\(A=x^2-8x+3\)

\(=x^2-8x+16-13\)

\(=\left(x-4\right)^2-13\)

\(A_{min}=-13\Leftrightarrow\left(x-4\right)^2=0\)

\(\Rightarrow x-4=0\Leftrightarrow x=4\)

Vậy \(A_{min}=-13\Leftrightarrow x=4\)

Ta có:

   A = x² - 8x + 3 = (x² - 8x + 16) - 13 = (x - 4)² - 13

Ta luôn có: (x - 4)² \(\ge\)0 \(\forall\)x

=> (x - 4)² - 13 \(\ge\)-13 \(\forall\)x

hay A \(\ge\)-13 \(\forall\)x

Dấu "=" xảy ra khi : x - 4 = 0 <=> x = 4

Vậy Min A = -13 tại x = 4

\(Pt\Leftrightarrow\left(x^3+4x^2+3x\right)+3\left(x+2-\sqrt[3]{2x^2+9x+8}\right)=0.\)

\(\Leftrightarrow\left(x^3+4x^2+3x\right)+3.\frac{\left(x+2\right)^3-2x^2-9x-8}{\left(x+2\right)^2+\left(x+2\right)\sqrt[3]{2x^2+9x+8}+\sqrt[3]{\left(2x^2+9x+8\right)^2}}=0\)

\(\Leftrightarrow\left(x^3+4x^2+3x\right)+3.\frac{x^3+4x^2+3x}{MS}=0\Leftrightarrow\left(x^3+4x^2+3x\right)\left(1+\frac{3}{MS}\right)=0\)

Dễ thấy MS >0 \(\Rightarrow PT\Leftrightarrow x^3+4x^2+x=0\Leftrightarrow x\left(x^2+4x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^2+4x+3=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\end{cases}}\)

\(\Rightarrow Pt\Leftrightarrow x^3+4x^2+3x=0\Leftrightarrow x\left(x^2+4x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^2+4x+3=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\end{cases}}\)\(\Rightarrow PT\Leftrightarrow x^3+4x^2+3x=0\)<=>\(x\in\left\{-3;-1;0\right\}\)

\(A=15x^2+\left(x^2-4x+4\right).\)

\(=15x^2+\left(x-2\right)^2\)

Vì \(15x^2\ge0\)với \(\forall x\)và \(\left(x-2\right)^2\ge0\)với \(\forall x\)

\(\Rightarrow A>0\)với mọi x 

\(B=x^2\left(x^2+6x+9\right)\)

\(=x^2\left(x+3\right)^2\)

Vì \(x^2\ge0\)với mọi x và \(\left(x+3\right)^2\ge0\)với mọi x 

\(\Rightarrow x^2\left(x+3\right)^2>0\)với mọi x

\(A=15x^2+\left(x^2-4x+4\right)\)

\(A=15x^2+x^2-4x+4\)

\(A=16x^2-4x+4\)

\(A=16x^2-4x+\frac{1}{4}+\frac{15}{4}\)

\(A=\left(16x^2-4x+\frac{1}{4}\right)+\frac{15}{4}\)

\(A=\left(4x+\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\)

Vậy A không âm

\(x^2+9x-400=0\)

\(\Leftrightarrow x^2-16x+25x-400=0\)

\(\Leftrightarrow x\left(x-16\right)+25\left(x-16\right)=0\)

\(\Leftrightarrow\left(x-16\right)\left(x+25\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=16\\x=-25\end{cases}}\)

\(a=1;b=9;c=-400\)

\(\Delta=b^2-4ac=9^2-4.1.\left(-400\right)=1681>0\)

Phương trình có 2 nghiệm phân biệt 

\(x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-9+\sqrt{1681}}{2.1}=16\)

\(x_2=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-9-\sqrt{1681}}{2.1}=-25\)

\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{x-3}{x+2\sqrt{x}+4}-\frac{7\sqrt{x}+10}{x\sqrt{x}-8}\right):\left(\frac{\sqrt{x}+7}{x+2\sqrt{x}+4}\right)\)

\(=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{x-3}{x+2\sqrt{x}+4}-\frac{7\sqrt{x}+10}{\sqrt{x}^3-8}\right):\left(\frac{\sqrt{x}+7}{x+2\sqrt{x}+4}\right)\)

\(=\left(\frac{\sqrt{x}\left(x+2\sqrt{x}+4\right)}{\sqrt{x}^3-8}-\frac{\left(x-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}^3-8}-\frac{7\sqrt{x}+10}{\sqrt{x}^3-8}\right)\)\(:\left(\frac{\sqrt{x}+7}{x+2\sqrt{x}+4}\right)\)

\(=\frac{\sqrt{x}^3+2x+4\sqrt{x}-\sqrt{x}^3+2x+3\sqrt{x}-6-7\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}.\frac{\left(x+2\sqrt{x}+4\right)}{\sqrt{x}+7}\)

\(=\)\(\frac{\left(4x-16\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}=\frac{4\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

Sai đề không ?

A= \(\left(\frac{\sqrt{x}\left(x+2\sqrt{x}+4\right)-\left(x-3\right)\left(\sqrt{x}-2\right)-7\sqrt{x}+10}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}\right)\)     .  \(\frac{x+2\sqrt{x}+4}{\sqrt{x}+7}\)

= \(\frac{x\sqrt{x}+2x+4\sqrt{x}-x\sqrt{x}+3\sqrt{x}-6+2x-7\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

= \(\frac{4x-16}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

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

= \(\frac{4\left(\sqrt{x}+2\right)}{\sqrt{x}+7}\)

= \(\frac{4\sqrt{x}+8}{\sqrt{x}+7}\)

#mã mã#

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

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

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

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)

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

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

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

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