\(ĐKXĐ:x\le0\)

\(\sqrt{-2x}-\sqrt{2-x}=0\)

\(\sqrt{-2x}=\sqrt{2-x}\)

\(\left|-2x\right|=\left|2-x\right|\)

\(-2x=2-x\)

\(-x=2\)

\(x=-2\left(TM\right)\)

bài tập hình thiếu câu hỏi nhé, hay bạn chỉ hỏi mỗi đại thôi ? 

a, \(A=2x^2+1\ge1\forall x\)

Dấu ''='' xảy ra khi x = 0 

Vậy GTNN A bằng 1 tại x = 0 

b, \(B=x^2-3x+2=x^2-2.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}+2\)

\(=\left(x-\frac{3}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\forall x\)

Dấu ''='' xảy ra khi x = 3/2 

Vậy GTNN B bằng -1/4 tại x = 3/2 

c, ĐK : x >= 0

\(C=2x-\sqrt{x}=2\left(x-\frac{1}{2}\sqrt{x}\right)=2\left(x-2.\frac{1}{4}\sqrt{x}+\frac{1}{16}-\frac{1}{16}\right)\)

\(=2\left(\sqrt{x}-\frac{1}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\forall x\)

Dấu ''='' xảy ra khi x = 1/16 

Vậy GTNN C bằng -1/8 tại 1/16 

d, \(D=3\sqrt{x}-x=-\left(x-3\sqrt{x}\right)=-\left(x-2.\frac{3}{2}\sqrt{x}+\frac{9}{4}-\frac{9}{4}\right)\)

\(=-\left(\sqrt{x}-\frac{3}{2}\right)^2+\frac{9}{4}\le\frac{9}{4}\forall x\)

Dấu ''='' xảy ra khi x = 9/4 

Vậy GTLN D bằng 9/4 tại x = 9/4 

bổ sung đk ý D hộ mình ĐK : x >= 0 

\(A=\left(\frac{x\sqrt{x}-1}{\sqrt{x}-1}+\sqrt{x}\right):\left(x-1\right)-\frac{2}{\sqrt{x}-1}\)ĐK : \(x\ge0;x\ne1\)

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

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

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

\(a,\sqrt{4x-20}-3\sqrt{\frac{x-5}{9}}=\sqrt{1-x}\)

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

\(2\sqrt{x-5}-\sqrt{x-5}=\sqrt{1-x}\)

\(\sqrt{x-5}=\sqrt{1-x}\)

\(ĐKXĐ:x\ge5;x\le1\)

vậy pt vô nghiệm

\(b,ĐKXĐ:x\ge2\)

\(b,\sqrt{50x-25}+\sqrt{8x-4}-3\sqrt{x}=\sqrt{72x}\)

\(5\sqrt{2x-1}+2\sqrt{2x-1}-3\sqrt{x}=6\sqrt{2x}\)

\(7\sqrt{2x+1}=6\sqrt{2x}+3\sqrt{x}\)

\(49\left(2x+1\right)=72x+9x+36\sqrt{2}x\)

\(98x+49=81x+36\sqrt{2}x\)

\(17x+49=36\sqrt{2}x\)

\(x\left(17+36\sqrt{2}\right)=-49\)

\(x=\frac{-49}{17+36\sqrt{2}}\left(KTM\right)\)

vậy pt vô nghiệm

\(c,ĐKXĐ:x\ge3\)

\(c,\sqrt{x^2-9}-\sqrt{4x-12}=0\)

\(\sqrt{x-3}\sqrt{x+3}-2\sqrt{x-3}=0\)

\(\sqrt{x-3}\left(\sqrt{x+3}-2\right)=0\)

\(\orbr{\begin{cases}\sqrt{x-3}=0\\\sqrt{x+3}-2=0\end{cases}}\orbr{\begin{cases}x=3\left(TM\right)\\x=1\left(KTM\right)\end{cases}}\)

\(d,ĐKXĐ:x\ge-1\)

\(\sqrt{9x+9}-2\sqrt{\frac{x+1}{4}}=4\)

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

\(2\sqrt{x+1}=4\)

\(x+1=4\)

\(x=3\left(TM\right)\)

\(K=x+2\sqrt{x-1}=x-1+2\sqrt{x-1}+1=\left(\sqrt{x-1}+1\right)^2\ge\left(0+1\right)^2=1\)

Dấu ''='' xảy ra khi \(\sqrt{x-1}=0\Leftrightarrow x=1\)

Vậy GTNN K bằng 1 tại x = 1 

bổ sung đk hộ mình x >= 1 

nó là a^5-a^2 nha

Áp dụng BĐT Cauchy Schwarz dạng Engel 

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

Theo BĐT : \(a^2+b^2+c^2\ge ab+ac+bc\)

\(2a^2+2b^2+2c^2\ge2ab+2ac+2bc\)

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)* đúng * 

\(=\frac{\left(1+1+1\right)^2}{2\left(ab+ac+bc\right)}=\frac{9}{2.3}=\frac{3}{2}\)

Dấu ''='' xảy ra khi a = b = c = 1 

ngược dấu rồi nó ở mẫu mà

ta có \(xy\le\frac{x^2+y^2}{2}\Rightarrow\frac{x^2}{x^2+xy+y^2}\ge\frac{2}{3}.\frac{x^2}{x^2+y^2}\)

Vậy \(A\ge\frac{2}{3}\left(\frac{x^2}{x^2+y^2}+\frac{y^2}{y^2+z^2}+\frac{z^2}{z^2+x^2}\right)=\frac{2}{3}\left(\frac{x^2+y^2+z^2}{x^2+y^2}-1+\frac{x^2+y^2+z^2}{y^2+z^2}-1+\frac{x^2+y^2+z^2}{x^2+z^2}-1\right)\)

hay \(A\ge\frac{2\left(x^2+y^2+z^2\right)}{3}\left(\frac{1}{x^2+y^2}+\frac{1}{y^2+z^2}+\frac{1}{z^2+x^2}\right)-2\)

\(\Rightarrow A\ge\frac{2}{3}\left(x^2+y^2+z^2\right).\frac{9}{2\left(x^2+y^2+z^2\right)}-2=3-2=1\)

Vậy ta có đpcm

dòng 4 sai rồi

ta có :

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

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

ta có 

\(S\ge\frac{1}{\frac{1+1998}{2}}+\frac{1}{\frac{2+1997}{2}}+..+\frac{1}{\frac{k+1998-k+1}{2}}+..+\frac{1}{\frac{1999}{2}}\)

hay \(S\ge\frac{2}{1999}+\frac{2}{1999}+..+\frac{2}{1999}=2.\frac{1998}{1999}\)

do dấu = không xảy ra nên \(S>2.\frac{1998}{1999}\)