Cho \(\dfrac{x^2\left(x^2+2y^2\right)}{1+y^2}=1-y^2\)
Tìm Min, Max của P = \(\dfrac{xy+1}{x^2-xy+3}\)
1) cho các số thực dương a,b thỏa mãn \(3a+b\le1\). Tìm Min của \(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\)
2) Với hai số thực a,b không âm thỏa mãn \(a^2+b^2=4\). Tìm Max \(M=\dfrac{ab}{a+b+2}\)
3) Cho x,y khác 0 thỏa mãn \(\left(x+y\right)xy=x^2+y^2-xy\). Tìm Max \(A=\dfrac{1}{x^3}+\dfrac{1}{y^3}\)
1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:
\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).
Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).
2.
\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)
Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)
\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )
\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)
\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)
Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)
3. Chia 2 vế giả thiết cho \(x^2y^2\)
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\ge\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)
\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)
\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)
Cho các số thực x, y, z thỏa mãn \(7\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)=6\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)=2016\).
Tìm max: \(P=\dfrac{1}{\sqrt{3\left(2x^2+y^2\right)}}+\dfrac{1}{\sqrt{3\left(2y^2+z^2\right)}}+\dfrac{1}{\sqrt{3\left(2z^2+x^2\right)}}\)
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 \(x=y=z=\sqrt{\frac{1}{672}}\)
Dễ có: \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\)
\(gt\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\le2016\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{1}{\sqrt{3\left(2x^2+y^2\right)}}=\frac{1}{\sqrt{\left(2+1\right)\left(2x^2+y^2\right)}}\le\frac{1}{2x+y}\)
\(\le\frac{1}{9}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{9}\left(\frac{2}{x}+\frac{1}{y}\right)\)
\(\Rightarrow P\le\frac{1}{3}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\le\frac{1}{3}\sqrt{3\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{Z^2}\right)}\le\sqrt{\frac{2016}{3}}\)
cho x,y,z>0 và x+y+z=\(\dfrac{3}{2}\)
tìm Min \(P=\dfrac{\sqrt{x^2+xy+y^2}}{\left(x+y\right)^2+1}+\dfrac{\sqrt{y^2+yz+z^2}}{\left(y+z\right)^2+1}+\dfrac{\sqrt{z^2+zx+x^2}}{\left(z+x\right)^2+1}\)
Đề bài sai, biểu thức này ko có min
Giải hệ
a) \(\left\{{}\begin{matrix}x^2+y^2-2y-6+2\sqrt{2y+3}=0\\\left(x-y\right)\left(x^2+xy+y^2+3\right)=3\left(x^2+y^2\right)+2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2y+2y+x=4xy\\\dfrac{1}{x^2}+\dfrac{1}{xy}+\dfrac{x}{y}=3\end{matrix}\right.\)
thuc hien phep tinh
a.\(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
b.\(\left(\dfrac{1}{x^2+1}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+1-2\right)\)
c.\(\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}.\left(\dfrac{1}{x^2-2x+1}+\dfrac{1}{1-x^2}\right)\)
d.\(\left(\dfrac{x^2+xy}{x^3+x^2y+xy^2+y^3}+\dfrac{y}{x^2+y^2}\right):\left(\dfrac{1}{x-y}-\dfrac{2xy}{x^3-x^2y+xy^2-y^3}\right)\)
a: \(=\dfrac{4x^2+4x+1-\left(4x^2-4x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{2x+1}\cdot\dfrac{5}{4x}=\dfrac{10}{2x+1}\)
c: \(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
1) cho x,y,z là các số thực thỏa mãn \(\left\{{}\begin{matrix}xyz=2\\2+x+xy\ne0\end{matrix}\right.\)
tính B= \(\dfrac{1}{1+y+yz}+\dfrac{2}{2+2z+xz}+\dfrac{2}{2+x+xy}\)
2) giải hpt \(\left\{{}\begin{matrix}\left(y^2-4y\right)\left(2y-x\right)=2\\y^2-2y-x=3\end{matrix}\right.\)
3)GPT \(x^2-2x=2\sqrt{2x-1}\)
4) tìm n nguyên dương để A=\(2^9+2^{13}+2^n\) là số chính phương
5) tìm Min của A=\(\dfrac{\left(x+y+1\right)^2}{xy+y+x}+\dfrac{xy+y+x}{\left(x+y+1\right)^2}\) (x;y dương )
Bài 5: Đặt \(t=\dfrac{\left(x+y+1\right)^2}{xy+x+y}\)
Ta đã biết bđt quen thuộc là \(x^2+y^2+1\ge xy+x+y\)
Vậy nên ta sẽ chứng minh \(t\geq 3\)
Thật vậy: \(t\geq 3\Leftrightarrow 2(x+y+1)^2\geq 6(x+y+xy)\)
\(\Leftrightarrow (x-y)^2+(x-1)^2+(y-1)^2\geq 0\)
Đẳng thức xảy ra khi và chỉ khi \(x=y=1\)
Ta có: \(A=\dfrac{8t}{9}+\left(\dfrac{t}{9}+\dfrac{1}{t}\right)\geq \dfrac{24}{9}+\dfrac{2}{3}=\dfrac{10}{3}\)
Dấu "=" xảy ra khi \(t=3\Leftrightarrow x=y=1\)
3)
x^2 = 2x + \(\sqrt{2x-1}\) \(\Rightarrow\) x^2 = ( 2x -1 ) + \(\sqrt{2x-1}\) +1
\(\Rightarrow\) x^2 = (\(\sqrt{2x-1}\) + 1)^2 chuyển vế rồi phân tích thành nhân tử là ok
Cho A = \(\dfrac{\left(x-y\right)^2+xy}{\left(x+y\right)^2-xy}.\left[1:\dfrac{x^5+y^5+x^3y^2+x^2y^3}{\left(x^3-y^3\right)\left(x^3+y^3+x^2y+xy^2\right)}\right]\)
B = x - y
Chứng minh đẳng thức A = B
Tính giá trị của A, B tại x = 0; y = 0 và giải thích vì sao A ≠ B
\(ĐK:x\ne y;x\ne-y;x^2+xy+y^2\ne0;x^2-xy+y^2\ne0\)
\(A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\left[1:\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2+y^2\right)}\right]\\ A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)\left(x^2+y^2\right)}\\ A=x-y=B\)
\(x=0;y=0\Leftrightarrow B=0\)
Giá trị của A không xác định vì \(x=y\) trái với ĐK:\(x\ne y\)
Vậy \(A\ne B\)
R/gọn: \(\left(\dfrac{x^2}{x+y}+y\right).\left(\dfrac{1}{x^2-xy}-\dfrac{3y^2}{x^4-xy^3}-\dfrac{y}{x^3+x^2y+xy^2}\right)\)
\(\left(\dfrac{x^2}{x+y}+y\right).\left(\dfrac{1}{x^2-xy}-\dfrac{3y^3}{x^4-xy^3}-\dfrac{y}{x^3+x^2y+xy^2}\right)\)
\(=\left(\dfrac{x^2+xy+y^2}{x+y}\right).\left(\dfrac{1}{x\left(x-y\right)}-\dfrac{3y^2}{x\left(x^3-y^3\right)}-\dfrac{y}{x\left(x^2+xy+y^2\right)}\right)\)\(=\left(\dfrac{x^2+xy+y^2}{x+y}\right).\left(\dfrac{x^2+xy+y^2}{x\left(x^3-y^3\right)}-\dfrac{3y^2}{x\left(x^3-y^3\right)}-\dfrac{xy-y^2}{x\left(x^3-y^3\right)}\right)\)
\(=\dfrac{x\left(x^3-y^3\right)}{x^3-xy^2}.\dfrac{x^2+xy+y^2-3y^2-xy+y^2}{x\left(x^3-y^3\right)}\\ =\dfrac{x^2-y^2}{x\left(x^2-y^2\right)}=\dfrac{1}{x}\)
mình viết trên máy tinh hơi xấu bạn thông cảm nhé!!!Nếu ko chê có thể xem cách giải này!
Thực hiện phép tính sau:
d) \(\left(\dfrac{x^2+xy}{x^3+x^2y+xy^2+y^3}+\dfrac{y}{x^2+y^2}\right):\left(\dfrac{1}{x-y}-\dfrac{2xy}{x^2-x^2y+xy^2-y^3}\right)\)
\(=\left(\dfrac{x\left(x+y\right)}{x^2\left(x+y\right)+y^2\left(x+y\right)}+\dfrac{y}{x^2+y^2}\right):\left(\dfrac{1}{x-y}-\dfrac{2xy}{x^2\left(x-y\right)+y^2\left(x-y\right)}\right)\)
\(=\dfrac{x+y}{x^2+y^2}:\left(\dfrac{1}{x-y}-\dfrac{2xy}{\left(x-y\right)\left(x^2+y^2\right)}\right)\)
\(=\dfrac{x+y}{x^2+y^2}:\dfrac{x^2+y^2-2xy}{\left(x-y\right)\left(x^2+y^2\right)}\)
\(=\dfrac{x+y}{x^2+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+y^2\right)}{\left(x-y\right)^2}\)
\(=\dfrac{x+y}{x-y}\)