giúp mik bài3B 4A vs ạ
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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:
$\frac{AD}{A'D'}=\frac{AB}{A'B'}=\frac{BC}{B'C'}$
$\Rightarrow AD.B'C'=BC.A'D'$
AK
1
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 1 2024
ĐKXĐ: \(\left|x-2\right|-1\ne0\)
\(\Rightarrow\left|x-2\right|\ne1\)
\(\Rightarrow\left\{{}\begin{matrix}x-2\ne1\\x-2\ne-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ne3\\x\ne1\end{matrix}\right.\)
DD
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 2 2023
Vì MN // BC theo Talet ta có:
\(\dfrac{y}{20}\) = \(\dfrac{10}{15}\) = \(\dfrac{x}{12}\) => x = \(\dfrac{10}{15}\) . 12 = 8; y = \(\dfrac{10}{15}\) . 20 = \(\dfrac{40}{3}\)
3B.
a.
\(\dfrac{6}{7-x}+\dfrac{-4}{x}+\dfrac{6}{x-7}+\dfrac{2}{x-2}+\dfrac{4}{x}+\dfrac{-5}{x+2}\)
\(=\dfrac{6}{7-x}-\dfrac{6}{7-x}+\dfrac{-4+4}{x}+\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{-5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=0+0+\dfrac{2x+4-5x+10}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-3x+14}{\left(x-2\right)\left(x+2\right)}\)
b.
\(\dfrac{x-1}{x^2-4}-\dfrac{1}{x+2}+\dfrac{x+2}{x}+\dfrac{1}{x+2}+\dfrac{x+2}{-x}-\dfrac{3}{x-2}\)
\(=\dfrac{x-1}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\left(\dfrac{1}{x+2}-\dfrac{1}{x+2}\right)+\left(\dfrac{x+2}{x}-\dfrac{x+2}{x}\right)\)
\(=\dfrac{x-1-3x-6}{\left(x-2\right)\left(x+2\right)}+0+0\)
\(=\dfrac{-2x-7}{\left(x-2\right)\left(x+2\right)}\)
4A.
\(\dfrac{3x+1}{\left(x-1\right)^2}-\dfrac{1}{x+1}+\dfrac{x+3}{1-x^2}\)
\(=\dfrac{\left(3x+1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)^2}-\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)^2}-\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)^2}\)
\(=\dfrac{3x^2+4x+1-\left(x^2-2x+1\right)-\left(x^2+2x-3\right)}{\left(x+1\right)\left(x-1\right)^2}\)
\(=\dfrac{x^2+4x+3}{\left(x+1\right)\left(x-1\right)^2}=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x-1\right)^2}\)
\(=\dfrac{x+3}{\left(x-1\right)^2}\)
b.
\(\dfrac{2x}{x^2-4}-\left(\dfrac{2}{x+5}+\dfrac{3-x}{x+1}\right)+\left(\dfrac{2}{x+5}-\left(\dfrac{4}{x^2-4}+\dfrac{x-3}{x+1}\right)\right)\)
\(=\dfrac{2x}{x^2-4}+\left(\dfrac{2}{x+5}-\dfrac{2}{x+5}\right)+\left(\dfrac{x-3}{x+1}-\dfrac{x-3}{x+1}\right)-\dfrac{4}{x^2-4}\)
\(=\dfrac{2x-4}{x^2-4}=\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2}{x+2}\)