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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
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)
\(P=\frac{x^2}{5x+25}+\frac{2x-10}{x}+\frac{50+5x}{x^2+5x}\)\(=\frac{x^2}{5\left(x+5\right)}+\frac{2\left(x-5\right)}{x}+\frac{5\left(x+10\right)}{x\left(x+5\right)}\)
\(=\frac{x^3}{5x\left(x+5\right)}+\frac{10\left(x-5\right)\left(x+5\right)}{5x\left(x+5\right)}+\frac{25\left(x+10\right)}{5x\left(x+5\right)}\)
\(=\frac{x^3+10\left(x-5\right)\left(x+5\right)+25\left(x+10\right)}{5x\left(x+5\right)}=\frac{x^3+10\left(x^2-25\right)+25x+250}{5x\left(x+5\right)}\)
\(=\frac{x^3+10x^2-250+25x+250}{5x\left(x+5\right)}=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}\)\(=\frac{x\left(x^2+10x+25\right)}{5x\left(x+5\right)}\)\(=\frac{\left(x+5\right)^2}{5\left(x+5\right)}=\frac{x+5}{5}\)
b) \(x^2-3x=0\)\(\Leftrightarrow x\left(x-3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
So sánh với ĐKXĐ, ta thấy \(x=0\)không thoả mãn
Thay \(x=3\)vào biểu thức ta được: \(P=\frac{3+5}{5}=\frac{8}{5}\)
c) Để \(P=-4\)thì \(\frac{x+5}{5}=-4\)\(\Leftrightarrow x+5=-20\)\(\Leftrightarrow x=-25\)( thoả mãn ĐKXĐ )
Vậy \(P=-4\)\(\Leftrightarrow x=-25\)
d) Để \(P\ge0\)thì \(\frac{x+5}{5}\ge0\)\(\Leftrightarrow x+5\ge0\)( vì \(5>0\))\(\Leftrightarrow x\ge-5\)
So sánh với ĐKXĐ, ta thấy x phải thoả mãn \(x>-5\)và \(x\ne0\)
Vậy \(P\ge0\)\(\Leftrightarrow\)\(x>-5\)và \(x\ne0\)
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) \(\frac{2ax-2x-3y+3ay}{4ax+\text{4x}+6y+6ay}=\frac{2x\left(a-1\right)+3y\left(a-1\right)}{\text{4x}\left(a+1\right)+6y\left(a+1\right)}\)
\(=\frac{\left(a-1\right)\left(2x+3y\right)}{2\left(a+1\right)\left(2x+3y\right)}=\frac{a-1}{2\left(a+1\right)}\)
\(P=\frac{2\left(x-2\right)\left(x+2\right)}{x^2+x+5}.\frac{5\left(x^2+x+5\right)}{\left(x-4\right)\left(x+3\right)}.\frac{\left(x-1\right)\left(x-4\right)}{10\left(x-2\right)\left(x+2\right)}=\frac{x-1}{x+3}\)
ĐK: \(x\ne\left\{4;-3;1;2;-2\right\}\)
b, \(P\in Z\Rightarrow\frac{x-1}{x+3}\in Z\Rightarrow x-1⋮\left(x+3\right)\Rightarrow-4⋮\left(x+3\right)\Rightarrow\left(x+3\right)\in\left\{-4;-2;-1;1;2;4\right\}\)
\(\Rightarrow x\in\left\{-7;-5;-4;-2;-1;1\right\}\)
\(\Rightarrow P\in\left\{2;3;5;-3;-1;0\right\}\)
Bài 1:
a) x≠2x≠2
Bài 2:
a) x≠0;x≠5x≠0;x≠5
b) x2−10x+25x2−5x=(x−5)2x(x−5)=x−5xx2−10x+25x2−5x=(x−5)2x(x−5)=x−5x
c) Để phân thức có giá trị nguyên thì x−5xx−5x phải có giá trị nguyên.
=> x=−5x=−5
Bài 3:
a) (x+12x−2+3x2−1−x+32x+2)⋅(4x2−45)(x+12x−2+3x2−1−x+32x+2)⋅(4x2−45)
=(x+12(x−1)+3(x−1)(x+1)−x+32(x+1))⋅2(2x2−2)5=(x+12(x−1)+3(x−1)(x+1)−x+32(x+1))⋅2(2x2−2)5
=(x+1)2+6−(x−1)(x+3)2(x−1)(x+1)⋅2⋅2(x2−1)5=(x+1)2+6−(x−1)(x+3)2(x−1)(x+1)⋅2⋅2(x2−1)5
=(x+1)2+6−(x2+3x−x−3)(x−1)(x+1)⋅2(x−1)(x+1)5=(x+1)2+6−(x2+3x−x−3)(x−1)(x+1)⋅2(x−1)(x+1)5
=[(x+1)2+6−(x2+2x−3)]⋅25=[(x+1)2+6−(x2+2x−3)]⋅25
=[(x+1)2+6−x2−2x+3]⋅25=[(x+1)2+6−x2−2x+3]⋅25
=[(x+1)2+9−x2−2x]⋅25=[(x+1)2+9−x2−2x]⋅25
=2(x+1)25+185−25x2−45x=2(x+1)25+185−25x2−45x
=2(x2+2x+1)5+185−25x2−45x=2(x2+2x+1)5+185−25x2−45x
=2x2+4x+25+185−25x2−45x=2x2+4x+25+185−25x2−45x
=2x2+4x+2+185−25x2−45x=2x2+4x+2+185−25x2−45x
=2x2+4x+205−25x2−45x=2x2+4x+205−25x2−45x
c) tự làm, đkxđ: x≠1;x≠−1
a) \(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne-3\end{cases}}\)
b) \(P=1+\frac{x+3}{x^2+5x+6}\div\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{1}{x+2}\right)\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{2x+4-x-x+2}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{1}{x+2}:\frac{6}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{\left(x-2\right)\left(x+2\right)}{6\left(x+2\right)}\)
\(\Leftrightarrow P=1+\frac{x-2}{6}\)
\(\Leftrightarrow P=\frac{x+4}{6}\)
c) Để P = 0
\(\Leftrightarrow\frac{x+4}{6}=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Để P = 1
\(\Leftrightarrow\frac{x+4}{6}=1\)
\(\Leftrightarrow x+4=6\)
\(\Leftrightarrow x=2\)
d) Để P > 0
\(\Leftrightarrow\frac{x+4}{6}>0\)
\(\Leftrightarrow x+4>0\)(Vì 6>0)
\(\Leftrightarrow x>-4\)
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 \(x\le-2;x\ge-1\)