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

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

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

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

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

\(=-2\)

\(\Rightarrow D=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)

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

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

\(D^2=2\)

=> \(D=\sqrt{2}\)

a/ \(P=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)

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

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

\(P=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(1-\sqrt{x}\right)^2\left(1+\sqrt{x}\right)^2}{2}\)

=> \(P=-\sqrt{x}\left(\sqrt{x}-1\right)\)

b/ Nếu 0<x<1 => \(\sqrt{x}-1< 0\); và \(\sqrt{x}>0\)

=> \(P=-\sqrt{x}\left(\sqrt{x}-1\right)>0\)

c/ \(P=-\sqrt{x}\left(\sqrt{x}-1\right)=-x+\sqrt{x}=-x+2.\frac{1}{2}\sqrt{x}-\frac{1}{4}+\frac{1}{4}\)

=> \(P=\frac{1}{4}-\left(\sqrt{x}-\frac{1}{2}\right)^2\le\frac{1}{4}\)

=> \(P_{max}=\frac{1}{4}\)

Đạt được khi x=1/4

a) ĐKXĐ : \(\hept{\begin{cases}\frac{1}{1-x^2}>0\\1-x^2\ne0\end{cases}}\)

Mà 1 > 0 nên \(\Leftrightarrow1-x^2>0\)

\(\Leftrightarrow x^2< 1\)

\(\Leftrightarrow-1< x< 1\)

Vậy ...

b) Có \(\frac{1}{1+x^2}>0\) với mọi x nên biểu thức XĐ với mọi x.

3 số thực dương nhé.

Áp dụng bất đẳng thức Cauchy Schwarz dạng Engel có :

\(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab}\ge\frac{\left(1+1+1\right)^2}{\left(a^2+2bc\right)+\left(b^2+2ca\right)+\left(c^2+2ab\right)}=\frac{9}{\left(a+b+c\right)^2}\ge\frac{9}{1^2}=9\)

Dấu bằng xảy ra \(\Leftrightarrow\frac{1}{a^2+2bc}=\frac{1}{b^2+2ca}=\frac{1}{c^2+2ab}\)và \(a+b+c=1\)

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

Mong có ai giúp mình từ đẳng thức trên giải ra a=b=c.

a=b=c ket hop voi a+b+c=<1 =>a=b=c=1/3 nhe

Đề thi vào lớp 10 môn toán chuyên Sư Phạm Hà Nội năm 2020-2021

ta có BĐT \(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(a+c-b\right)\)(chứng minh = AM-GM)

\(abc\ge\left(2-2a\right)\left(2-2b\right)\left(2-2c\right)=8\left(1-a\right)\left(1-b\right)\left(1-c\right)\)

\(abc\ge8\left[1-\left(a+b+c\right)+\left(ab+bc+ca\right)-abc\right]\)

\(\Leftrightarrow9abc\ge-8+8\left(ab+bc+ca\right)\)

do đó \(VT\ge4\left(a^2+b^2+c^2\right)+8\left(ab+bc+ca\right)-8\)

\(VT\ge4\left(a+b+c\right)^2-8=16-8=8\)

Dấu = xảy ra khi \(a=b=c=\frac{2}{3}\)