Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a.
\(A=\left(\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x\left(x-1\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}+\dfrac{x-2}{x}\right):\dfrac{x+1}{x}\)
\(=\left(\dfrac{x^2+x+1}{x}+\dfrac{x+2}{x}+\dfrac{x-2}{x}\right):\dfrac{x+1}{x}\)
\(=\left(\dfrac{x^2+3x+1}{x}\right).\dfrac{x}{x+1}\)
\(=\dfrac{x^2+3x+1}{x+1}\)
2.
\(x^3-4x^3+3x=0\Leftrightarrow x\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=1\left(loại\right)\\x=3\end{matrix}\right.\)
Với \(x=3\Rightarrow A=\dfrac{3^2+3.3+1}{3+1}=\dfrac{19}{4}\)
Bài 4:
a. Vì $\triangle ABC\sim \triangle A'B'C'$ nên:
$\frac{AB}{A'B'}=\frac{BC}{B'C'}=\frac{AC}{A'C'}(1)$ và $\widehat{ABC}=\widehat{A'B'C'}$
$\frac{DB}{DC}=\frac{D'B'}{D'C}$
$\Rightarrow \frac{BD}{BC}=\frac{D'B'}{B'C'}$
$\Rightarrow \frac{BD}{B'D'}=\frac{BC}{B'C'}(2)$
Từ $(1); (2)\Rightarrow \frac{BD}{B'D'}=\frac{BC}{B'C'}=\frac{AB}{A'B'}$
Xét tam giác $ABD$ và $A'B'D'$ có:
$\widehat{ABD}=\widehat{ABC}=\widehat{A'B'C'}=\widehat{A'B'D'}$
$\frac{AB}{A'B'}=\frac{BD}{B'D'}$
$\Rightarrow \triangle ABD\sim \triangle A'B'D'$ (c.g.c)
b.
Từ tam giác đồng dạng phần a và (1) suy ra:
$\frac{AD}{A'D'}=\frac{AB}{A'B'}=\frac{BC}{B'C'}$
$\Rightarrow AD.B'C'=BC.A'D'$
Đặt \(a=\dfrac{1}{x};b=\dfrac{1}{y};c=\dfrac{1}{z}\Rightarrow xyz=1\) và \(x;y;z>0\)
Gọi biểu thức cần tìm GTNN là P, ta có:
\(P=\dfrac{1}{\dfrac{1}{x^3}\left(\dfrac{1}{y}+\dfrac{1}{z}\right)}+\dfrac{1}{\dfrac{1}{y^3}\left(\dfrac{1}{z}+\dfrac{1}{x}\right)}+\dfrac{1}{\dfrac{1}{z^3}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)}\)
\(=\dfrac{x^3yz}{y+z}+\dfrac{y^3zx}{z+x}+\dfrac{z^3xy}{x+y}=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)
\(P\ge\dfrac{\left(x+y+z\right)^2}{y+z+z+x+x+y}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\)
\(P_{min}=\dfrac{3}{2}\) khi \(x=y=z=1\) hay \(a=b=c=1\)
\(Bài.1:\\ a,0,125.\left(-3,7\right).2^3=0,125.\left(-3.7\right).8\\ =\left(0,125.8\right).\left(-3,7\right)=1.\left(-3,7\right)=-3,7\\ b,\sqrt{36}.\sqrt{\dfrac{25}{16}}+\dfrac{1}{4}=6.\dfrac{5}{4}+\dfrac{1}{4}=\dfrac{15}{2}+\dfrac{1}{4}=\dfrac{30}{4}+\dfrac{1}{4}=\dfrac{31}{4}\\ c,\sqrt{\dfrac{4}{81}}.\sqrt{\dfrac{25}{81}}-\dfrac{12}{5}\\ =\dfrac{2}{9}.\dfrac{5}{9}-\dfrac{12}{5}=\dfrac{10}{81}-\dfrac{12}{5}=\dfrac{10.5-12.81}{420}=-\dfrac{461}{210}\\ d,0,1.\sqrt{225}.\sqrt{\dfrac{1}{4}}=0,1.15.\dfrac{1}{2}=0,75\)
Bài 2:
\(a,\dfrac{1}{5}+x=\dfrac{2}{3}\\ x=\dfrac{2}{3}-\dfrac{1}{5}=\dfrac{10}{15}-\dfrac{3}{15}=\dfrac{7}{15}\\ ---\\ b,-\dfrac{5}{8}+x=\dfrac{4}{9}\\ x=\dfrac{4}{9}-\left(-\dfrac{5}{8}\right)=\dfrac{4}{9}+\dfrac{5}{8}=\dfrac{4.8+5.9}{72}=\dfrac{77}{72}\\ ---\\ c,\dfrac{13}{4}x+1\dfrac{1}{2}=-\dfrac{4}{5}\\ \dfrac{13}{4}x+\dfrac{3}{2}=-\dfrac{4}{5}\\ \dfrac{13}{4}x=-\dfrac{4}{5}-\dfrac{3}{2}=\dfrac{-4.2-3.5}{10}=-\dfrac{23}{10}\\ x=-\dfrac{23}{10}:\dfrac{13}{4}=-\dfrac{23}{10}.\dfrac{4}{13}=-\dfrac{46}{65}\\ ---\\ d,\dfrac{1}{4}+\dfrac{3}{4}x=\dfrac{3}{4}\\ \dfrac{3}{4}x=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{1}{2}\\ x=\dfrac{1}{2}:\dfrac{3}{4}=\dfrac{1}{2}.\dfrac{4}{3}=\dfrac{4}{6}=\dfrac{2}{3}\)