CMR: \(\frac{x^2+x+1}{x^2-x+1}>0\)
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Xho x+y=1 và x,y khác 0
CMR \(\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2+y^2+3}=0\)
Bạn vt đề bài rõ ra nhé, mk RG trc rùi phần câu hỏi xem sau( P là j z?)
\(=\frac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}-2\)
\(=x-\sqrt{x}-3\)
câu 1.Ta có:
\(\frac{x^2}{x+3y}+\frac{x+3y}{16}\ge2\sqrt{\frac{x^2}{x+3y}.\frac{x+3y}{16}}=\frac{x}{2}\)
\(\frac{y^2}{y+3x}+\frac{y+3x}{16}\ge2\sqrt{\frac{y^2}{y+3x}.\frac{y+3x}{16}}=\frac{y}{2}\)
\(\frac{x^2}{x+3y}+\frac{y^2}{y+3x}+\frac{x+y+3x+3y}{16}\ge\frac{x+y}{2}\)
\(\frac{x^2}{x+3y}+\frac{y^2}{y+3x}+\frac{1}{4}\ge\frac{1}{2}\)
\(\frac{x^2}{x+3y}+\frac{y^2}{y+3x}\ge\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\left(đpcm\right)\)
Câu 2:
điều kiện \(a^2+b^2+c^2+d^2=4\)(đúng ko)
Ta có:
\(\frac{1}{a^2+1}+\frac{a^2+1}{4}\ge2\sqrt{\frac{1}{a^2+1}.\frac{a^2+1}{4}}=1\)
\(\frac{1}{b^2+1}.\frac{b^2+1}{4}\ge2\sqrt{\frac{1}{b^2+1}.\frac{b^2+1}{4}}=1\)
\(\frac{1}{c^2+1}+\frac{c^2+1}{4}\ge2\sqrt{\frac{1}{c^2+1}.\frac{c^2+1}{4}}=1\)
\(\frac{1}{d^2+1}+\frac{d^2+1}{4}\ge2\sqrt{\frac{1}{d^2+1}.\frac{d^2+1}{4}}=1\)
\(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}+\frac{1}{d^2+1}+\frac{a^2+b^2+c^2+d^2+4}{4}\ge4\)
\(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}+\frac{1}{d^2+1}\ge4-\frac{8}{4}=2\left(đpcm\right)\)
a)\(B=\left(\frac{x-2}{x^2+2x}+\frac{1}{x+2}\right).\frac{x+1}{x-1}=\left(\frac{x^2-2}{x^2+2x}+\frac{x}{x^2+2x}\right).\frac{x+1}{x-1}=\frac{x^2+x-2}{x^2+2x}.\frac{x+1}{x-1}\)
\(=\frac{x^2-x+2x-2}{x\left(x+2\right)}.\frac{x+1}{x-1}=\frac{x\left(x-1\right)+2\left(x-1\right)}{x\left(x+2\right)}.\frac{x+1}{x-1}=\frac{\left(x-1\right)\left(x+2\right)}{x\left(x+2\right)}.\frac{x+1}{x-1}=\frac{x+1}{x}\)
b)\(2B=2x+5\Leftrightarrow2.\frac{x+1}{x}=2x+5\Leftrightarrow\frac{2x+2}{x}=2x+5\Leftrightarrow2x+2=2x^2+5x\)
\(\Leftrightarrow0=2x^2+3x-2\Leftrightarrow2x^2+4x-x-2=0\Leftrightarrow2x\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+2\right)=0\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-2\end{cases}}\)
Xin lỗi bạn, mình mới học lớp 6 nén không bít trả lời cau hỏi này.
Xin lỗi nha!
Ta có : \(x^2+x+1\)
\(=x^2+2.\frac{1}{2}x+\frac{1}{4}-\frac{1}{4}+1\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
CMTT \(x^2-x+1>0\)
\(\RightarrowĐPCM\)