Gọi cái biểu thức đó là P nha

Trước tiên chứng minh:

\(\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}-\left(\frac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\right)=0\)

\(\Leftrightarrow\frac{x^4-y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4-z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4-x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)

\(\Leftrightarrow x-y+y-z+z-x=0\)( đúng )

Giờ ta quay lại bài toán ban đầu 

Ta có:

\(\Leftrightarrow2P=\frac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4+z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4+x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)

\(\ge\frac{\left(x^2+y^2\right)^2}{2\left(x^2+y^2\right)\left(x+y\right)}+\frac{\left(y^2+z^2\right)^2}{2\left(y^2+z^2\right)\left(y+z\right)}+\frac{\left(z^2+x^2\right)^2}{2\left(z^2+x^2\right)\left(z+x\right)}\)

\(=\frac{x^2+y^2}{2\left(x+y\right)}+\frac{y^2+z^2}{2\left(y+z\right)}+\frac{z^2+x^2}{2\left(z+x\right)}\)

\(\ge\frac{\left(x+y\right)^2}{4\left(x+y\right)}+\frac{\left(y+z\right)^2}{4\left(y+z\right)}+\frac{\left(z+x\right)^2}{4\left(z+x\right)}\)

\(=\frac{x+y}{4}+\frac{y+z}{4}+\frac{z+x}{4}=\frac{1}{2}\)

\(\Rightarrow P\ge\frac{1}{4}\)

= -1,4 

HT

l:à

√0.7

chawxc vậy thôi nhé

\(\sqrt{6-2\sqrt{5}}+\sqrt{8+2\sqrt{15}}-\sqrt{3}\)

\(=\sqrt{5}-1+\sqrt{5}+\sqrt{3}-\sqrt{3}=2\sqrt{5}-1\)

a, \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\)

\(=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\)

\(=\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}=\sqrt{2}\)

b, \(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\frac{15\left(\sqrt{6}-1\right)}{5}+\frac{4\left(\sqrt{6}+2\right)}{2}-\frac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)=6-121=-115\)

ĐỂ phép tính k bị lẻ lên thay \(\frac{5}{4}.\sqrt{\frac{4}{5}}=\frac{5}{2}.\sqrt{\frac{4}{5}}\)

\(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}.\sqrt{20}-\frac{5}{2}.\sqrt{\frac{4}{5}}+\sqrt{5}\right):2\sqrt{5}\)

\(=\left(\sqrt{5}+\sqrt{5}-\sqrt{5}+\sqrt{5}\right):2\sqrt{5}=2\sqrt{5}:2\sqrt{5}=1\)

ai kết bẠN đi

Ta có \(B=\frac{a^2+b^2}{a-b}=\frac{a^2+b^2-4+4}{a-b}=\frac{a^2+b^2-2ab}{a-b}+\frac{4}{a-b}=\left(a-b\right)+\frac{4}{a-b}\)

Áp dụng bất đẳng thức Cauchy cho 2 số không âm ta có : 

\(B=\left(a-b\right)+\frac{4}{a-b}=2\sqrt{\left(a-b\right).\frac{4}{a-b}}=4\)

Dấu "=" xảy ra <=> \(a-b=\frac{4}{a-b}\)

Kết hợp giả thiết => \(\hept{\begin{cases}a=\frac{\sqrt{12}+2}{2}\\b=\frac{\sqrt{12}-2}{2}\end{cases}}\)

= có lm thì mới có ăn nha bn

Theo tính chất hai tiếp tuyến cắt nhau ta có:

    DM = DB, EM = EC, AB = AC

Chu vi ΔADE:

    CΔADE = AD + DE + AE = AD + DM + ME + AE = AD + DB + EC + AE = AB + AC = 2AB (đpcm)

2ab (đpcm)

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