KB
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 1 2024
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 \(x=3\Rightarrow A=\dfrac{3^2+3.3+1}{3+1}=\dfrac{19}{4}\)
AH
Akai Haruma
Giáo viên
3 tháng 2 2024
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:
$\frac{AD}{A'D'}=\frac{AB}{A'B'}=\frac{BC}{B'C'}$
$\Rightarrow AD.B'C'=BC.A'D'$
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 1 2024
ĐKXĐ: \(\left|x-2\right|-1\ne0\)
\(\Rightarrow\left|x-2\right|\ne1\)
\(\Rightarrow\left\{{}\begin{matrix}x-2\ne1\\x-2\ne-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ne3\\x\ne1\end{matrix}\right.\)
1: \(\left(2x+1\right)^2-2\left(2x+1\right)\left(3-x\right)+\left(x-3\right)^2\)
\(=\left(2x+1\right)^2+2\left(2x+1\right)\left(x-3\right)+\left(x-3\right)^2\)
\(=\left(2x+1+x-3\right)^2=\left(3x-2\right)^2=9x^2-12x+4\)
2: \(\left(x-1\right)^3-\left(x+1\right)\left(x^2-x+1\right)-\left(1-3x\right)\left(1+3x\right)\)
\(=x^3-3x^2+3x-1-\left(x^3+1\right)-\left(1-9x^2\right)\)
\(=x^3-3x^2+3x-1-x^3-1-1+9x^2\)
\(=6x^2+3x-3\)
3: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-1\right)\left(x+1\right)+3x\)
\(=x^3+8-x\left(x^2-1\right)+3x\)
\(=x^3+8-x^3+x+3x=4x+8\)
4: \(\left(3x-2\right)^2-3\left(x-4\right)\left(x+4\right)+\left(x-3\right)^2-\left(x+1\right)\left(x^2-x+1\right)\)
\(=9x^2-12x+4-3\left(x^2-16\right)+x^2-6x+9-\left(x^3+1\right)\)
\(=10x^2-18x+13-3x^2+48-x^3-1\)
\(=-x^3+7x^2-18x+12\)
5: \(\left(x+1\right)^2-\left(x-1\right)^2-3\left(x-3\right)\left(x+3\right)\)
\(=x^2+2x+1-x^2+2x-1-3\left(x^2-9\right)\)
\(=4x-3x^2+27\)
6: \(\left(x-1\right)^3-x\left(x-2\right)^2+x-1\)
\(=x^3-3x^2+3x-1-x\left(x^2-4x+4\right)+x-1\)
\(=x^3-3x^2+4x-2-x^3+4x^2-4x\)
\(=x^2-2\)
7: \(\left(x+2\right)^3-x^2\left(x+6\right)-8\)
\(=x^3+6x^2+12x+8-x^3-6x^2-8\)
=12x
8: \(\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x^3-3x^2y+3xy^2-y^3\right)-2y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)
\(=6x^2y\)
9: \(\left(x-1\right)^3-\left(x+1\right)^3+6\left(x-1\right)\left(x+1\right)\)
\(=x^3-3x^2+3x-1-x^3-3x^2-3x-1+6\left(x^2-1\right)\)
\(=-6x^2-2+6x^2-6=-8\)
10: \(4x\left(3x-5\right)-2\left(4x+1\right)-x-7\)
\(=12x^2-20x-8x-2-x-7\)
\(=12x^2-29x-9\)
11: \(\left(3x+1\right)^2-2\left(3x+1\right)\left(5x+5\right)+\left(5x+5\right)^2\)
\(=\left(5x+5-3x-1\right)^2\)
\(=\left(2x+4\right)^2=4x^2+16x+16\)
12: \(\left(2x+3\right)^2+\left(2x+3\right)\left(2x-6\right)+\left(x-3\right)^2\)
\(=\left(2x+3\right)^2+2\cdot\left(2x+3\right)\left(x-3\right)+\left(x-3\right)^2\)
\(=\left(2x+3+x-3\right)^2=\left(3x\right)^2=9x^2\)
13: \(\left(x^2-2x+4\right)\left(x+2\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=x^3+8-x^3-3x^2-3x-1+3\left(x^2-1\right)\)
\(=-3x^2-3x+7+3x^2-3=-3x+4\)
14: \(\left(x-2\right)^2+2\left(x-2\right)\left(2x+2\right)+4\left(x+1\right)^2\)
\(=\left(x-2\right)^2+2\left(x-2\right)\left(2x+2\right)+\left(2x+2\right)^2\)
\(=\left(x-2+2x+2\right)^2=\left(3x\right)^2=9x^2\)