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a) Ta có
Biến đổi tử phân số A
x^3-x^2-10x-8=(x^3-4x^2)+(3x^2-12x)+(2x-8)
=x^2(x-4)+3x(x-4)+2(x-4)=(x^2+3x+2)(x-4)
=(x^2+x+2x+2)(x-4)=[x(x+1)+2(x+1)](x-4)
=(x+1)(x+2)(x+4) (1)
Biến đổi mẫu của phân số A:
x^3-4x^2+5x-20=x^2(x-4)+5(x-4)=(x^2+5)(x-4) (2)
Từ (1) và (2) suy ra:
A=(x+1)(x+2)/x^2+5
\(A=\dfrac{x^3-x^2-10x-8}{x^3-4x^2+5x-20}\\ ĐKXĐ:x\ne4\)
a) Với \(x\ne4\)
\(\text{Ta có : }A=\dfrac{x^3-x^2-10x-8}{x^3-4x^2+5x-20}\\ =\dfrac{x^3+x^2-2x^2-2x-8x-8}{\left(x^3-4x^2\right)+\left(5x-20\right)}\\ =\dfrac{\left(x^3+x^2\right)-\left(2x^2+2x\right)-\left(8x+8\right)}{x^2\left(x-4\right)+5\left(x-4\right)}\\ =\dfrac{x^2\left(x+1\right)-2x\left(x+1\right)-8\left(x+1\right)}{\left(x^2+5\right)\left(x-4\right)}\\ =\dfrac{\left(x^2-2x-8\right)\left(x+1\right)}{\left(x^2+5\right)\left(x-4\right)}\\ = \dfrac{\left(x^2-4x+2x-8\right)\left(x+1\right)}{\left(x^2+5\right)\left(x-4\right)}\\ =\dfrac{\left[\left(x^2-4x\right)+\left(2x-8\right)\right]\left(x+1\right)}{\left(x^2+5\right)\left(x-4\right)}\\ =\dfrac{\left[x\left(x-4\right)+2\left(x-4\right)\right]\left(x+1\right)}{\left(x^2+5\right)\left(x-4\right)}\\ =\dfrac{\left(x+2\right)\left(x-4\right)\left(x+1\right)}{\left(x^2+5\right)\left(x-4\right)}\\ =\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+5}\)
Vậy \(A=\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+5}\) với \(x\ne4\)
b) Với \(x\ne4\)
Để \(A\ge0\) thì \(\Rightarrow\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+5}\ge0\) \(\Rightarrow\left(x+2\right)\left(x+1\right)\ge0\left(\text{Vì }x^2+5>0\right)\) Lập bảng xét dấu: x+2 x+1 (x+1)(x+2) (x+1)(x+2) x -2 -1 0 0 0 0 _ + + _ _ + + _ + \(\Rightarrow\left[{}\begin{matrix}x\le-2\\x\ge-1\end{matrix}\right.\) Vậy để \(A\ge0\) thì \(x\le-2;x\ge-1\)
a) \(\frac{x^2-y^2}{\left(x+y\right)\left(ay-\text{ax}\right)}=\frac{\left(x+y\right)\left(x-y\right)}{-a\left(x+y\right)\left(x-y\right)}=\frac{-1}{a}\)
b) \(\frac{2ax-2x-3y+3ay}{4ax+\text{4x}+6y+6ay}=\frac{2x\left(a-1\right)+3y\left(a-1\right)}{\text{4x}\left(a+1\right)+6y\left(a+1\right)}\)
\(=\frac{\left(a-1\right)\left(2x+3y\right)}{2\left(a+1\right)\left(2x+3y\right)}=\frac{a-1}{2\left(a+1\right)}\)
BÀI 1:
a) \(ĐKXĐ:\) \(\hept{\begin{cases}x-2\ne0\\x+2\ne0\end{cases}}\) \(\Leftrightarrow\)\(\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}\)
b) \(A=\left(\frac{2}{x-2}-\frac{2}{x+2}\right).\frac{x^2+4x+4}{8}\)
\(=\left(\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right).\frac{\left(x+2\right)^2}{8}\)
\(=\frac{2x+4-2x+4}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x+2\right)^2}{8}\)
\(=\frac{x+2}{x-2}\)
c) \(A=0\) \(\Rightarrow\)\(\frac{x+2}{x-2}=0\)
\(\Leftrightarrow\) \(x+2=0\)
\(\Leftrightarrow\)\(x=-2\) (loại vì ko thỏa mãn ĐKXĐ)
Vậy ko tìm đc x để A = 0
p/s: bn đăng từng bài ra đc ko, mk lm cho
a) ĐK: \(x\ne-3;x\ne-2;x\ne1\)
\(A=\left(\frac{2-x}{x+3}+\frac{x-3}{x+2}+\frac{2-x}{\left(x+2\right)\left(x+3\right)}\right):\frac{x-1-x}{x-1}\)
\(=\frac{\left(2-x\right)\left(x+2\right)+\left(x-3\right)\left(x+3\right)+2-x}{\left(x+3\right)\left(x+2\right)}:\frac{-1}{x-1}\)
\(=\frac{4-x^2+x^2-9+2-x}{\left(x+2\right)\left(x+3\right)}.\left(1-x\right)\)
\(=\frac{-x-3}{\left(x+2\right)\left(x+3\right)}.\left(1-x\right)=\frac{-1}{x+2}.\left(1-x\right)=\frac{x-1}{x+2}\)
b) A = 0 \(\Leftrightarrow\)\(\frac{x-1}{x+2}=0\)
Do x khác -2 nên x - 1 = 0 hay x = 1 (loại vì ko thỏa ĐK)
A = 0 \(\Leftrightarrow\)\(\frac{x-1}{x+2}>0\)Xét 2 TH:
- TH1: x - 1 > 0 và x + 2 > 0 suy ra x > 1 và x > -2 nên ta chọn x > 1.
- TH1: x - 1 < 0 và x + 2 < 0 suy ra x < 1 và x < -2 nên ta chọn x < -2. Và x khác -3
Vậy để A > 0 thì x > 1 hoặc x < -2 \(\left(x\ne-3\right)\)
bài này dễ mà mk gợi ý rồi cậu tự làm ha . tách mẫu x^2 + 5x + 6 sau đó đặt nhân tử chung rồi tính con ve sau thì quy đồng lên rồi tính . mk goi y thế chắc cậu ko hiểu lắm đúng ko nhưg hiện h mk bạn làm chưa có ai thèm giải hộ mk có cậu làm đc phần đó thì giải hộ mk đi . Làm ơn !
\(a,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\) (x khác -3; khác 0)
\(=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x}{2x.\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x-x+6}{2x.\left(x+3\right)}=\frac{2x+6}{x.\left(2x+6\right)}=\frac{1}{x}\)
\(b,\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\) (x khác 0 , khác 1/2 khác -1/2 )
\(=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)
\(=\left(\frac{4x^2+4x+1}{\left(2x-1\right)\left(2x+1\right)}-\frac{4x^2-4x+1}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)
\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{5.\left(2x-1\right)}{4x}=\frac{10}{2x+1}\)
a: \(A=\left(\dfrac{2\left(2x+1\right)}{2\left(2x+4\right)}-\dfrac{x}{3x-6}-\dfrac{2x^3}{3x^3-12x}\right):\dfrac{6x+13x^2}{24x-12x^2}\)
\(=\left(\dfrac{2x+1}{2\left(x+2\right)}-\dfrac{x}{3\left(x-2\right)}-\dfrac{2x^3}{3x\left(x^2-4\right)}\right):\dfrac{x\left(13x+6\right)}{x\left(24-12x\right)}\)
\(=\left(\dfrac{2x+1}{2\left(x+2\right)}-\dfrac{x}{3\left(x-2\right)}-\dfrac{2x^2}{3\left(x-2\right)\left(x+2\right)}\right):\dfrac{13x+6}{-12\left(x-2\right)}\)
\(=\dfrac{3\left(2x+1\right)\left(x-2\right)-2x\left(x+2\right)-4x^2}{6\left(x+2\right)\left(x-2\right)}\cdot\dfrac{-12\left(x-2\right)}{13x+6}\)
\(=\dfrac{3\left(2x^2-3x-2\right)-2x^2-4x-4x^2}{x-2}\cdot\dfrac{-2}{13x+6}\)
\(=\dfrac{6x^2-9x-6-6x^2-4x}{x-2}\cdot\dfrac{-2}{13x+6}\)
\(=\dfrac{-\left(13x+6\right)\cdot\left(-2\right)}{\left(13x+6\right)\left(x-2\right)}=\dfrac{2}{x-2}\)
b: Để A>0 thì x-2>0
hay x>2
Để A>-1 thì A+1>0
\(\Leftrightarrow\dfrac{2+x-2}{x-2}>0\)
=>x/x-2>0
=>x>2 hoặc x<0
Bài 2 :
a) Phân thức A xác định \(\Leftrightarrow\hept{\begin{cases}x-2\ne0\\x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}}\)
b) \(A=\left(\frac{1}{x-2}-\frac{1}{x+2}\right)\cdot\frac{x^2-4x+4}{4}\)
\(A=\left(\frac{x+2}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{\left(x-2\right)^2}{4}\)
\(A=\left(\frac{x+2-x+2}{\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{\left(x-2\right)^2}{4}\)
\(A=\frac{4}{\left(x-2\right)\left(x+2\right)}\cdot\frac{\left(x-2\right)^2}{4}\)
\(A=\frac{4\cdot\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)\cdot4}\)
\(A=\frac{x-2}{x+2}\)
c) Thay x = 4 ta có :
\(A=\frac{4-2}{4+2}=\frac{2}{6}=\frac{1}{3}\)
Vậy.........
\(4x^2y^3.\frac{2}{4}x^3y=4x^2y^3.\frac{1}{2}x^3y=2x^5y^4\)
\(\left(5x-2\right)\left(25x^2+10x+4\right)\)
\(=\left(5x-2\right)\left[\left(5x\right)^2+5x.2+2^2\right]\)
\(=\left(5x\right)^3-2^3\)
\(=125x^3-8\)
Có: A=\(\frac{x^3-x^2-10x-8}{x^3-4x^2+5x-20}\)
A=\(\frac{\left(x^3-4x^2\right)+\left(3x^2-10x-8\right)}{x^2\left(x-4\right)+5\left(x-4\right)}\)
A=\(\frac{x^2\left(x-4\right)+\left(3x^2-12x+2x-8\right)}{\left(x^2+5\right)\left(x-4\right)}\)
A=\(\frac{x^2\left(x-4\right)+3x\left(x-4\right)+2\left(x-4\right)}{\left(x^2+5\right)\left(x-4\right)}\) ĐKXĐ:\(x\ne4\)
A=\(\frac{\left(x^2+3x+2\right)\left(x-4\right)}{\left(x^2+5\right)\left(x-4\right)}\) A=\(\frac{\left(x^2+x+2x+2\right)\left(x-4\right)}{\left(x^2+5\right)\left(x-4\right)}\) A=\(\frac{\left[x\left(x+1\right)+2\left(x+1\right)\right]\left(x-4\right)}{\left(x^2+5\right)\left(x-4\right)}\) A=\(\frac{\left(x+1\right)\left(x+2\right)\left(x-4\right)}{\left(x^2+5\right)\left(x-4\right)}\) A=\(\frac{\left(x+1\right)\left(x+2\right)}{x^2+5}\)Vậy A=\(\frac{\left(x+1\right)\left(x+2\right)}{x^2+5}\)với \(x\ne4\)
b) Có A=\(\frac{\left(x+1\right)\left(x+2\right)}{x^2+5}\text{với x}\ne4\)
A=0⇔\(\frac{\left(x+1\right)\left(x+2\right)}{x^2+5}=0\)
⇔(x+1)(x+2)=0 (vì \(x^2+5\ne0\))
⇔\(\left[{}\begin{matrix}x+1=0\\x+2=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)(Thoả mãn ĐKXĐ)
Vậy với x=1 hoặc x=2 thì A=0