a: ĐKXĐ: (x-1)(x-3)>=0

=>x>=3 hoặc x<=1

b: ĐKXĐ: (x-4)(x-3)>=0

=>x>=4 hoặc x<=3

c: ĐKXĐ: (x-5)(x-4)>=0

=>x>=5 hoặc x<=4

a: \(3x^2-2x\left(5+1,5x\right)+10\)

\(=3x^2-10x-3x^2+10\)

=-10x+10

b: \(7x\left(4y-x\right)+4y\left(y-7x\right)-2\left(2y^2-3,5x^2\right)\)

\(=28xy-7x^2+4y^2-28xy-4y^2+7x^2\)

=0

a) ĐKXĐ: \(x\notin\left\{3;-3;-2\right\}\)

Ta có: \(P=\left(\dfrac{2x-1}{x+3}-\dfrac{x}{3-x}-\dfrac{3-10x}{x^2-9}\right):\dfrac{x+2}{x-3}\)

\(=\left(\dfrac{\left(2x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3-10x}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{x+2}{x-3}\)

\(=\dfrac{2x^2-6x-x+3+x^2+3x-3+10x}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+2}{x-3}\)

\(=\dfrac{3x^2+6x}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+2}{x-3}\)

\(=\dfrac{3x\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{x+2}\)

\(=\dfrac{3x}{x+3}\)

b) Ta có: \(x^2-7x+12=0\)

\(\Leftrightarrow x^2-3x-4x+12=0\)

\(\Leftrightarrow x\left(x-3\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(loại\right)\\x=4\left(nhận\right)\end{matrix}\right.\)

Thay x=4 vào biểu thức \(P=\dfrac{3x}{x+3}\), ta được: 

\(P=\dfrac{3\cdot4}{4+3}=\dfrac{12}{7}\)

Vậy: Khi \(x^2-7x+12=0\) thì \(P=\dfrac{12}{7}\)

a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)

b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)

c) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5}{3}\left(x\ne y\right)\)

d) \(\dfrac{x^2-y^2}{x+y}=x-y\left(đk:x\ne-y\right)\)

e) \(\dfrac{x^3-x^2+x-1}{x^2-1}=\dfrac{x^2+1}{x+1}\left(đk:x\ne\pm1\right)\)

f) \(\dfrac{x^2+4x+4}{2x+4}=\dfrac{x+2}{2}\left(đk:x\ne-2\right)\)

[2x-2=0=>x=1

x-1=0=>x=1

x+1=0=>x=-1

5=0=>x=5

a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)

b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)

\(Q=\frac{\left(x-4\right)\left(x+3\right)}{x^2+1}\)

a: ĐKXĐ: \(\frac23x-\frac15\ge0\)

=>\(\frac23x\ge\frac15\)

=>\(x\ge\frac15:\frac23=\frac15\cdot\frac32=\frac{3}{10}\)

b; ĐKXĐ: \(x^2+2\ge0\)

mà \(x^2+2\ge2>0\forall x\)

nên x∈R

c: ĐKXĐ: \(\frac{1+x}{2x-3}\ge0\)

=>\(\frac{x+1}{2x-3}\ge0\)

=>x>3/2 hoặc x<=-1

d: ĐKXĐ: \(\begin{cases}3x-5\ge0\\ \frac{2}{x-4}\ge0\end{cases}\Rightarrow\begin{cases}x\ge\frac53\\ x-4>0\end{cases}\Rightarrow x>4\)

Lời giải:

a. ĐKXĐ: 

\(\left\{\begin{matrix} x-1\geq 0\\ 2\geq \sqrt{x-1}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ 4\geq x-1\end{matrix}\right. \Leftrightarrow 5\geq x\geq 1\)

b. ĐKXĐ:

\(\left\{\begin{matrix} x\geq 0\\ 3\geq \sqrt{x}\end{matrix}\right.\Leftrightarrow 0\leq x\leq 9\)

\(a,ĐK:x\ne\pm1\\ b,B=\dfrac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2\left(x+1\right)}\\ c,B=-\dfrac{1}{2}\Leftrightarrow2\left(x+1\right)=-2\Leftrightarrow x+1=-1\Leftrightarrow x=-2\left(tm\right)\)