Bài giải:
ab x 7 = 1ab
b x 7 = b.Vậy b = 0
a x7 = 1a.Vậy a = 2
Vậy:20 x 7 = 140
Bài giải:
ab x 7 = 1ab
b x 7 = b.Vậy b = 0
a x7 = 1a.Vậy a = 2
Vậy:20 x 7 = 140
\(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\frac{\left(1-x\right)^2}{2}\)
\(P=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\frac{\left(1-x\right)^2}{2}\)
\(P=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right)\frac{\left(x-1\right)^2}{2}\)
\(P=\left(\frac{\left(x-\sqrt{x}-2\right)-\left(x+\sqrt{x}-2\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right)\frac{\left(x-1\right)^2}{2}\)
\(P=\frac{2\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}+1\right)}\frac{\left(x-1\right)^2}{2}\)
\(P=\frac{\sqrt{x}\left(x-1\right)}{\sqrt{x}+1}=\sqrt{x}\left(\sqrt{x}-1\right)=x-\sqrt{x}\)
giải hệ pt :
\(\hept{\begin{cases}3x^2+6xy+9y^2+\left(x+2y\right)^2\sqrt{x+2y}-3\left(x+2y\right)\sqrt{x+2y}-4\left(x+2y\right)+4\sqrt{x+2y}=0\\\left(\frac{\sqrt[3]{x^2-y^2}}{\sqrt[4]{x}}+\sqrt[4]{\frac{x}{y}}\right)^{2017}+\left(\sqrt[3]{\frac{x}{y}}-\sqrt[4]{\frac{y}{x}}\right)^{2018}=1\end{cases}}\)
\(x-1-2-3-4=0\sqrt{\sqrt[]{}\dfrac{ }{ }}\)
\(\hept{\begin{cases}ab\\6\\1ab\end{cases}}x\)Thay các chữ thành số phù hợp.Ý của mình là:ab x 6 = 1ab
lời mời kb nữa nhé
\(2\sqrt{x}\ge2\sqrt{y}\)
https://olm.vn/hoi-dap/detail/240862214307.html và
https://d3.violet.vn//uploads/previews/present/4/447/16/preview.swf
\(\frac{-3+\sqrt{17}}{4}\frac{-3-\sqrt{17}}{4}\)
$\frac{a^2.sin 60^{o}}{2}=\frac{3\sqrt{3}}{4}$ suy ra $a=\sqrt{3}$
\(y=\frac{1}{x^2+\sqrt{x}}\)
\(y=\frac{1}{x^2+\sqrt{x}}\)