kết bạn nha

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

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

\(=\sqrt{4-3}=1\)

ĐẶT \(A=\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}+\sqrt{6}\)

\(\Rightarrow\sqrt{2}A=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}+2\sqrt{3}\)

\(\Leftrightarrow\sqrt{2}A=\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}+2\sqrt{3}\)

\(\Leftrightarrow\sqrt{2}A=\left|\sqrt{3}-\sqrt{5}\right|-\left|\sqrt{3}+\sqrt{5}\right|+2\sqrt{3}\)

\(\Leftrightarrow\sqrt{2}A=\sqrt{5}-\sqrt{3}-\sqrt{3}-\sqrt{5}+2\sqrt{3}=0\Rightarrow A=0\)

VẬY A = 0

\(P=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)

\(P=\frac{2\sqrt{x}+2-3}{\sqrt{x}+1}\)

\(P=2-\frac{3}{\sqrt{x}+1}\)

\(\sqrt{x}+1\ge1\)

\(\frac{3}{\sqrt{x}+1}\le3\)

\(P=2-\frac{3}{\sqrt{x}+1}\ge-1\)

dấu "=" xảy ra khi \(x=0\)

\(MIN=-1\)

Lại phải cảm ơn bạn lần nữa r :D

Trả lời:

\(\sqrt{11-6\sqrt{2}}\)

\(=\sqrt{2-6\sqrt{2}+9}\)

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

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

\(=\left|\sqrt{2}-3\right|\)

\(=3-\sqrt{2}\) ( vì \(2< \sqrt{3}\Leftrightarrow2-\sqrt{3}< 0\))

\(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{\left(a+b+c\right)^2}{2a+2b+2c}\)

\(=\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\)

dấu "=" xảy ra khi \(\frac{a}{a+b}=\frac{b}{b+c}=\frac{c}{c+a}\)

ÁP DỤNG BẤT ĐẲNG THỨC BUNYAKOVSKY DẠNG PHÂN THỨC TA CÓ :

\(\frac{A^2}{B+C}+\frac{B^2}{A+C}+\frac{C^2}{A+B}\ge\frac{\left(A+B+C\right)^2}{B+C+A+C+A+B}=\frac{\left(A+B+C\right)^2}{2\left(A+B+C\right)}=\frac{A+B+C}{2}\)

=> ĐPCM. ĐẲNG THỨC XẢY RA <=> A = B = C

Câu 1 : tự làm nhé

Câu 2 : 

a, \(\sqrt{2x-15}=3\)ĐK : \(x\ge\frac{15}{2}\)

bình phương 2 vế : \(2x-15=9\Leftrightarrow x=12\)( tmđk )

b, \(\sqrt{x^2-6x+9}=5\Leftrightarrow\sqrt{\left(x-3\right)^2}=5\Leftrightarrow\left|x-3\right|=5\)

TH1 : \(x-3=5\Leftrightarrow x=8\)

TH2 : \(x-3=-5\Leftrightarrow x=-2\)

Câu 3 : Với \(a\ge0;a\ne1\)

a, \(B=\left(\frac{a+\sqrt{a}}{\sqrt{a}+1}+\frac{3-3\sqrt{a}}{1-\sqrt{a}}\right):\frac{3-\sqrt{a}}{9-a}\)

\(=\left(\sqrt{a}+3\right):\frac{1}{3+\sqrt{a}}=\left(\sqrt{a}+3\right)^2\)

b, Ta có : \(B=\left(\sqrt{a} +3\right)^2=21\Leftrightarrow a+6\sqrt{a}-12=0\)

dùng tam thức bậc 2 nhé =)) 

\(\frac{\sqrt{x}+3}{\sqrt{x}+2}\)

\(\frac{\sqrt{x}+2+1}{\sqrt{x}+2}=1+\frac{1}{\sqrt{x}+2}\)

\(\sqrt{x}+2\ge2\)

\(1+\frac{1}{\sqrt{x}+2}\le1+\frac{1}{2}=\frac{3}{2}\)

dấu "=" xảy ra khi \(x=0\)

\(< =>MAX=\frac{3}{2}\)

Đặt \(\hept{\begin{cases}-a+2b+2c=x\\2a-b+2c=y\\2a+2b-c=z\end{cases}}\)\(\Rightarrow\hept{\begin{cases}a=\frac{2y+2z-x}{9}\\b=\frac{2z+2x-y}{9}\\c=\frac{2x+2y-z}{9}\end{cases}}\)

Vì a,b,c là độ dài 3 cạnh của 1 tam giác nên x,y,z>0

Khi đó : \(VT=\frac{2y+2z-x}{9x}+\frac{2z+2x-y}{9y}+\frac{2x+2y-z}{9z}\)

\(=\frac{2}{9}\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{2}{9}\left(\frac{y}{z}+\frac{z}{y}\right)+\frac{2}{9}\left(\frac{z}{x}+\frac{x}{z}\right)-\frac{1}{3}\)

\(\ge\frac{2}{9}.2+\frac{2}{9}.2+\frac{2}{9}.2-\frac{1}{3}\)(BĐT Cauchy cho 2 số không âm)

\(=\frac{4}{9}.3-\frac{1}{3}=\frac{4}{3}-\frac{1}{3}=1\)

\(\frac{a}{2b+2c-a}+\frac{b}{2a+2c-b}+\frac{c}{2a+2b-c}\)

\(\frac{a^2}{2ab+2ac-a^2}+\frac{b^2}{2ab+2bc-b^2}+\frac{c^2}{2ac+2bc-c^2}\)

đặt pt là P

\(P\ge\frac{\left(a+b+c\right)^2}{2ab+2ac-a^2+2ab+2bc-b^2+2ac+2bc-c^2}\)

\(P\ge\frac{\left(a+b+c\right)^2}{4ab+4ac+4bc-a^2-b^2-c^2}\)

\(a^2+b^2+c^2\ge2ab+2bc+2ca\)(BĐT tương đương)

\(P\ge\frac{\left(a+b+c\right)^2}{4ab+4ac+4bc-a^2-b^2-c^2}\ge\frac{\left(a+b+c\right)^2}{2ab+2ac+2bc}\)

\(\left(a+b+c\right)^2\ge2ab+2ac+2bc\)(BĐT tương đương)

\(P\ge1\)

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