Bài 1: CMR: với mọi \(x,y\in R^+\)ta có:
a, \(\left(a^2+b^2\right)\left(a+b\right)\ge4ab\left(a+b-\sqrt{ab}\right)\)
b, \(\left(a^2+b^2\right)\left(a+b\right)\ge4ab\left[2\left(a+b\right)-3\sqrt{ab}\right]\)
Bài 1: CMR: với mọi \(x,y\in R^+\)ta có:
a, \(\left(a^2+b^2\right)\left(a+b\right)\ge4ab\left(a+b-\sqrt{ab}\right)\)
b, \(\left(a^2+b^2\right)\left(a+b\right)\ge4ab\left[2\left(a+b\right)-3\sqrt{ab}\right]\)
Mọi người giúp mình đi maaaaaaaaaaaaaaà!
từ giả thiết, ta có \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
đặt \(\left(\dfrac{1}{xy};\dfrac{1}{yz};\dfrac{1}{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c=1\) =>\(\left(\dfrac{ac}{b};\dfrac{ab}{c};\dfrac{bc}{a}\right)=\left(\dfrac{1}{x^2};\dfrac{1}{y^2};\dfrac{1}{z^2}\right)\)
ta có VT=\(\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{y^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{z^1}}}=\sqrt{\dfrac{1}{1+\dfrac{ac}{b}}}+\sqrt{\dfrac{1}{1+\dfrac{ab}{c}}}+\sqrt{\dfrac{1}{1+\dfrac{bc}{a}}}\)
=\(\dfrac{1}{\sqrt{\dfrac{b+ac}{b}}}+\dfrac{1}{\sqrt{\dfrac{a+bc}{a}}}+\dfrac{1}{\sqrt{\dfrac{c+ab}{c}}}=\sqrt{\dfrac{a}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{b}{\left(b+c\right)\left(b+a\right)}}+\sqrt{\dfrac{c}{\left(c+a\right)\left(c+b\right)}}\)
\(\le\sqrt{3}\sqrt{\dfrac{ac+ab+bc+ba+ca+cb}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=\sqrt{3}.\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
ta cần chứng minh \(\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\le\dfrac{3}{2}\Leftrightarrow\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9}{4}\Leftrightarrow8\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
<=>\(8\left(a+b+c\right)\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (luôn đúng )
^_^
Rút gọn:\(a,\sqrt{\left(x+2\sqrt{x+1}\right)\left(x+3+4\sqrt{x-1}\right)}\left(x>1\right)\)
\(b,\sqrt{\left(a^2+b^2+c^2+2\left(ab+bc+ac\right)\right)\left(a+b-2\sqrt{ab}\right)}\)
\(c,\dfrac{2+a-2\sqrt{a}}{3+a-3\sqrt{a}}\)
1. \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=1\end{matrix}\right.\). Cmr: \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}+\frac{bc}{\sqrt{\left(1-a\right)^2\left(1+a\right)}}+\frac{ca}{\sqrt{\left(1-b\right)^3\left(1+b\right)}}\le\frac{3\sqrt{2}}{8}\)
2. \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c\le1\end{matrix}\right.\). Cmr: \(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab\left(a+b\right)}+\frac{1}{bc\left(b+c\right)}+\frac{1}{ac\left(a+c\right)}\ge\frac{87}{2}\)
3. \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca=2abc\end{matrix}\right.\). Cmr: \(\frac{1}{a\left(2a-1\right)^2}+\frac{1}{b\left(2b-1\right)^2}+\frac{1}{c\left(2c-1\right)^2}\ge\frac{1}{2}\)
4. \(\left\{{}\begin{matrix}x,y,z>0\\x+y+z=2015\end{matrix}\right.\). Tìm min \(A=\frac{x^4+y^4}{x^3+y^3}+\frac{y^4+z^4}{y^3+z^3}+\frac{z^4+x^4}{z^2+x^2}\)
Mn giúp mk với ạ! Thanks nhiều
Mới nghĩ ra 3 câu:
a/ \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}=\frac{ab}{\sqrt{\left(a+b\right)^2\left(1+c\right)}}\le\frac{ab}{2\sqrt{ab\left(1+c\right)}}=\frac{1}{2}\sqrt{\frac{ab}{1+c}}\)
\(\sum\sqrt{\frac{ab}{1+c}}\le\sqrt{2\sum\frac{ab}{1+c}}\)
\(\sum\frac{ab}{1+c}=\sum\frac{ab}{a+c+b+c}\le\frac{1}{4}\sum\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)=\frac{1}{4}\)
c/ \(ab+bc+ca=2abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(\left(x;y;z\right)=\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\Rightarrow x+y+z=2\)
\(VT=\sum\frac{x^3}{\left(2-x\right)^2}\)
Ta có đánh giá: \(\frac{x^3}{\left(2-x\right)^2}\ge x-\frac{1}{2}\) \(\forall x\in\left(0;2\right)\)
\(\Leftrightarrow2x^3\ge\left(2x-1\right)\left(x^2-4x+4\right)\)
\(\Leftrightarrow9x^2-12x+4\ge0\Leftrightarrow\left(3x-2\right)^2\ge0\)
d/ Ta có đánh giá: \(\frac{x^4+y^4}{x^3+y^3}\ge\frac{x+y}{2}\)
\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)
Akai Haruma, Nguyễn Ngọc Lộc , @tth_new, @Băng Băng 2k6, @Trần Thanh Phương, @Nguyễn Việt Lâm
Mn giúp e vs ạ! Thanks!
Bài 1 : NĂNG KHIẾU 2016-2017
A) Tính S=a+b biết a;b>0, a \(\ne\)b và \(\left(\dfrac{a\left(a-4b\right)+b\left(b+2a\right)}{a+b}\right):\left[\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\right]=2016\)
B) Giải: \(x\sqrt{x+5}=2x^2-5x\left(1\right)và\left\{{}\begin{matrix}\left(\sqrt{y}+x-3\right)\left(y+\sqrt{x}\right)=0\\x^2+y=5\end{matrix}\right.\)
Cho a,b,c đôi một khác nhau và ab+bc+ca=1
Tính
a) \(A=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\)
b)\(B=\frac{\left(a^2+2bc-1\right)\left(b^2+2ac-1\right)\left(c^2+2ba-1\right)}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}\)
c)\(C=x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{\left(1+y^2\right)}}+z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\)
Nhiều quá làm 1 bài tiêu biểu thôi nhé:
a/ \(A=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\)
\(=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(ab+bc+ca+a^2\right)\left(ab+bc+ca+b^2\right)\left(ab+bc+ca+c^2\right)}\)
\(=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)\left(c+a\right)\left(b+c\right)\left(a+b\right)\left(c+a\right)\left(b+c\right)}=1\)
Nhiều quá! Làm bài tiêu biểu nhé!
a) Đặt \(a;b;c=0\)
\(\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\Leftrightarrow\frac{\left(0+0\right)^2\left(0+0\right)^2\left(0+0\right)^2}{\left(1+0^2\right)\left(1+0^2\right)\left(1+0^2\right)}\)
\(\Leftrightarrow\frac{0^2+0^2+0^2}{1^2+1^2+1^2}=\frac{0}{3}=0\)
alibaba nguyễn: Hình như bạn làm sai rồi! Vì mình bấm máy tính ra kết quả 0 mà! Cô mình cũng nói kết quả bằng 0.
Trần Hoàng Việt : Mấy bài kia y chang.
Cho a,b,c ko âm. CMR:
\(3\left(a^2+b^2+c^2\right)\ge P\ge\left(a+b+c\right)^2\)
với \(P=\left(a+b+c\right)\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)
Cmr \(\left(a+b\right)^2\ge4ab\) với mọi a,b
Ta có : \(\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow\left(a+b\right)^2-4ab\ge0\)
\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\) ( luôn đúng với mọi a , b )
\(\Rightarrow\left(a+b\right)^2\ge4ab\) ( đpcm )
Các bn xem bài này mk làm đúng không
a)\(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)
VT=\(\left(\frac{a\sqrt{a}+b\sqrt{b}-\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
=\(\left(\frac{a\sqrt{a}+b\sqrt{b}-a\sqrt{b}-b\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
=\(\left(\frac{\left(a\sqrt{a}-a\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
=\(\left(\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}{\sqrt{a+\sqrt{b}}}\right)\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)
= \(\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{a-b}=\frac{a-b}{a-b}=1\Rightarrow\left(=VP\right)\)
b)\(\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=a-b\)
VT=\(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\sqrt{a}+\sqrt{b}=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)\)
=\(a+\sqrt{ab}-\sqrt{ab}-b=a-b\Rightarrow\left(=VP\right)\)
Đây là đề chứng minh hả !
Phần a , b đúng r
Nhưng phần b chỗ \(\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)=\left(\sqrt{a}\right)^2-\left(\sqrt{b}\right)^2\) = a - b
Dùng hằng đẳng thức thức 3 như vậy sẽ hay hơn !
Chúc bạn học tốt!