\sqrt{2y^2+z^2}}\right)^2+\left(\frac{1}{\sqrt{2z^2+x^2}}\right)^2\right)\)
\(\Leftrightarrow P^2\le\frac{1}{2x^2+y^2}+\frac{1}{2y^2+z^2}+\frac{1}{2z^2+x^2}\)
Mặt khác ta có:
\(\frac{1}{2x^2+y^2}=\frac{1}{x^2+x^2+y^2}\le\frac{1}{9}\left(\frac{1}{x^2}+\frac{1}{x^2}+\frac{1}{y^2}\right)\)
\(\frac{1}{2y^2+z^2}\le\frac{1}{9}\left(\frac{1}{y^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)
\(\frac{1}{2z^2+x^2}\le\frac{1}{9}\left(\frac{1}{z^2}+\frac{1}{z^2}+\frac{1}{x^2}\right)\)
\(\Rightarrow P^2\le\frac{1}{3}\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\le\frac{1}{3}.2016=672\)
\(\Rightarrow P\le4\sqrt{42}\)
Dấu '=' xảy ra khi \(x=y=z=\sqrt{\frac{1}{672}}\)
4) Áp dụng bất đẳng thức Bunyakovsky
\(\Rightarrow\left(x^4+yz\right)\left(1+1\right)\ge\left(x^2+\sqrt{yz}\right)^2\)
\(\Rightarrow\dfrac{x^2}{x^4+yz}\le\dfrac{2x^2}{\left(x^2+\sqrt{yz}\right)^2}\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{y^2}{y^4+xz}\le\dfrac{2y^2}{\left(y^2+\sqrt{xz}\right)^2}\\\dfrac{z^2}{z^4+xy}\le\dfrac{2z^2}{\left(z^2+\sqrt{xy}\right)^2}\end{matrix}\right.\)
\(\Rightarrow VT\le2\left[\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\)
Chứng minh rằng \(2\left[\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\le\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)^2}\le\dfrac{3}{4}\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow x^2+\sqrt{yz}\ge2\sqrt{x^2\sqrt{yz}}=2x\sqrt{\sqrt{yz}}\)
\(\Rightarrow\left(x^2+\sqrt{yz}\right)^2\ge4x^2\sqrt{yz}\)
\(\Rightarrow\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}\le\dfrac{x^2}{4x^2\sqrt{yz}}=\dfrac{1}{4\sqrt{yz}}\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}\le\dfrac{1}{4\sqrt{xz}}\\\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)^2}\le\dfrac{1}{4\sqrt{xy}}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)^2}\le\dfrac{1}{4}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\)
Chứng minh rằng \(\dfrac{1}{4}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\le\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le3\)
Theo đề bài ta có \(x^2+y^2+z^2=3xyz\)
\(\Rightarrow\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}=3\)
\(\Rightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le3\)
\(\Leftrightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\dfrac{1}{\sqrt{xy}}\le\dfrac{\dfrac{1}{x}+\dfrac{1}{y}}{2}\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{xz}}\le\dfrac{\dfrac{1}{x}+\dfrac{1}{z}}{2}\\\dfrac{1}{\sqrt{yz}}\le\dfrac{\dfrac{1}{z}+\dfrac{1}{y}}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) (1)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\dfrac{x}{yz}+\dfrac{y}{xz}\ge2\sqrt{\dfrac{1}{z^2}}=\dfrac{2}{z}\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{y}{xz}+\dfrac{z}{xy}\ge\dfrac{2}{x}\\\dfrac{x}{zy}+\dfrac{z}{xy}\ge\dfrac{2}{y}\end{matrix}\right.\)
\(\Rightarrow2\left(\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}\right)\ge2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(\Leftrightarrow\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}\ge\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) (2)
Từ (1) và (2)
\(\Rightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le3\) ( đpcm )
Vậy \(\dfrac{1}{4}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\le\dfrac{3}{4}\)
\(\Rightarrow2\left[\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\le\dfrac{3}{2}\)
Mà \(VT\le2\left[\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\)
\(\Rightarrow VT\le\dfrac{3}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(x=y=z=1\)
3. Ta có :\(x^2\left(1-2x\right)=x.x.\left(1-2x\right)\le\dfrac{\left(x+x+1-2x\right)^3}{27}=\dfrac{1}{27}\)(bđt cô si)
Dấu "=" xảy ra khi :x=1-2x\(\Leftrightarrow x=\dfrac{1}{3}\)
Vậy max của Qlaf 1/27 khi x=1/3
Do vai trò của x,y bình đẳng như nhau,giả sử \(x\ge y\),khi đó:
\(\frac{x+y}{x^2+y^2}=\frac{7}{25}\)
\(\Rightarrow7\left(x^2+y^2\right)=25\left(x+y\right)\)
\(\Rightarrow7x^2+7y^2=25x+25y\)
\(\Rightarrow7x^2-25x=25y-7y^2\)
\(\Rightarrow x\left(7x-25\right)=y\left(25-7y\right)\)
\(\Rightarrow7x-25\)và \(25-7y\)cùng dấu vì \(x,y\inℕ\)
Nếu \(\hept{\begin{cases}7x+25< 0\\25-7y< 0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x< 4\\y< 4\end{cases}}\)(trái với giả sử)
Nếu \(\hept{\begin{cases}7x-25\ge0\\25-7y\ge0\end{cases}}\)
\(\Rightarrow x\ge4,y< 4\)
Thử y là các số tự nhiên từ 0 đến 3 ta được \(x=4,y=3\)
Vậy các cặp số (x,y) cần tìm là:\(\left(3;4\right)\)và các hoán vị của chúng
hỏi trước tí, bạn biết giải cái hệ này chứ?
\(\left\{{}\begin{matrix}2x+y=3\\2x-3y=1\end{matrix}\right.\)
Bài 3 \(\hept{\begin{cases}x+y+xy=2+3\sqrt{2}\\x^2+y^2=6\end{cases}}\)
\(\hept{\begin{cases}\left(x+y\right)+xy=2+3\sqrt{2}\\\left(x+y\right)^2-2xy=6\end{cases}}\)
\(\hept{\begin{cases}S+P=2+3\sqrt{2}\left(1\right)\\S^2-2P=6\left(2\right)\end{cases}}\)
Từ (1)\(\Rightarrow P=2+3\sqrt{2}-S\)Thế P vào (2) rồi giải tiếp nhé. Mình lười lắm ^.^
Bài 1:
\((x,y,z)=(\frac{2a^2}{bc}; \frac{2b^2}{ca}; \frac{2c^2}{ab})\) (\(a,b,c>0\) )
Khi đó:
\(\text{VT}=\frac{\frac{4a^4}{b^2c^2}}{\frac{4a^4}{b^2c^2}+\frac{4a^2}{bc}+1}+\frac{\frac{4b^4}{c^2a^2}}{\frac{4b^4}{c^2a^2}+\frac{4b^2}{ca}+4}+\frac{\frac{4c^4}{a^2b^2}}{\frac{4c^4}{a^2b^2}+\frac{4c^2}{ab}+4}\)
\(=\frac{a^4}{a^4+a^2bc+b^2c^2}+\frac{b^4}{b^4+b^2ac+a^2c^2}+\frac{c^4}{c^4+c^2ab+a^2b^2}\)
\(\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+a^2bc+b^2ac+c^2ab+(a^2b^2+b^2c^2+c^2a^2)}\)
(Áp dụng BĐT Cauchy_Schwarz)
Theo BĐT Cauchy dễ thấy:
\(a^2b^2+b^2c^2+c^2a^2\geq a^2bc+b^2ca+c^2ab\)
\(\Rightarrow \text{VT}\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)}=\frac{(a^2+b^2+c^2)^2}{(a^2+b^2+c^2)^2}=1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=2$
Bài 2:
Đặt \((x,y,z)=\left(\frac{a}{b};\frac{b}{c}; \frac{c}{a}\right)\)
Ta có:
\(\text{VT}=\left(\frac{a}{b}+\frac{c}{b}-1\right)\left(\frac{b}{c}+\frac{a}{c}-1\right)\left(\frac{c}{a}+\frac{b}{a}-1\right)\)
\(=\frac{(a+c-b)(b+a-c)(c+b-a)}{abc}\)
Áp dụng BĐT Cauchy:
\((a+c-b)(b+a-c)\leq \left(\frac{a+c-b+b+a-c}{2}\right)^2=a^2\)
\((b+a-c)(c+b-a)\leq \left(\frac{b+a-c+c+b-a}{2}\right)^2=b^2\)
\((a+c-b)(c+b-a)\leq \left(\frac{a+c-b+c+b-a}{2}\right)^2=c^2\)
Nhân theo vế:
\(\Rightarrow [(a+c-b)(b+a-c)(c+b-a)]^2\leq (abc)^2\)
\(\Rightarrow (a+c-b)(b+a-c)(c+b-a)\leq abc\)
\(\Rightarrow \text{VT}\leq 1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=1$
a, \(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}=\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}=1\)
b, Đặt \(B=\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(\sqrt{x}=a,\sqrt{y}=b\)
Ta có: \(B=\dfrac{a^3-b^3}{a-b}=\dfrac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a-b}=a^2+ab+b^2\)
\(\Rightarrow B=x+\sqrt{xy}+y\)
Vậy...
c, \(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}}=\dfrac{a}{\left(b-2\right)^2}.\dfrac{\left(b-2\right)^2}{a}=1\)
d, \(2x+\dfrac{\sqrt{1-6x+9x^2}}{3x-1}=2x+\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}=2x+1\)
a:b(a−4)2.√(a−4)4b2(b>0;a≠4)b(a−4)2.(a−4)4b2(b>0;a≠4)
= \(\dfrac{b}{\left(a-4\right)}.\dfrac{\sqrt{\left[\left(a-4\right)^2\right]^2}}{\sqrt{b^2}}\)
=\(\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}\)
= 1 ( nhân tử với tử mẫu với mẫu rồi rút gọn)
b:x√x−y√y√x−√y(x≥0;y≥0;x≠0)xx−yyx−y(x≥0;y≥0;x≠0)
=\(\dfrac{\sqrt{x^3}-\sqrt{y^3}}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}-\sqrt{y}\right).\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}\)(áp dụng hằng đẳng thức )
= (x+\(\sqrt{xy}\)+y)
c:a(b−2)2.√(b−2)4a2(a>0;b≠2)a(b−2)2.(b−2)4a2(a>0;b≠2)
Tương tự câu a
d:x(y−3)2.√(y−3)2x2(x>0;y≠3)x(y−3)2.(y−3)2x2(x>0;y≠3)
tương tự câu a
e:2x +√1−6x+9x23x−1
= \(2x+\dfrac{\sqrt{\left(3x\right)^2-6x+1}}{3x-1}\)
= 2x+\(\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}\)(hằng đẳng thức)
=2x+\(\dfrac{3x-1}{3x-1}\)
=2x+1
Ta thấy \(2x^2< 4\) \(\Leftrightarrow x^2< 2\) \(\Leftrightarrow x^2=1\) (do \(x\ne0\))
Thế vào pt đề bài, ta có \(3+\dfrac{y^2}{4}=4\)
\(\Leftrightarrow\dfrac{y^2}{4}=1\)
\(\Leftrightarrow y^2=4\)
\(\Leftrightarrow y=\pm2\)
Vậy, các cặp số (x; y) thỏa ycbt là \(\left(1;2\right);\left(-1;-2\right);\left(1;-2\right);\left(-1;2\right)\)
a