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

a,\(A\) xác định \(\Leftrightarrow\left[{}\begin{matrix}x+1\ne0\\1-x^2\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ne-1\\x\ne1\end{matrix}\right.\)

Vậy...

b,\(A=\dfrac{x}{x+1}:\dfrac{1-3x^2}{1-x^2}\)

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

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

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

\(=\dfrac{x}{1}.\dfrac{x-1}{3x^2-1}\)

\(=\dfrac{x^2-x}{3x^2-1}\)

Câu a :

Để phân thức được xác định thì :

\(\left\{{}\begin{matrix}x+1\ne0\\1-x^2\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne1\end{matrix}\right.\)

Câu b :

\(\dfrac{x}{x+1}:\dfrac{1-3x^2}{1-x^2}\)

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

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

\(=\dfrac{x}{x+1}\times\dfrac{\left(x-1\right)\left(x+1\right)}{3x^2-1}\)

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

Thay x=-y-z

suy ra x^2-y^2-z^2=(-y-z)^2-y^2-z^2=2yz

Vậy M=x^+y^+z^3/2xyz(quy đồng)

Tiếp tuc thay: x=-y-z

M=(-y-z)^3+y^3+z^3/2(-y-z)yz

M=-3x^2y-3xy^2/-2x^y-2xy^2=3/2

Ta có:

Ta có:

(b-c)(a-c)(a+b)+(a-c)(a+b)(c-b)=0

suy ra

(b-c)(a^2+ab-ac-bc)+(a-c)(ac+bc-ab-b^2)=0

Ta có: c^2 +2(ab-ac-bc)=0

suy ra:

ab-ac-bc=-c^2+ab+bc

c^2+ab-bc-ac=ac+bc-ab

Vậy (b-c)(a^2-c^2+ab+bc+ac)+(a-c)(ab-bc-ac+c^2-b^2)=0

suy ra:(b-c)[a^2-(a-c)(b-c)]+(a-c)[(a-c)(b-c)-b^2]=0

suy ra a^2(b-c)-(a-c)(b-c)^2+(a-c)^2.(b-c)-(a-c).b^2=0

Suy ra a^2(b-c)+(a-c)^2.(b-c)=(a-c).b^2+(a-c)(b-c)^2

suy ra:(b-c)(a^2+(a-c)^2)=(a-c)(b^2+(b-c)^2)

suy ra a-c/b-c=a^2+(a-c)^2/b^2+(b-c)^2(đpcm)

tớ làm đc kết quá là -4x^2/3-x

(xin lỗi nha, h tớ trình bày mất thời gian lắm,bạn thử đối chiếu với kq của bạn xem tớ làm đúng k)

a: ĐKXĐ: \(x\notin\left\{0;1;-1\right\}\)

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

\(=\dfrac{x^2+3x+2+x^2-3x+2}{x}\cdot\dfrac{1}{x^2+2}\)

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

b: Để B là số nguyên thì \(x\inƯ\left(2\right)\)

hay \(x\in\left\{2;-2\right\}\)

1: \(B=\left(\dfrac{4x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{4\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{16}{x+2}\cdot\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+x+1}\)

\(=\left(\dfrac{4x}{x+2}-\dfrac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\right)\cdot\dfrac{x+2}{16}\cdot\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+x+1}\)

\(=\dfrac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)^2}\cdot\dfrac{\left(x+2\right)^2\cdot\left(x+1\right)}{16\left(x^2+x+1\right)}\)

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

2: Để B=0 thì -x-1=0

hay x=-1(nhận)

4a^2+b^2=5ab

=>4a^2 -5ab +b^2=0

=>4a^2-4ab+b^2-ab=0

=>4a(a-b)+b(b-a)=0

=>(4a-b)(a-b)=0\(\begin{matrix}\\\end{matrix}\)

=>\(\left[{}\begin{matrix}4a-b=0\\a-b=0\end{matrix}\right.\)=>\(\begin{matrix}4a=b\\a=b\end{matrix}\)

thay vào bt ta tính được 2 trường hợp là \(\dfrac{1}{3}\)và\(\dfrac{-1}{3}\)

a) M xác định \(\Leftrightarrow4x^3-9x\ne0\)

\(\Leftrightarrow x\left(4x^2-9\right)\ne0\\ \Leftrightarrow\left[{}\begin{matrix}x\ne0\\x\ne\pm\dfrac{3}{2}\end{matrix}\right.\)

b)

\(M=\dfrac{\left(2x^3+3x^2\right)\left(2x+1\right)}{4x^3-9x}=\dfrac{4x^4+2x^3+6x^3+3x^2}{4x^3-9x}\\ =\dfrac{4x^4+8x^3+3x^2}{4x^3-9x}\\ =\dfrac{x\left(4x^3+8x^2+3x\right)}{x\left(x^2-9\right)}\\ =\dfrac{4x^3+8x^2+3x}{x^2-9}\)

c)

\(M=0\\ \Leftrightarrow\left(2x^3+3x^2\right)\left(2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x^3+3x^2=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2\left(2x+3\right)=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

a) điều kiện xát định : \(x\ne\pm1\)

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

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

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

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

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

\(P=\dfrac{\left(x-1\right)\left(\left(x-2\right)\left(x+1\right)-\left(x+2\right)\left(x-1\right)\right)}{4}\)

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

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

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

b) ta có : \(P-4=5x\Leftrightarrow\dfrac{-2x^2+2x}{4}-4=5x\)

\(\Leftrightarrow\dfrac{-2x^2+2x-16}{4}=5x\Leftrightarrow-2x^2+2x-16=20x\)

\(\Leftrightarrow20x-\left(-2x^2+2x-16\right)=0\Leftrightarrow2x^2+18x+16=0\)

\(\Leftrightarrow2x^2+2x+16x+16=0\Leftrightarrow2x\left(x+1\right)+16\left(x+1\right)=0\)

\(\Leftrightarrow\left(2x+16\right)\left(x+1\right)\Leftrightarrow\left[{}\begin{matrix}2x+16=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(tmđk\right)\\x=-1\left(loại\right)\end{matrix}\right.\)

vậy \(x=-8\) thỏa mãng điều kiện bài toán