Xem lại biểu thức P.

Mình phải đi ăn nên chiều mình làm nốt câu d nhé

a) Điều kiện để P được xác định là: \(x\ne1;x\ne-1\)

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

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

\(P=0:\dfrac{2x}{5x-5}x-\dfrac{x-1}{x+1}\)

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

c) Theo đề ta có:

\(P=2\)

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

\(\Leftrightarrow-\left(x-1\right)=2x+2\)

\(\Leftrightarrow-x-2x=2-1\)

\(\Leftrightarrow-3x=1\)

\(\Leftrightarrow x=-\dfrac{1}{3}\)

d) \(P=-\dfrac{x-1}{x+1}\) nguyên khi:

\(\Leftrightarrow x-1⋮-\left(x+1\right)\)

\(\Leftrightarrow\left(x+1\right)-2⋮-\left(x+1\right)\)

\(\Leftrightarrow-2⋮-\left(x+1\right)\)

\(\Leftrightarrow2⋮x+1\)

\(\Rightarrow x+1\inƯ\left(2\right)\)

Vậy \(P\) nguyên khi \(x\in\left\{-2;0;-3;1\right\}\)

câu d

\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)

`a,ĐKXĐ:x-4 ne 0,2x+2 ne 0`

`<=>x ne 4,x me -1`

`b,ĐKXĐ:4x^2-25 ne 0`

`<=>(2x-5)(2x+5) ne 0`

`<=>x ne +-5/2`

`c,ĐKXĐ:8x^3+27 ne 0`

`<=>8x^3 ne -27`

`<=>2x ne -3`

`<=>x ne -3/2`

`d,2x+2 ne 0,4y^2-9 ne 0`

`<=>2x ne -2,(2y-3)(2y+3) ne 0`

`<=>x ne -1,y ne +-3/2`

b) ĐKXĐ: \(x\notin\left\{\dfrac{5}{2};-\dfrac{5}{2}\right\}\)

c) ĐKXĐ: \(x\ne-\dfrac{3}{2}\)

d) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\notin\left\{\dfrac{3}{2};-\dfrac{3}{2}\right\}\end{matrix}\right.\)

1) 

a) Biểu thức \(\dfrac{x-2}{x^2+8x}\) vô nghĩa khi \(x^2+8x=0\)

\(\Leftrightarrow x\left(x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

Vậy: Khi \(x\in\left\{0;-8\right\}\) thì biểu thức \(\dfrac{x-2}{x^2+8x}\) vô nghĩa

b) Biểu thức \(\dfrac{25x^2-1}{16x^2-25}\) vô nghĩa khi \(16x^2-25=0\)

\(\Leftrightarrow\left(4x-5\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-5=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=5\\4x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=-\dfrac{5}{4}\end{matrix}\right.\)

Vậy: Khi \(x\in\left\{\dfrac{5}{4};-\dfrac{5}{4}\right\}\) thì biểu thức \(\dfrac{25x^2-1}{16x^2-25}\) vô nghĩa

c) Biểu thức \(\dfrac{x^2+1}{2x^2-28x+98}\) vô nghĩa khi \(2x^2-28x+98=0\)

\(\Leftrightarrow2\left(x^2-14x+49\right)=0\)

\(\Leftrightarrow\left(x-7\right)^2=0\)

\(\Leftrightarrow x-7=0\)

hay x=7

Vậy: Khi x=7 thì biểu thức \(\dfrac{x^2+1}{2x^2-28x+98}\) vô nghĩa

d) Để biểu thức \(\dfrac{2x+3}{9-\left(x+3\right)^2}\) vô nghĩa thì \(9-\left(x+3\right)^2=0\)

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

\(\Leftrightarrow-x\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

Vậy: Khi \(x\in\left\{0;-6\right\}\) thì biểu thức \(\dfrac{2x+3}{9-\left(x+3\right)^2}\) vô nghĩa

2) 

a) ĐKXĐ: \(x\notin\left\{0;-8\right\}\)

b) ĐKXĐ: \(x\notin\left\{\dfrac{5}{4};-\dfrac{5}{4}\right\}\)

c) ĐKXĐ: \(x\ne7\)

d) ĐKXĐ: \(x\notin\left\{0;-6\right\}\)

3) 

a) Để phân thức \(\dfrac{x-2}{x^2+8x}=0\) thì x-2=0

hay x=2(nhận)

Vậy: Khi x=2 thì phân thức \(\dfrac{x-2}{x^2+8x}=0\)

b) Để phân thức \(\dfrac{25x^2-1}{16x^2-25}=0\) thì \(25x^2-1=0\)

\(\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\5x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=1\\5x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\left(nhận\right)\\x=-\dfrac{1}{5}\left(nhận\right)\end{matrix}\right.\)

Vậy: Khi \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\) thì phân thức \(\dfrac{25x^2-1}{16x^2-25}=0\)

c) Để phân thức \(\dfrac{x^2+1}{2x^2-28x+98}=0\) thì \(x^2+1=0\)

mà \(x^2+1>0\forall x\) thỏa mãn ĐKXĐ

nên \(x\in\varnothing\)

Vậy: Không có giá trị nào của x để \(\dfrac{x^2+1}{2x^2-28x+98}=0\)

d) Để phân thức \(\dfrac{2x+3}{9-\left(x+3\right)^2}=0\) thì 2x+3=0

\(\Leftrightarrow2x=-3\)

hay \(x=-\dfrac{3}{2}\)(nhận)

Vậy: Khi \(x=-\dfrac{3}{2}\) thì phân thức \(\dfrac{2x+3}{9-\left(x+3\right)^2}=0\)

mình chỉ làm 1 câu thôi nhé các câu khác làm tương tự

1. biểu thức vô nghĩa <=> mẫu thức = 0 

\(x^2+8x=0< =>\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

vậy ...

2. tập xác định là tập hợp các giá trị làm phân thức có nghĩa (trong căn thì ≥ 0 ; dưới mẫu thì ≠ 0)

\(x^2+8x\ne0< =>\left[{}\begin{matrix}x\ne0\\x\ne-8\end{matrix}\right.\)

vậy ...

3. để phân thức = 0 => tử bằng không và mẫu khác không

\(\left\{{}\begin{matrix}x-2=0\\x^2+8x\ne0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=2\left(tm\right)\\\left[{}\begin{matrix}x\ne0\\x\ne-8\end{matrix}\right.\end{matrix}\right.\)

1: \(MTC=2\left(x-y\right)\left(x+y\right)\)

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

\(\dfrac{x+y}{2x^2+4xy+2y^2}\)

\(=\dfrac{x+y}{2\left(x^2+2xy+y^2\right)}\)

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

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

2: \(\dfrac{1}{x^2+8x+15}=\dfrac{1}{\left(x+3\right)\left(x+5\right)}=\dfrac{x+3}{\left(x+3\right)^2\cdot\left(x+5\right)}\)

\(\dfrac{1}{x^2+6x+9}=\dfrac{1}{\left(x+3\right)^2}=\dfrac{x+5}{\left(x+3\right)^2\cdot\left(x+5\right)}\)

3: \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}=\dfrac{1\cdot\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{a-c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(\dfrac{1}{\left(c-b\right)\left(c-a\right)}=\dfrac{1}{\left(b-c\right)\left(a-c\right)}=\dfrac{a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(\dfrac{1}{\left(b-a\right)\left(a-c\right)}=\dfrac{-1}{\left(a-b\right)\left(a-c\right)}=\dfrac{-\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

Bài 2:

a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)

b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)

\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)

\(=x^4-22x^3+108x^2-45x\)

c: \(=12x^5-18x^4+30x^3-24x^2\)

d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)

Nguyễn Huy Tú :v

a,\(\dfrac{3}{x-3}\) - \(\dfrac{6x}{9-x^2}\) + \(\dfrac{x}{x+3}\) (*)

đkxđ: x khác 3, x khác -3

(*) \(\dfrac{3(x+3)}{\left(x-3\right).\left(x+3\right)}\)- \(\dfrac{6x}{\left(x-3\right).\left(x+3\right)}\) + \(\dfrac{x\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}\)

=>3x+9 -6x + x²+3x

<=>x² + 3x-6x+3x + 9

<=>x² +9

<=>(x-3).(x+3)

help me

a: ĐKXĐ: x<>0; x<>-3

b: \(=\dfrac{x^2+6x+9}{x\left(x+3\right)}\cdot\dfrac{2}{x+3}=\dfrac{2}{x}\)

c: Khi x=1/5 thì A=2:1/5=10

a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)

b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)

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

d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)

e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)