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Ta có:\(A=1+19^{19}+93^{199}+1993^{1994}\)

Dễ thấy:

\(19^2\equiv1\left(mod10\right)\Rightarrow19^{18}\equiv1\left(mod10\right)\Rightarrow19^{19}\equiv9\left(mod10\right)\)

\(93^4\equiv1\left(mod10\right)\Rightarrow93^{196}\equiv1\left(mod10\right)\Rightarrow93^{199}\equiv7\left(mod10\right)\)

\(1993\equiv3\left(mod10\right)\Rightarrow1993^4\equiv1\left(mod10\right)\Rightarrow1993^{1992}\equiv1\left(mod10\right)\Rightarrow1993^{1994}\equiv9\left(mod10\right)\)

\(\Rightarrow1+19^{19}+93^{199}+1993^{1994}\equiv1+9+7+9\equiv6\left(mod10\right)\)

Cho bạn 1 ý tưởng làm bài này nhưng không khả thi lắm :v

\(\text{Ta có:}\frac{x}{y+1}+\frac{y}{x+1}=\frac{x^2+x+y^2+y}{\left(x+1\right)\left(y+1\right)}\)

\(=\frac{\left(x+y\right)^2-2xy+1}{xy+x+y+1}=\frac{1-2xy+1}{xy+2}\)

\(=\frac{2-2xy}{2+xy}\)

\(\text{Vì }2-2xy\le2+xy\left(do\text{ x,y không âm}\right)\text{ nên }\frac{2-2xy}{2+xy}\le1\)

\(=>\frac{x}{y+1}+\frac{y}{x+1}\le1\)

\(a\)

\(\sqrt{2,7}\)\(.\)\(\sqrt{1,2}\)

\(=\)\(\sqrt{2,7.1,2}\)

\(=\)\(\sqrt{3,24}\)

\(=\)\(1,8\)

\(b\)

\(\sqrt{85}.\sqrt{125}.\sqrt{68}\)

\(=\)\(\sqrt{85.125.68}\)

\(=\)\(\sqrt{722500}\)

\(=\)\(850\)

học tốt!!!

Đề sai rồi nha bạn ơi

Vào TKHĐ để xem bằng chứng :3

\(a+\frac{4\: a^3}{\left(a\: -1\right)\left(a+1\right)^3}\)>3

Mình nhầm

Bài làm:

đk: \(x\ge0;x\ne1\)

Ta có:

\(A=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\frac{\left(1-x\right)^2}{2}\)

\(A=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\cdot\frac{\left(x-1\right)^2}{2}\)

\(A=\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\cdot\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)

\(A=\frac{-2\sqrt{x}\cdot\left(\sqrt{x}-1\right)}{2}\)

\(A=\left(1-\sqrt{x}\right)\sqrt{x}=\sqrt{x}-x\)

ai giúp vs