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a. Để \(\frac{\sqrt{x-3}}{2x+1}\)có nghĩa thì 2x+1 \(\ne\)0
\(\Leftrightarrow\)2x \(\ne\)-1
\(\Leftrightarrow\)x \(\ne\)\(\frac{-1}{2}\)
b. Để \(\frac{\sqrt{1-2x}}{x^2-6x+9}\) có nghĩa thì x2-6x+9\(\ne\)0
\(\Leftrightarrow\)(x-3)2 \(\ne\)0
\(\Leftrightarrow\)x-3 \(\ne\)0
\(\Leftrightarrow\)x \(\ne\)3
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{\sqrt{x}-1}\)
Biểu thức \(A\) có nghĩa khi \(\hept{\begin{cases}\sqrt{x}+1\ne0;\text{ }x\ge0\\\sqrt{x}-1\ne0\end{cases}}\) \(\Leftrightarrow\) \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
Ta có:
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{\sqrt{x}-1}=\frac{\sqrt{x}\left(\sqrt{x}+1\right)-2\left(\sqrt{x}-1\right)-2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\frac{x+\sqrt{x}-2\sqrt{x}+2-2\sqrt{x}-2}{x-1}=\frac{x-3\sqrt{x}}{x-1}\)
Vậy, \(A=\frac{x-3\sqrt{x}}{x-1}\)
5x - y = 1 => y=5x-1
Do đó 2x + 3y = 17x - 3
Vì 17x -3 > 5x -1 ( do x >0 )
Để y > 0 thì 5x -1 > 0 => x > 1/5
Suy ra m > 2/5
Vậy m > 2/5 thì nghiệm x , y thỏa mãn đề bài
P = ( x + 9 + 6cănx - 6cănx - 18 + 25 ) / (cănx + 3)
áp dụng bđt Côsi suy ra min P = 4 khi x = 4
Ta có: \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2ab+2bc+2ac}\)
Mặt khác : \(a^2+b^2+c^2\ge ab+bc+ac\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)\(\Rightarrow\frac{\left(a+b+c\right)^2}{2ab+2bc+2ac}\ge\frac{3}{2}\)
Dự đoán \(MinL=\frac{3}{2}\)khi a = b = c
Ta cần chứng minh \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\ge\frac{3}{2}\Leftrightarrow\left(\frac{a}{a+b}-\frac{1}{2}\right)+\left(\frac{b}{b+c}-\frac{1}{2}\right)+\left(\frac{c}{c+a}-\frac{1}{2}\right)\ge0\)\(\Leftrightarrow\frac{a-b}{2\left(a+b\right)}+\frac{b-c}{2\left(b+c\right)}+\frac{c-a}{2\left(c+a\right)}\ge0\Leftrightarrow\frac{a-b}{2\left(a+b\right)}-\frac{\left(a-b\right)+\left(c-a\right)}{2\left(b+c\right)}+\frac{c-a}{2\left(c+a\right)}\ge0\)\(\Leftrightarrow\frac{a-b}{2\left(a+b\right)}-\frac{a-b}{2\left(b+c\right)}-\frac{c-a}{2\left(b+c\right)}+\frac{c-a}{2\left(c+a\right)}\ge0\)\(\Leftrightarrow\frac{a-b}{2}\left(\frac{1}{a+b}-\frac{1}{b+c}\right)-\frac{c-a}{2}\left(\frac{1}{b+c}-\frac{1}{c+a}\right)\ge0\)\(\Leftrightarrow\frac{a-b}{2}.\frac{c-a}{\left(a+b\right)\left(b+c\right)}-\frac{c-a}{2}.\frac{a-b}{\left(b+c\right)\left(c+a\right)}\ge0\)\(\Leftrightarrow\frac{\left(a-b\right)\left(c-a\right)\left(c+a\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}-\frac{\left(a-b\right)\left(c-a\right)\left(a+b\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)\(\Leftrightarrow\frac{\left(a-b\right)\left(c-a\right)\left(c-b\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\Leftrightarrow\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)(đúng do \(a\ge b\ge c>0\))
Đẳng thức xảy ra khi a = b = c
Đặt \(x=a;\frac{1}{y}=b\Rightarrow a,b>0;a^2+b^2=1\). Quy về tìm Min \(A=ab+\frac{1}{ab}\)
Ta có: \(A=\left(4ab+\frac{1}{ab}\right)-3ab\ge2\sqrt{4ab.\frac{1}{ab}}-\frac{3}{2}\left(a^2+b^2\right)=4-\frac{3}{2}=\frac{5}{2}\)
Đẳng thức xảy ra khi \(\hept{\begin{cases}4ab=\frac{1}{ab}\\a=b\end{cases}}\Leftrightarrow\hept{\begin{cases}2ab=1\\a=b\end{cases}}\Rightarrow a=b=\frac{1}{\sqrt{2}}\) (thỏa mãn \(a^2+b^2=1\))
\(\Rightarrow x=\frac{1}{\sqrt{2}};y=\sqrt{2}\)
Vậy...
Áp dụng bđt Cauchy :
\(B=\frac{x^3+200}{x}=x^2+\frac{200}{x}=x^2+\frac{100}{x}+\frac{100}{x}\ge3.\sqrt[3]{x^2.\frac{100}{x}.\frac{100}{x}}=30\sqrt[3]{10}\)
Dấu "=" xảy ra khi \(x^2=\frac{100}{x}\)=> ..................
Vậy Min B = ............... tại x = .......................