a) đk: \(x\ge0\)

Ta có: \(M< -\frac{1}{2}\)

\(\Leftrightarrow-\frac{3}{\sqrt{x}+3}+\frac{1}{2}< 0\)

\(\Leftrightarrow\frac{-6+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\)

\(\Leftrightarrow\frac{\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}< 0\)

Mà \(2\left(\sqrt{x}+3\right)>0\left(\forall x\right)\Rightarrow\sqrt{x}-3< 0\)

\(\Leftrightarrow\sqrt{x}< 3\Rightarrow x< 9\)

Vậy \(0\le x< 9\)

b) Ta có: \(M=\frac{-3}{\sqrt{x}+3}\ge-\frac{3}{0+3}=-1\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\sqrt{x}=0\Rightarrow x=0\)

Vậy Min(M) = -1 khi x = 0

Dạ sai đề ạ? 1 : 1 và 3 : 1 là tỉ lệ ở lai một cặp tính trạng

Xét: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}=\frac{bc\left(b-c\right)+ca\left(c-a\right)+ab\left(a-b\right)}{abc}\)

\(=\frac{b^2c-bc^2+ca\left(c-a\right)+a^2b-ab^2}{abc}=\frac{b^2\left(c-a\right)+ca\left(c-a\right)-b\left(c^2-a^2\right)}{abc}\)
\(=\frac{\left(c-a\right)\left(b^2+ca\right)-b\left(c-a\right)\left(c+a\right)}{abc}=\frac{\left(c-a\right)\left(b^2+ca-bc-ba\right)}{abc}\)

\(=\frac{\left(c-a\right)\left(b-a\right)\left(b-c\right)}{abc}=-\frac{\left(b-c\right)\left(c-a\right)\left(a-b\right)}{abc}=-\frac{1}{xyz}\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{-1}{xyz}\Leftrightarrow xy+yz+zx=-1\)

\(xy+yz+zx=\frac{a}{b-c}.\frac{b}{c-a}+\frac{b}{c-a}.\frac{c}{a-b}+\frac{c}{a-b}.\frac{a}{b-c}\)\(=\frac{ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

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

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

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

ĐKXĐ : \(x\ge\pm5\)

\(\sqrt{x-5}-3\sqrt{x^2-25}=0\)

\(\Leftrightarrow\sqrt{x-5}\left(1-3\sqrt{x+5}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-5}=0\\1-3\sqrt{x+5}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3\sqrt{x+5}=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\\sqrt{x+5}=\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x+5=\frac{1}{9}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=-\frac{44}{9}\end{cases}\left(tm\right)}\)

Vậy ....

đk: \(x\ge5\)

Ta có: \(\sqrt{x-5}-3\sqrt{x^2-25}=0\)

\(\Leftrightarrow\sqrt{x-5}=3\sqrt{x^2-25}\)

\(\Leftrightarrow x-5=9\left(x^2-25\right)\)

\(\Leftrightarrow9x^2-x-220=0\)

\(\Leftrightarrow\left(x-5\right)\left(9x+44\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\left(tm\right)\\x=-\frac{44}{9}\left(ktm\right)\end{cases}}\)

Vậy x = 5

ĐKXĐ: \(x\ge1\)hoặc  \(x< -3\)

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

\(\Leftrightarrow\frac{x-1}{x+3}=25\)

\(\Leftrightarrow x-1=25x+75\)

\(\Leftrightarrow x=-\frac{19}{6}\)(TM)

Vậy nghiệm của pt là: \(x=-\frac{19}{6}\)

ĐK: \(x\ge0;x\ne1\)

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

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

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

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

Để  \(P=\sqrt{x}\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}=\sqrt{x}\Leftrightarrow\sqrt{x}+1=\sqrt{x}\left(\sqrt{x}-1\right)\)\(\sqrt{x}+1\Leftrightarrow x-\sqrt{x}\Leftrightarrow-x+2\sqrt{x}+1=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1=\sqrt{2}\\\sqrt{x}-1=-\sqrt{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=\sqrt{2}+1\\\sqrt{x}=-\sqrt{2}+1\end{cases}\Leftrightarrow}x=3\pm2\sqrt{2}}\)

b) Với \(x>1\)thì \(P>0\)

Ta dễ thấy \(P=\frac{\sqrt{x}+1}{\sqrt{x}-1}>1\)

Ta có: \(P>0;P>1\)\(\Rightarrow P\left(P-1\right)>0\Leftrightarrow P^2>P\Leftrightarrow P>\sqrt{P}\)

Chứng minh cấm copy mạng nha.

3-1=2 1+1=2

2-1=1 1+1=2

\(A=\sqrt{7-2\sqrt{10}}+\sqrt{7+2\sqrt{10}}\)

\(A^2=\left(7+2\sqrt{10}+7-2\sqrt{10}\right)+2\sqrt{\left(7-2\sqrt{10}\right)\left(7+2\sqrt{10}\right)}\)

\(=14+2\sqrt{49-40}=14+6=20\)

Khi đó:\(A=\sqrt{20}\)

Các câu còn lại bạn làm nốt nhé

\(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\Leftrightarrow a+b+2\sqrt{ab}\ge a+b\)

\(\Leftrightarrow2\sqrt{ab}\ge0\)hiển nhiên đúng \(\forall a,b\ge0\) ---> ĐPCM

Có \(\sqrt{3}>\sqrt{1}\)

=> \(1-\sqrt{3}< 0\)

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

\(=\sqrt{2\left(1-\sqrt{3}\right)^2}=\sqrt{2}.\sqrt{\left(1-\sqrt{3}\right)^2}=\sqrt{2}.\left(\sqrt{3}-1\right)=\sqrt{6}-\sqrt{2}\)

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

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