Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(P=\frac{4x^3+8x^2+x-2}{4x^2+4x+1}=\frac{\left(x+2\right)\left(2x-1\right)\left(2x+1\right)}{\left(2x+1\right)^2}\)
ĐKXĐ :\(\left(2x+1\right)^2\ne0=>2x+1\ne0=>x\ne-\frac{1}{2}\)
b) \(P=\frac{3}{2}\Leftrightarrow\frac{\left(x+2\right)\left(2x-1\right)\left(2x+1\right)}{\left(2x+1\right)^2}=\frac{3}{2}\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(2x-1\right)}{2x+1}=\frac{3}{2}\Leftrightarrow4x^2-2x+8x-4=6x+3\)
\(\Rightarrow4x^2=7=>x^2=\frac{7}{4}=>x=\pm\sqrt{\frac{7}{4}}\)
c) \(P=\frac{\left(x+2\right)\left(2x-1\right)}{\left(2x+1\right)}=\frac{\left(x+2\right)\left(2x+1-2\right)}{2x+1}=\frac{\left(x+2\right)\left(2x+1\right)-2\left(x+2\right)}{2x+1}\)
\(=x+2-\frac{2x+2}{2x+1}=x+2-1-\frac{1}{2x+1}\)
để P nguyền khi zà chỉ khi
\(1⋮2x+1\)
\(=>2x+1\inƯ\left(1\right)=\pm1\)
=>\(\orbr{\begin{cases}2x+1=1\\2x+1=-1\end{cases}=>\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
\(\text{Đk:}x\ne-\frac{1}{2}\Rightarrow P=\frac{4x^2\left(x+2\right)-\left(x+2\right)}{\left(2x+1\right)^2}=\frac{\left(4x^2-1\right)\left(x+2\right)}{\left(2x+1\right)^2}=\frac{\left(2x-1\right)\left(x+2\right)}{2x+1}\)
\(=\frac{2x^2+4x-x-2}{2x+1}=\frac{3}{2}\Rightarrow2x^2+3x-2=3x+\frac{3}{2}\Leftrightarrow2x^2-\frac{7}{2}=0......\)
\(P\text{ nguyên }\Rightarrow2x^2+3x-2⋮2x+1\Leftrightarrow2x^2+3x-2-\left(x+1\right)\left(2x+1\right)⋮2x+1\Leftrightarrow-3⋮2x+1....\)
a) ĐKXĐ của phương trình : \(4x^2+4x+1\ne0\)\(\Rightarrow x\ne-\frac{1}{2}\)
b) \(P=\frac{4x^3+8x^2-x-2}{4x^2+4x+1}\)
\(\Rightarrow P=\frac{\left(4x^3-x\right)+\left(8x^2-2\right)}{\left(2x+1\right)^2}\)
\(\Rightarrow P=\frac{x\left(4x^2-1\right)+2\left(4x^2-1\right)}{\left(2x+1\right)^2}\)
\(\Rightarrow P\left(x\right)=\frac{\left(x+2\right)\left(2x-1\right)\left(2x+1\right)}{\left(2x+1\right)^2}\)
\(\Rightarrow P\left(x\right)=\frac{\left(x+2\right)\left(2x-1\right)}{\left(2x+1\right)}=\frac{3}{2}\)\(\Rightarrow P\left(x\right)=2\left(x+2\right)\left(2x-1\right)=3\left(2x+1\right)\)
\(\Rightarrow P\left(x\right)=4x^2+6x-6-\left(6x+3\right)=0\)
\(\Rightarrow P\left(x\right)=4x^2-9=0\)\(\Rightarrow P\left(x\right)=x^2=\frac{9}{4}\)
\(\Rightarrow P\left(x\right)=x^2=\sqrt{\frac{9}{4}}\)\(\Rightarrow P\left(x\right)=\frac{3}{2}\)
câu c) cx tương tự
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
\(B=\frac{x^2-2}{x^2+1}=\frac{x^2+1-3}{x^2+1}=1-\frac{3}{x^2+1}\)
\(B_{min}\Rightarrow\left(\frac{3}{x^2+1}\right)_{max}\Rightarrow\left(x^2+1\right)_{min}\)
\(x^2+1\ge1\). dấu = xảy ra khi x2=0
=> x=0
Vậy \(B_{min}\Leftrightarrow x=0\)
ta có: \(x^2+2x-2=x^2+2x+1^2-3=\left(x+1\right)^2-3\ge-3\)
dấu = xảy ra khi \(x+1=0\)
\(\Rightarrow x=-1\)
Vậy\(\left(x^2+2x-2\right)_{min}\Leftrightarrow x=-1\)
Câu 1 :
a) ĐKXĐ : \(\hept{\begin{cases}x+1\ne0\\2x-6\ne0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)
b) Để \(P=1\Leftrightarrow\frac{4x^2+4x}{\left(x+1\right)\left(2x-6\right)}=1\)
\(\Leftrightarrow\frac{4x^2+4x-\left(x+1\right)\left(2x-6\right)}{\left(x+1\right)\left(2x-6\right)}=0\)
\(\Rightarrow4x^2+4x-2x^2+4x+6=0\)
\(\Leftrightarrow2x^2+8x+6=0\)
\(\Leftrightarrow x^2+4x+4-1=0\)
\(\Leftrightarrow\left(x+2-1\right)\left(x+2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=-3\left(TMĐKXĐ\right)\end{cases}}\)
Vậy : \(x=-3\) thì P = 1.
\(B=\sqrt{x^2+8x+14}+\sqrt{9-x^2}\)
ĐKXĐ :
\(\hept{\begin{cases}x^2+8x+14\ge0\\9-x^2\ge0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x\ge-4-\sqrt{2}\\x\le3\end{cases}}\)
\(\Leftrightarrow-4-\sqrt{2}\le x\le3\)