giải pt (1) ta có:

\(\sqrt{2x-y-1}\)- \(\sqrt{x+2y}\)+ \(\sqrt{3y+1}\)- \(\sqrt{x}\)=0

\(\frac{2x-y-1-x-2y}{\sqrt{2x-y-1}+\sqrt{x+2y}}\)+\(\frac{3y+1-x}{\sqrt{3y+1}+\sqrt{x}}\)=0

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

=> x=3y+1 thay vào (2) => x=1; y=0

trường hợp 2:

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

=> \(\sqrt{3y+1}+\sqrt{x}\)=\(\sqrt{x+2y}+\sqrt{2x-y-1}\)

=> \(\sqrt{x}\)- \(\sqrt{2x-y-1}\)+ \(\sqrt{3y+1}\)- \(\sqrt{x+2y}\)=0

=> \(\frac{x-2x+y+1}{\sqrt{x}+\sqrt{2x-y-1}}\)+\(\frac{3y+1-x-2y}{\sqrt{3y+1}+\sqrt{x+2y}}\)=0

=>(-x + y + 1)(\(\frac{1}{\sqrt{x}+\sqrt{2x-y-1}}\)+ \(\frac{1}{\sqrt{3y+1}+\sqrt{x+2y}}\))=0

mà \(\frac{1}{\sqrt{x}+\sqrt{2x-y-1}}\)+\(\frac{1}{\sqrt{3y+1}+\sqrt{x+2y}}\)>0

=> x=y+1 thay vào 2 => \(\hept{\begin{cases}x=1\\y=0\end{cases}}\)

để đấy ku

\(\frac{a^2+2}{\sqrt{a^2+1}}=\frac{a^2+1+1}{\sqrt{a^2+1}}=\sqrt{a^2+1}+\frac{1}{\sqrt{a^2+1}}\ge2\)

\(\forall a\inℝ\)

ta có: a² + 2 \(\ge\)\(2\sqrt{a^2+1}\)

\(\Rightarrow\)a² + 1 -\(2\sqrt{a^2+1}\)+ 1 \(\ge\)0

\(\Rightarrow\)(\(\sqrt{a^2+1}\)-  1)² \(\ge\)0 (luôn đúng)

Chỉ cần bỏ chữ cmr là đc

làm hộ mn với,vì đây là đề dành cho học sinh giỏi,mn sao chép về

ta có: (a+b)⁴\(\ge\)16ab(a-b)²

\(\Leftrightarrow\)a⁴ + 4ab³ + 4a³b + b⁴\(\ge\)16ab(a² - 2ab + b²)

\(\Leftrightarrow\)a⁴ + 4ab³ + 4a³b + b⁴\(\ge\)16a³b - 32a²b² + 16ab³

\(\Leftrightarrow\)a⁴ - 12a³b + 38a²b² - 12ab³ + b⁴ \(\ge\)0

\(\left(a+b\right)^4\ge16ab\left(a-b\right)^2\)

\(\Leftrightarrow a^4+4ab^3+6a^2b^2+4a^3b+b^4\ge16ab\left(a^2-2ab+b^2\right)\)

​​\(\Leftrightarrow a^4+4ab^3+6a^2b^2+4a^3b+b^4\ge16a^3b-32a^2b^2+16ab^3\)

\(\Leftrightarrow a^4-12a^3b+38a^2b^2-12ab^3+b^4\ge0\)

\(\Leftrightarrow\left(a^2\right)^2-\left(b^2\right)^2+\left(6ab\right)^2+2a^2b^2-2.6aba^2-2.6abb^2\ge0\) 

\(\Leftrightarrow\left(a^2-6ab+b^2\right)^2\ge0\)( luôn đúng )

Vậy ....

Ta có: \(ab+\frac{a}{b}+\frac{b}{a}\)

\(=\frac{a^2b^2+a^2+b^2}{ab}\ge\frac{ab\cdot a+ab\cdot b+a\cdot b}{ab}=\frac{ab\left(a+b+1\right)}{ab}=a+b+1\)

Dấu "=" xảy ra khi: a = b = 1

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

vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\)

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

Vậy P\(\ge1\)

1) Với m = 1 thì ta có:

\(x^2-2\left(1-1\right)x+2\cdot1-3=0\)

\(\Leftrightarrow x^2-1=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)

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

2) Ta có: \(\Delta^'=\left[-\left(m-1\right)\right]^2-\left(2m-3\right)\cdot1=m^2-2m+1-2m+3\)

\(=m^2-4m+4=\left(m-2\right)^2\ge0\left(\forall m\right)\)

=> PT luôn có nghiệm với mọi m

Theo hệ thức viet ta có:

\(\hept{\begin{cases}x_1+x_2=2\left(m-1\right)\\x_1x_2=2m-3\end{cases}}\Leftrightarrow\hept{\begin{cases}x_1+x_2-1=2m-3\\x_1x_2=2m-3\end{cases}}\)

\(\Rightarrow x_1+x_2-1=x_1x_2\)

\(\Leftrightarrow x_1x_2-\left(x_1+x_2\right)+1=0\)