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\(P=\frac{\frac{1}{a^2}}{\frac{1}{b}+\frac{1}{c}}+\frac{\frac{1}{b^2}}{\frac{1}{a}+\frac{1}{c}}+\frac{\frac{1}{c^2}}{\frac{1}{a}+\frac{1}{b}}\)
Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\Rightarrow xyz=1\Rightarrow P=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(P\ge\frac{\left(x+y+z\right)^2}{y+z+x+z+x+y}=\frac{x+y+z}{2}\ge\frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z\Leftrightarrow a=b=c=1\)
Cần cách khác thì nhắn cái
\(A=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\left(\sqrt{3}-1\right)^2\left(2+\sqrt{3}\right)=\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)\)
\(=2\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2\)
\(B=\frac{\left(\sqrt{a}-1\right)\left(\sqrt{6}-\sqrt{2}\right)\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}{a\left(\sqrt{a}-1\right)\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\frac{\sqrt{6}-\sqrt{2}}{a+\sqrt{ab}}\)
1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)
\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\) (1)
áp dụng (x2 +y2 +z2)(m2+n2+p2) \(\ge\left(xm+yn+zp\right)^2\)
(2a2bc +b2c2 + 2b2ac+a2c2 + 2c2ab+a2b2). VT\(\ge\left(bc+ca+ab\right)^2\) <=> (ab+bc+ca)2. VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\) ( vậy (1) đúng)
dấu '=' khi a=b=c
1)
Ta có: \(M=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\sqrt{3\left(a+b\right)\left(a+b+4c\right)}}\ge\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\frac{3\left(a+b\right)+\left(a+b+4c\right)}{2}}=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{2\left(a+b+c\right)}=3\sqrt{3}\)
Dấu "=" xảy ra khi a=b=c
2)
\(\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}=\Sigma_{cyc}\frac{2a}{\sqrt[3]{2a\left(ab+1\right)^2}}\ge\Sigma_{cyc}\frac{2a}{\frac{2a+\left(ab+1\right)+\left(ab+1\right)}{3}}=3\Sigma_{cyc}\frac{a}{ab+a+1}\)
Ta có bổ đề: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)
\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)
Bài làm:
a) \(A=\left(\sqrt{3}+1\right)^2+\frac{5}{4}\sqrt{48}-\frac{2}{\sqrt{3+1}}\)
\(A=3+2\sqrt{3}+1+\sqrt{\frac{25.48}{16}}-\frac{2}{\sqrt{4}}\)
\(A=4+2\sqrt{3}+\sqrt{25.3}-\frac{2}{2}\)
\(A=4+2\sqrt{3}+5\sqrt{3}-1\)
\(A=3+7\sqrt{3}\)
b) \(\frac{4}{3-\sqrt{5}}-\frac{3}{\sqrt{5}+\sqrt{2}}-\frac{1}{\sqrt{2}-1}\)
\(=\frac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}-\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(A=\frac{4\left(3+\sqrt{5}\right)}{9-5}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{5-2}-\frac{\sqrt{2}+1}{2-1}\)
\(A=3+\sqrt{5}-\sqrt{5}+\sqrt{2}-\sqrt{2}-1\)
\(A=2\)
Phần b mình viết nhầm tên thành A, bn sửa thành B nhé
c) \(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)
\(C=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(C=\sqrt{3}-1-2-\sqrt{3}\)
\(C=-3\)
mk làm luôn
a)\(A=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right).\)
=\(\frac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}-1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}\)
=\(\frac{\left(3x+3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right).3}\)
=\(\frac{3x+3\sqrt{x}-1}{9\sqrt{x}-3}\)
=
Cảm ơn chú đã kb giờ thì t sẽ làm hộ chú :V
\(P=\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}+\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\right):\left(\frac{\sqrt{a}+1}{\sqrt{ab}+1}-\frac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+1\right)\)
\(P=\left[\frac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)}{\left(\sqrt{ab}+1\right)\left(\sqrt{ab}-1\right)}+\frac{\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}-\frac{ab-1}{ab-1}\right]\)
\(:\left[\frac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)}{\left(\sqrt{ab}+1\right)\left(\sqrt{ab-1}\right)}\right]-\frac{\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}+\frac{ab-1}{ab-1}\)
\(P=\frac{\left(a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1\right)+\left(ab+\sqrt{ab}+a\sqrt{b}+\sqrt{a}\right)-\left(ab-1\right)}{ab-1}\)
\(:\frac{\left(a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1\right)-\left(ab+\sqrt{ab}+a\sqrt{b}+\sqrt{a}\right)+\left(ab-1\right)}{ab-1}\)
\(P=\frac{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1+ab+\sqrt{ab}+a\sqrt{b}+\sqrt{a}-ab+1}{ab-1}\)
\(:\frac{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1-ab-\sqrt{ab}-a\sqrt{b}-\sqrt{a}+ab-1}{ab-1}\)
\(P=\frac{2a\sqrt{b}+2\sqrt{ab}}{ab-1}:\frac{-2\sqrt{a}-2}{ab-1}\)
\(P=\frac{2\sqrt{ab}\left(\sqrt{a}+1\right)}{ab-1}.\frac{ab-1}{-2\left(\sqrt{a}+1\right)}=-\sqrt{ab}\)
P/s: :V bài này tính toán kĩ nhưng chưa chắc đúng :VVVV
Mr.Fuff cảm ơn ạ >_<!!!!