Bạn cần hỗ trợ bài nào nhỉ?

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 \(x=3\Rightarrow A=\dfrac{3^2+3.3+1}{3+1}=\dfrac{19}{4}\)

ko bt

4.linda sometimes brings her home made after the class

Linh 6A3(THCS Mai Đình) à

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:
$\frac{AD}{A'D'}=\frac{AB}{A'B'}=\frac{BC}{B'C'}$

$\Rightarrow AD.B'C'=BC.A'D'$

Hình bài 4:

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 \(a=b=c=1\)