\(a+b+c=3\)

\(\Leftrightarrow\left(a+b+c\right)^2=9\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=9\)

\(\Rightarrow a^2+c^2+b^2=9-2\left(ab+bc+ca\right)\)

Nếu \(ab+bc+ca>3\) thì \(a^2+b^2+c^2< 3\left(vl\right)\)

\(\Rightarrow ab+bc+ca\le3\)

\(\approx103680.0018\)

\(\frac{5^7}{9^8}+5\frac{6^8}{3^4}=\frac{5^7}{9^8}+\frac{5.3^4.6^8}{3^4}\)

\(=\frac{5^7}{9^8}+\frac{5.3^{16}.6^8}{9^2.9^6}=\frac{5^7}{9^8}+\frac{5.3^{24}.2^8}{9^8}=\frac{5\left(5^6+3^{24}.2^8\right)}{9^8}\)

\(A=\left(\frac{\sqrt{x}-4x}{1-4x}-1\right):\left(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\right)\)

\(=\left(\frac{\sqrt{x}-4x-1+4x}{1-4x}\right):\left(\frac{1+2x-2\sqrt{x}-2\sqrt{x}\left(2\sqrt{x}+1\right)-1+4x}{1-4x}\right)\)

\(=\frac{\sqrt{x}-1}{1-4x}:\frac{2x-4\sqrt{x}}{1-4x}=\frac{\sqrt{x}-1}{1-4x}.\frac{1-4x}{2\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{1}{2\sqrt{x}}\)

b, \(A>A^2\Rightarrow\frac{1}{2\sqrt{x}}>\left(\frac{1}{2\sqrt{x}}\right)^2\Rightarrow\frac{1}{2\sqrt{x}}>\frac{1}{4x}\Rightarrow\frac{1}{2\sqrt{x}}-\frac{1}{4x}>0\Rightarrow\frac{2\sqrt{x}-1}{4x}>0\)

\(2\sqrt{x}-1>0\);\(4x>0\)

\(\Rightarrow x>0\)thì \(A>A^2\)

Vì c là cạnh huyền 

=> \(c>a;c>b\)=> \(c^{n-2}>a^{n-2};c^{n-2}>b^{n-2}\)

Ta có \(c^2=a^2+b^2\)

=> \(c^n=a^2.c^{n-2}+b^2.c^{n-2}>a^2.a^{n-2}+b^2.b^{n-2}=a^n+b^n\)với n>2 (ĐPCM)

Vậy \(c^n>a^n+b^n\)

a. Phải là nhỏ hơn hẳn nhé, ko có dấu = đâu

CM:

a,b,c là 3 cạnh 1 tam giác\(\Rightarrow\left(a-b\right)^2< c^2\Rightarrow a^2+b^2< c^2+2ab\Rightarrow\sqrt{a^2+b^2}< \sqrt{c^2+2ab}\)

cm tương tự ta có: \(VT< \sqrt{c^2+2ab}+\sqrt{b^2+2ac}+\sqrt{a^2+2bc}\)

Theo BĐT Bunhia \(\Rightarrow VT< \sqrt{a^2+2bc}+\sqrt{b^2+2ac}+\sqrt{c^2+2ab}\)\(\le\sqrt{\left(1+1+1\right)\left(a^2+b^2+c^2+2ab+2bc+2ac\right)}=\sqrt{3\left(a+b+c\right)^2}=\sqrt{3}.\left(a+b+c\right)\)

2, (cần cù bù thông minh) Quy đồng

\(\left|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\right|=...=\left|\frac{\left(b-c\right)\left(a-c\right)\left(a-b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\right|\)        (chỗ ba chấm là bước quy đồng tự làm)

                                                                       \(=\frac{\left|a-b\right|}{a+b}.\frac{\left|b-c\right|}{b+c}.\frac{\left|a-c\right|}{a+c}\)

                                                                         \(\le\frac{ \left|a-b\right|}{2\sqrt{ab}}.\frac{\left|b-c\right|}{2\sqrt{bc}}.\frac{\left|a-c\right|}{2\sqrt{ca}}\left(Cauchy\right)\)

                                                                            \(< \frac{c}{2\sqrt{ab}}.\frac{a}{2\sqrt{bc}}.\frac{b}{2\sqrt{ca}}\left(Bđt\Delta\right)\)

                                                                              \(=\frac{1}{8}\left(đpcm\right)\)

Các đường trung tuyến BD và CE vuông góc với nhau nên tam giác ABC vuông tại A

=> BC^2 = AB^2 + AC^2

=> BC^2 = 6^2 + 8^2

=> BC = 10 cm

\(\sqrt{13-4\sqrt{3}}+\sqrt{13+4\sqrt{3}}=2\sqrt{3}-1+2\sqrt{3}+1=4\sqrt{3}\)

\(\sqrt{19-6\sqrt{2}}-\sqrt{19+6\sqrt{2}}=3\sqrt{2}-1-3\sqrt{2}-1=-2\)

\(\left(2-\sqrt{3}\right)^x+\left(7-4\sqrt{3}\right)\left(2+\sqrt{3}\right)^x=4\left(2-\sqrt{3}\right)\)

Ta có: \(2-\sqrt{3}=\frac{1}{2+\sqrt{3}}\)

\(7-4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)

\(\left(2-\sqrt{3}\right)^x+\left(7-4\sqrt{3}\right)\left(2+\sqrt{3}\right)^x=4\left(2-\sqrt{3}\right)\)

<=> \(\frac{1}{\left(2+\sqrt{3}\right)^x}+\left(2-\sqrt{3}\right)^2\left(2+\sqrt{3}\right)^x=4\left(2-\sqrt{3}\right)\)

<=> \(1+\left(2-\sqrt{3}\right)^2\left(2+\sqrt{3}\right)^x\left(2+\sqrt{3}\right)^x=4\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^x\)

<=> \(1+\left(2-\sqrt{3}\right)^2\left(2+\sqrt{3}\right)^{2x}=4\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^x\)

Đặt:  \(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^x=t\)

Ta có pt ẩn t: \(1+t^2=4t\)

<=> \(t^2-4t+1=0\Leftrightarrow\orbr{\begin{cases}t=2-\sqrt{3}\\t=2+\sqrt{3}\end{cases}}\)

+) Với \(t=2+\sqrt{3}\), ta có: 

\(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^x=2+\sqrt{3}\)

<=> \(\left(2+\sqrt{3}\right)^x=\frac{2+\sqrt{3}}{2-\sqrt{3}}=\left(2+\sqrt{3}\right)^2\)

<=> x=2 

Trường hợp còn lại em làm tương tự

bạn ơi khu 7−4√3=(2+√3)2 nó phải là 7−4√3=(2-√3)2 mới đúng chứ?