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a) ĐKXĐ: \(x\ne1\)
Ta có: \(x^2-8x+7=0\)
\(\Leftrightarrow x^2-x-7x+7=0\)
\(\Leftrightarrow x\left(x-1\right)-7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(loại\right)\\x=7\left(nhận\right)\end{matrix}\right.\)
Thay x=7 vào B, ta được:
\(B=\dfrac{1}{7-1}=\dfrac{1}{6}\)
Vậy: Khi \(x^2-8x+7=0\) thì \(B=\dfrac{1}{6}\)
b) Ta có: \(A=\dfrac{x^2+2}{x^3-1}+\dfrac{x+1}{x^2+x+1}\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}\)
\(=\dfrac{x^2+2+x^2-1}{x^3-1}\)
\(=\dfrac{2x^2+1}{x^3-1}\)
2: Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=\dfrac{a\left(a+b+c\right)}{b+c}+\dfrac{b\left(a+b+c\right)}{c+a}+\dfrac{c\left(a+b+c\right)}{a+b}-a-b-c=\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c-a-b-c=0\)
1: Sửa đề: Cho \(x,y,z\ne0\) và \(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}=\dfrac{2}{2x+y+2z}\).
CM:....
Đặt 2x = x', 2z = z'.
Ta có: \(\dfrac{2}{x'}+\dfrac{2}{y}+\dfrac{2}{z'}=\dfrac{2}{x'+y+z'}\)
\(\Leftrightarrow\dfrac{1}{x'}+\dfrac{1}{y}+\dfrac{1}{z'}=\dfrac{1}{x'+y+z'}\)
\(\Leftrightarrow\dfrac{1}{x'}-\dfrac{1}{x'+y+z'}+\dfrac{1}{y}+\dfrac{1}{z'}=0\)
\(\Leftrightarrow\dfrac{y+z'}{x'\left(x'+y+z'\right)}+\dfrac{y+z'}{yz'}=0\)
\(\Leftrightarrow\dfrac{\left(y+z'\right)\left(yz'+x'^2+x'y+x'z'\right)}{x'yz'\left(x'+y+z'\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x'+y\right)\left(y+z'\right)\left(z'+x'\right)}{x'yz'\left(x'+y+z'\right)}=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(2z+2x\right)=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(z+x\right)=0\left(đpcm\right)\)
a.
Giả sử: \(\dfrac{a^2+b^2}{2}\ge ab\) ( đúng )
\(\Leftrightarrow a^2+b^2\ge2ab\)
\(\Leftrightarrow a^2+b^2-2ab\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\) ( đúng )
Vậy \(\dfrac{a^2+b^2}{2}\ge ab\)
b.Giả sử: \(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\) ( đúng )
\(\Leftrightarrow\left(a+b\right)\left(\dfrac{a+b}{ab}\right)\ge4\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{ab}\ge4\)
\(\Leftrightarrow\left(a+b\right)^2-4ab\ge0\)
\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\) ( luôn đúng )
Vậy \(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)
ab=1
⇒ \(a=\dfrac{1}{b}\)
⇒ \(a^2=\dfrac{1}{b^2}\)
Thay vào P:
\(P=\dfrac{1}{\dfrac{1}{b^2}}+\dfrac{1}{b^2}+\dfrac{2}{\dfrac{1}{b^2}+b^2}\)
\(=\left(b^2+\dfrac{1}{b^2}\right)+\dfrac{2}{b^2+\dfrac{1}{b^2}}\)
Áp dụng BĐT Cô Si cho 2 số dương
⇒ \(P\) ≥ \(2\sqrt{\left(b^2+\dfrac{1}{b^2}\right).\dfrac{2}{b^2+\dfrac{1}{b^2}}}\)
\(=2\sqrt{2}\)
Min P= \(2\sqrt{2}\) ⇔ \(b^2=\dfrac{1}{b^2}\) ⇔b=1
