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Câu 1:
a/ Biểu thức không tồn tại GTNN.
Bạn cứ thử với vài giá trị âm có trị tuyệt đối lớn, ví dụ \(a=-10^3\) và \(b=-\frac{1}{10^3}\) sẽ thấy
b/
\(x^3+3x^2+3x+1+y^3+3y^2+3y+1+x+y+2=0\)
\(\Leftrightarrow\left(x+1\right)^3+\left(y+1\right)^3+x+y+2=0\)
\(\Leftrightarrow\left(x+y+2\right)\left[\left(x+1\right)^2+\left(y+1\right)^2-\left(x+1\right)\left(y+1\right)\right]+x+y+2=0\)
\(\Leftrightarrow\left(x+y+2\right)\left[\left(x+1-\frac{y+1}{2}\right)^2+\frac{3\left(y+1\right)^2}{4}+1\right]=0\)
\(\Rightarrow x+y=-2\Rightarrow\left\{{}\begin{matrix}x< 0\\y< 0\end{matrix}\right.\)
\(\Rightarrow-x+\left(-y\right)=2\)
\(M=\frac{1}{x}+\frac{1}{y}=-\left(\frac{1}{-x}+\frac{1}{-y}\right)\le-\frac{4}{-x+\left(-y\right)}=-\frac{4}{2}=-2\)
\(\Rightarrow M_{max}=-2\) khi \(x=y=-1\)
1c/
\(T=\sum\frac{a}{2a+a+b+c}=\frac{1}{25}\sum\frac{a\left(2+3\right)^2}{2a+a+b+c}\le\frac{1}{25}\sum\left(\frac{4a}{2a}+\frac{9a}{a+b+c}\right)\)
\(\Rightarrow T\le\frac{1}{25}\left(6+\frac{9\left(a+b+c\right)}{a+b+c}\right)=\frac{15}{25}=\frac{3}{5}\)
Dấu "=" xảy ra khi \(a=b=c\)
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
bài 2
ta có \(\left(\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\right)^2\)
\(=\left(\sqrt{a}.\sqrt{\frac{8a^2+1}{a}}+\sqrt{b}.\sqrt{\frac{8b^2+1}{b}}+\sqrt{c}.\sqrt{\frac{8c^2+1}{c}}\right)^2\)\(=\left(A\right)\)
Áp dụng bất đẳng thức Bunhiacopxki ta có;
\(\left(A\right)\le\left(a+b+c\right)\left(8a+\frac{1}{a}+8b+\frac{1}{b}+8c+\frac{8}{c}\right)\)
\(=\left(a+b+c\right)\left(9a+9b+9c\right)=9\left(a+b+c\right)^2\)
\(\Rightarrow3\left(a+b+c\right)\ge\sqrt{8a^2+1}+\sqrt{8b^2+1}+\sqrt{8c^2+1}\)(đpcm)
Dấu \(=\)xảy ra khi \(a=b=c=1\)
7. \(S=9y^2-12\left(x+4\right)y+\left(5x^2+24x+2016\right)\)
\(=9y^2-12\left(x+4\right)y+4\left(x+4\right)^2+\left(x^2+8x+16\right)+1936\)
\(=\left[3y-2\left(x+4\right)\right]^2+\left(x-4\right)^2+1936\ge1936\)
Vậy \(S_{min}=1936\) \(\Leftrightarrow\) \(\hept{\begin{cases}3y-2\left(x+4\right)=0\\x-4=0\end{cases}}\) \(\Leftrightarrow\) \(\hept{\begin{cases}x=4\\y=\frac{16}{3}\end{cases}}\)
8. \(x^2-5x+14-4\sqrt{x+1}=0\) (ĐK: x > = -1).
\(\Leftrightarrow\) \(\left(x+1\right)-4\sqrt{x+1}+4+\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow\) \(\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2=0\)
Với mọi x thực ta luôn có: \(\left(\sqrt{x+1}-2\right)^2\ge0\) và \(\left(x-3\right)^2\ge0\)
Suy ra \(\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2\ge0\)
Đẳng thức xảy ra \(\Leftrightarrow\) \(\hept{\begin{cases}\left(\sqrt{x+1}-2\right)^2=0\\\left(x-3\right)^2=0\end{cases}}\) \(\Leftrightarrow\) x = 3 (Nhận)
5/ĐK: \(\left[{}\begin{matrix}x\le-1\\x\ge5\end{matrix}\right.\)
PT \(\Leftrightarrow2\left(x^2-4x-6\right)+\sqrt{x^2-4x-5}-1=0\)
\(\Leftrightarrow\left(x^2-4x-6\right)\left(2+\frac{1}{\sqrt{x^2-4x-5}+1}\right)=0\)
\(\Leftrightarrow x^2-4x-6=0\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{10}\\x=2-\sqrt{10}\end{matrix}\right.\)
Vậy..
Câu 2 a chắc là bình phương hai vế lên,đặt ẩn phụ rồi... chăng?