Bài 7 :

2)

a) \(5x^2y-10xy^2\)

\(=5xy\left(x-2y\right)\)

b) \(3\left(x+3\right)-x^2+9\)

\(=3\left(x+3\right)-\left(x^2-3^2\right)\)

\(=3\left(x+3\right)-\left(x-3\right)\left(x+3\right)\)

\(=\left(x+3\right)\left[3-\left(x-3\right)\right]\)

\(=\left(x+3\right)\left(3-x+3\right)\)

\(=\left(x+3\right)\left(6-x\right)\)

c) \(x^2-y^2+xz-yz\)

\(=\left(x^2-y^2\right)+\left(xz-yz\right)\)

\(=\left(x-y\right)\left(x+y\right)+z\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y+z\right)\)

3)

a) \(A=\dfrac{x^2}{x^2-4}-\dfrac{x}{x-2}+\dfrac{2}{x+2}\)

\(\Leftrightarrow A=\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x}{x-2}+\dfrac{2}{x+2}\)

Điều kiện xác định là: \(\left\{{}\begin{matrix}x-2\ne0\Rightarrow x\ne2\\x+2\ne0\Rightarrow x\ne-2\end{matrix}\right.\)

b) \(A=\dfrac{x^2}{x^2-4}-\dfrac{x}{x-2}+\dfrac{2}{x+2}\)

\(\Leftrightarrow A=\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x}{x-2}+\dfrac{2}{x+2}\) MTC: \(\left(x-2\right)\left(x+2\right)\)

\(\Leftrightarrow A=\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow A=\dfrac{x^2-x\left(x+2\right)+2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow A=\dfrac{x^2-x^2-2x+2x-4}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow A=\dfrac{-4}{\left(x-2\right)\left(x+2\right)}\)

c) Thay \(x=1\) và biểu thức A ta được:

\(\dfrac{-4}{\left(1-2\right)\left(1+2\right)}=\dfrac{-4}{\left(-1\right).3}=\dfrac{-4}{-3}=\dfrac{4}{3}\)

Vậy giá trị của biểu thức A tại \(x=1\) là \(\dfrac{4}{3}\)

a) Phân thức B xác định \(\Leftrightarrow\hept{\begin{cases}2x-2\ne0\\x^2-1\ne0\\2x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne\left\{\pm1\right\}\\x\ne-1\end{cases}\Leftrightarrow}x\ne\left\{\pm1\right\}}\)

b) \(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)\cdot\frac{4x^2-4}{5}\)

\(B=\left[\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{3\cdot2}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right]\cdot\frac{\left(2x\right)^2-2^2}{5}\)

\(B=\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\frac{\left(2x-2\right)\left(2x+2\right)}{5}\)

\(B=\frac{10\cdot2\left(x-1\right)\cdot2\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)\cdot5}\)

\(B=\frac{40\left(x-1\right)\left(x+1\right)}{10\left(x-1\right)\left(x+1\right)}\)

\(B=4\)

Vậy với mọi giá trị của x thì B luôn bằng 4

Vậy giá trị của B không phụ thuộc vào biến ( đpcm )

\(Giải:\)

\(ĐKXĐ:x\ne\pm1\)
\(B=\left[\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right]=\left[\frac{x+1}{2x-2}+\frac{12}{4x^2-4}-\frac{x+3}{2x+2}\right]\)

\(=\left[\frac{x+1}{2x-2}+\frac{12}{\left(2x+2\right)\left(2x-2\right)}-\frac{x+3}{2x+2}\right]\)

\(=\left[\frac{\left(x+1\right)\left(2x+2\right)}{\left(2x+2\right)\left(2x-2\right)}+\frac{12}{\left(2x+2\right)\left(2x-2\right)}-\frac{\left(x+3\right)\left(2x-2\right)}{\left(2x-2\right)\left(2x+2\right)}\right]\)

\(=\frac{2x^2+4x+14-2x^2+2x-6x+6}{\left(2x-2\right)\left(2x+2\right)}\)

\(=\frac{6}{\left(2x-2\right)\left(2x+2\right)}\)

a,ĐK:  \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)

\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)

\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)

c, Với x = 4 thỏa mãn ĐKXĐ thì

\(A=\frac{-3}{4-3}=-3\)

d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)

\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)

Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)

1.

a) \(x\left(x+4\right)+x+4=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)

b) \(x\left(x-3\right)+2x-6=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

Bài 1:

a, \(x\left(x+4\right)+x+4=0\)

\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)

Vậy \(x=-4\) hoặc \(x=-1\)

b, \(x\left(x-3\right)+2x-6=0\)

\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy \(x=3\) hoặc \(x=-2\)

B3;a,ĐKXĐ:\(x\ne\pm4\)

A=\(\left(\dfrac{4}{x-4}-\dfrac{4}{x+4}\right)\dfrac{x^2+8x+16}{32}=\left(\dfrac{4x+16}{x^2-16}-\dfrac{4x-16}{x^2-16}\right)\dfrac{x^2+2.4x+4^2}{32}=\left(\dfrac{4x+16-4x+16}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\left(\dfrac{32}{x^2-16}\right)\dfrac{\left(x+4\right)^2}{32}=\dfrac{32\left(x+4\right)^2}{32.\left(x-4\right)\left(x+4\right)}=\dfrac{x+4}{x-4}\\ \\ \\ \\ \\ \\ b,Tacó\dfrac{x+4}{x-4}=\dfrac{1}{3}\Leftrightarrow3x+12=x-4\Leftrightarrow x=-8\left(TM\right)c,TAcó\dfrac{x+4}{x-4}=3\Leftrightarrow x+4=3x-12\Leftrightarrow x=8\left(TM\right)\)

a) giải phương trình

\(\dfrac{2x^2-3x-2^{ }}{_{ }x^2-4}\) = 2

=>\(\dfrac{2x^2-3x-2}{x^2-4}\) = \(\dfrac{2\left(x^2-4\right)}{x^2-4}\)

=>2x² - 3x - 2 = 2(x² - 4)

<=>2x² -3x - 2 = 2x² - 8

<=>2x² - 2x² - 3x = -8 + 2

<=>-3x = -6

<=> x = 2

Vậy không tồn tại giá trị nào của x thỏa mãn điều kiện của bài toán

b) Ta phải giải phương trình

\(\dfrac{6x-1}{3x+2}\) = \(\dfrac{2x+5}{x-3}\)

=>x = \(\dfrac{-7}{38}\)

c) Ta phải giải phương trình

\(\dfrac{y+5}{y-1}\) - \(\dfrac{y+1}{y-3}\) = \(\dfrac{-8}{\left(y-1\right)\left(y+1\right)}\)

không tồn tại giá trị nào của y thỏa mãn điều kiện của bài toán

Bài 3:

a: DKDXĐ: x<>1

b: \(=\dfrac{x^2+2+x^2-x-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{2}{x-1}=\dfrac{x^2-2x+1}{\left(x-1\right)^2}\cdot\dfrac{2}{x^2+x+1}=\dfrac{2}{x^2+x+1}\)

c: Để C lớn nhất thì \(A=x^2+x+1_{MIN}\)

\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>=\dfrac{3}{4}\)

Dấu = xảy ra khi x=-1/2