Bài 9: Biến đổi các biểu thức hữu tỉ. Giá trị của phân thức

a: ĐKXĐ: x<>2; x<>-2

b: A=(x-2)^2/(x+2)(x-2)=(x-2)/(x+2)

c: Để A=3 thì 3x+6=x-2

=>2x=-8

=>x=-4

\(a,đk:9x^2-16\ne0\\ \Rightarrow\left\{{}\begin{matrix}3x-4\ne0\\3x+4\ne0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{4}{3}\\x\ne-\dfrac{4}{3}\end{matrix}\right.\\ b,đk:x^2-4x+4\ne0\\ \Rightarrow\left(x-2\right)^2\ne0\\ \Rightarrow x\ne2\\ c,đk\left\{{}\begin{matrix}x+1\ne0\\x-3\ne0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)

a)

\(ĐKXĐ:9x^2-16\ne0\\ \left(3x-4\right)\left(3x+4\right)\ne0\\ \left[{}\begin{matrix}3x-4\ne0\\3x+4\ne0\end{matrix}\right.\left[{}\begin{matrix}x\ne\dfrac{4}{3}\\x\ne-\dfrac{4}{3}\end{matrix}\right.\)

b)

\(ĐKXĐ:x^2-4x+4\ne0\\ \left(x-2\right)^2\ne0\\ =>x-2\ne0\\ x\ne2\)

c)

\(ĐKXĐ:\left(x+1\right)\left(x-3\right)\ne0\\ \left[{}\begin{matrix}x+1\ne0\\x-3\ne0\end{matrix}\right.\left[{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)

\(a,đk:x\ne3;-3\\ b,P=\dfrac{3}{x+3}+\dfrac{1}{x-3}-\dfrac{-18}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3\left(x-3\right)+\left(x+3\right)+18}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{3x-9+x+3+18}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{4x+12}{\left(x-3\right)\left(x+3\right)}\\ =\dfrac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\\ =\dfrac{4}{x-3}\)

\(c,P=4\\ \Rightarrow\dfrac{4}{x-3}=4\\ \Rightarrow4=\left(x-3\right).4\\ \Rightarrow4=4x-12\\ \Rightarrow-4x=-16\\ \Rightarrow x=4\left(thoamanđk\right)\)

a. \(x\ne\pm3\)

b. \(=\dfrac{3\left(x-3\right)}{x^2-9}+\dfrac{x+3}{x^2-9}+\dfrac{18}{x^2-9}\)

\(=\dfrac{3x-9+x+3+18}{x^2-9}\)

\(=\dfrac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{4}{x-3}\)

c. Với \(P=4\Leftrightarrow\dfrac{4}{x-3}=4\)

\(\Leftrightarrow4x-12=4\)

\(\Leftrightarrow4x=16\)

\(\Leftrightarrow x=4\left(tm\right)\)

a: ĐKXĐ: x<>0; x<>-5

b: \(P=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}=\dfrac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}\)

\(=\dfrac{x-1}{2}\)

Để P=1 thì x-1=2

=>x=3

Để P=-3 thì x-1=-6

=>x=-5(loại)

\(P=\left(\dfrac{x}{x-2}+\dfrac{1}{x^2-4}\right):\dfrac{x+1}{x+2}\left(x\ne-1;x\ne\pm2\right)\)

a. \(=\left[\dfrac{x\left(x+2\right)}{x^2-4}+\dfrac{1}{x^2-4}\right].\dfrac{x+2}{x+1}\)

\(=\dfrac{x^2+2x+1}{x^2-4}.\dfrac{x+2}{x+1}\)

\(=\dfrac{\left(x+1\right)^2}{\left(x+2\right)\left(x-2\right)}.\dfrac{x+2}{x+1}\)

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

b. Thay x = 1/2 vào P, ta đc:

\(P=\dfrac{\dfrac{1}{2}+1}{\dfrac{1}{2}-2}=\dfrac{\dfrac{3}{2}}{-\dfrac{3}{2}}=-1\)

\(a.a\ne0;a\ne\pm3\)

\(b.P=\dfrac{\left(a+3\right)^2}{2a^2+6a}.\left(1-\dfrac{6a-18}{a^2-9}\right)\)

\(=\dfrac{\left(a+3\right)^2}{2a\left(a+3\right)}.\left[1-\dfrac{6\left(a-3\right)}{\left(a+3\right)\left(a-3\right)}\right]\)

\(=\dfrac{a+3}{2a}.\left(1-\dfrac{6}{a+3}\right)\)

\(=\dfrac{a+3}{2a}-\dfrac{a+3}{2a}.\dfrac{6}{a+3}\)

\(=\dfrac{a+3}{2a}-\dfrac{6}{2a}\)

\(=\dfrac{a-3}{2a}\)

c. Với: \(P=0\Leftrightarrow\dfrac{a-3}{2a}=0\)

\(\Leftrightarrow a-3=0\)

\(\Leftrightarrow a=3\left(loại\right)\)

Vậy ko có a thỏa mãn để P = 0

Với \(P=1\Leftrightarrow\dfrac{a-3}{2a}=1\)

\(\Leftrightarrow2a=a-3\)

\(\Leftrightarrow a=-3\) (loại)

Vậy ......

a: ĐKXĐ: x<>-1; x<>3

b: \(P=\dfrac{3x\left(x+1\right)}{\left(x+1\right)\left(2x-6\right)}=\dfrac{3x}{2x-6}\)

Để P=1 thì 3x=2x-6

=>x=-6

a: ĐKXĐ: x<>0; x<>5

b \(P=\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}=\dfrac{x-5}{x}\)

Để P=0 thì x-5=0

=>x=5(loại)

Để P=5/2 thì \(\dfrac{x-5}{x}=\dfrac{5}{2}\)

=>2x-10=5x

=>-3x=10

=>x=-10/3(nhận)

c: Để P nguyên thì x-5 chia hết cho x

=>\(x\in\left\{1;-1;-5\right\}\)

a: ĐKXĐ: x<>1; x<>-1

b: \(B=\dfrac{x^2-2x+1-x^2-2x-1+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{-4}{x+1}\)

c: Để B>=0 thì x+1<0

=>x<-1