Giải hệ phương trình : \(\hept{\begin{cases}\sqrt{2x-y-1}+\sqrt{3y+1}=\sqrt{x}+\sqrt{x+2y}\\x^3-3x+2=2y^3-y^2\end{cases}}\)
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\(\frac{a^2+2}{\sqrt{a^2+1}}=\frac{a^2+1+1}{\sqrt{a^2+1}}=\sqrt{a^2+1}+\frac{1}{\sqrt{a^2+1}}\ge2\)
\(\forall a\inℝ\)
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)
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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
\(\Leftrightarrow\)(a2 - 6ab + b2)2 \(\ge\)0 (luôn đúng)Vậy\(\left(a+b\right)^4\ge16ab\left(a-b\right)^2\)
\(\Leftrightarrow a^4+4ab^3+6a^2b^2+4a^3b+b^4\ge16ab\left(a^2-2ab+b^2\right)\)
\(\Leftrightarrow a^4+4ab^3+6a^2b^2+4a^3b+b^4\ge16a^3b-32a^2b^2+16ab^3\)
\(\Leftrightarrow a^4-12a^3b+38a^2b^2-12ab^3+b^4\ge0\)
\(\Leftrightarrow\left(a^2\right)^2-\left(b^2\right)^2+\left(6ab\right)^2+2a^2b^2-2.6aba^2-2.6abb^2\ge0\)
\(\Leftrightarrow\left(a^2-6ab+b^2\right)^2\ge0\)( luôn đúng )
Vậy ....
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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
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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\(\ge1\)
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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:
\(\hept{\begin{cases}x_1+x_2=2\left(m-1\right)\\x_1x_2=2m-3\end{cases}}\Leftrightarrow\hept{\begin{cases}x_1+x_2-1=2m-3\\x_1x_2=2m-3\end{cases}}\)
\(\Rightarrow x_1+x_2-1=x_1x_2\)
\(\Leftrightarrow x_1x_2-\left(x_1+x_2\right)+1=0\)
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