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\(ĐKXĐ:x\ne\pm1\)
a) \(P=\frac{2x+3}{x+1}-\frac{x+2}{x-1}+\frac{3x+5}{x^2-1}\)
\(\Leftrightarrow P=\frac{\left(2x+3\right)\left(x-1\right)-\left(x+2\right)\left(x+1\right)+3x+5}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow P=\frac{2x^2+x-3-x^2-3x-2+3x+5}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow P=\frac{x^2+x}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow P=\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow P=\frac{x}{x-1}\)
b) Để \(P\inℤ\)
\(\Leftrightarrow x⋮x-1\)
\(\Leftrightarrow x-1+1⋮x-1\)
\(\Leftrightarrow1⋮x-1\)
\(\Leftrightarrow x-1\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Leftrightarrow x\in\left\{0;2\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{0;2\right\}\)
a, Ta có : \(A=\frac{1}{x+2}-\frac{2x}{4-x^2}+\frac{3}{x-2}\)
\(=\frac{1}{x+2}-\frac{2x}{\left(2-x\right)\left(x+2\right)}+\frac{3}{x-2}\)
\(=\frac{x-2}{\left(x+2\right)\left(x-2\right)}+\frac{2x}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x-2+2x+3x+6}{\left(x-2\right)\left(x+2\right)}=\frac{6x+4}{\left(x-2\right)\left(x+2\right)}\)
Suy ra : \(M=\frac{6x+4}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{3x+2}\)
\(=\frac{2\left(3x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)\left(3x+2\right)}=\frac{2}{x-2}\)
a ) ĐKXĐ : \(\hept{\begin{cases}x+1\ne0\\x-1\ne0\\x^2-1\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne-1\\x\ne1\end{cases}}}\)
b ) \(P=\frac{2x+3}{x+1}-\frac{x+2}{x-1}+\frac{3x+5}{x^2-1}\)
\(=\frac{\left(2x+3\right)\left(x-1\right)-\left(x+2\right)\left(x+1\right)+\left(3x+5\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{\left(2x^2+x-3\right)-\left(x^2+3x+2\right)+\left(3x+5\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2-x}{\left(x-1\right)\left(x+1\right)}=\frac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x}{x+1}\)
Sr còn thiếu
Để \(P\in Z\Leftrightarrow\frac{x}{x+1}=\frac{x+1-1}{x+1}=1-\frac{1}{x+1}\in Z\Rightarrow x+1\inƯ\left(1\right)\)
\(\Rightarrow x+1=\left\{-1;1\right\}\Rightarrow x=\left\{-2;0\right\}\)
a) \(A = \frac{2x^2 - 16x+43}{x^2-8x+22}\) = \(\frac{2(x^2-8x+22)-1}{x^2-8x+22}\) = \(2 - \frac{1}{x^2-8x+22}\)
Ta có : \(x^2-8x+22 \) = \(x^2-8x+16+6 = ( x-4)^2 +6 \)
Vì \((x-4)^2 \ge 0 \) với \( \forall x\in R\) Nên \(( x-4)^2 +6 \ge 6 \)
\(\Rightarrow \) \(x^2-8x+22 \) \( \ge 6\)\(\Rightarrow \) \(\frac{1}{x^2-8x+22} \) \(\le \frac{1}{6}\) \(\Rightarrow \) - \(\frac{1}{x^2-8x+22} \) \(\ge - \frac{1}{6}\)
\(\Rightarrow \) A = \(2 - \frac{1}{x^2-8x+22}\) \( \ge 2-\frac{1}{6}\) = \(\frac{11}{6}\) Dấu "=" xảy ra khi và chỉ khi x=4
Vậy GTNN của A = \(\frac{11}{6}\) khi và chỉ khi x=4
a) \(A=\left(\dfrac{1}{3}+\dfrac{3}{x^2-3x}\right):\left(\dfrac{x^2}{27-3x^2}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3\left(x^2-3x\right)}:\left(\dfrac{x^2}{3\left(9-x^2\right)}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\left(\dfrac{x^2}{3.\left(3-x\right).\left(3+x\right)}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\dfrac{x^2+3.\left(3-x\right)}{3.\left(3-x\right).\left(3+x\right)}\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\dfrac{x^2+9-3x}{3.\left(3-x\right).\left(3+x\right)}\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}.\dfrac{3.\left(3x-x\right).\left(3+x\right)}{x^2+9-3x}\)
\(\Rightarrow A=\dfrac{1}{x.\left(x-3\right)}.\left(-\left(x-3\right)\right).\left(3+x\right)\)
\(\Rightarrow A=\dfrac{1}{x}.\left(-1\right).\left(3+x\right)\)
\(\Rightarrow A=-\dfrac{1}{x}.\left(3+x\right)\)
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
B1:dài quá :vv
B2:\(Q=\frac{x^2}{x^4+x^2+1}=\frac{x^2}{x^4+2x^2+1-x^2}=\frac{x^2}{\left(x^2+1\right)-x^2}=\frac{x^2}{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)
\(=\frac{x}{x^2-x+1}.\frac{x}{x^2+x+1}=\frac{2}{3}.\frac{x}{x^2+x+1}\)
\(\frac{x}{x^2-x+1}=\frac{2}{3}\Rightarrow\frac{x^2-x+1}{x}=\frac{3}{2}\Rightarrow\frac{x^2-x+1}{x}+2=\frac{3}{2}+2\Rightarrow\frac{x^2+x+1}{x}=\frac{7}{2}\)
\(\Rightarrow\frac{x}{x^2+x+1}=\frac{2}{7}\Rightarrow Q=\frac{2}{3}.\frac{2}{7}=\frac{4}{21}\)
3.
Ta có: \(a^5-a=a\left(a^4-1\right)=a\left(a^2-1\right)\left(a^2+1\right)=a\left(a+1\right)\left(a-1\right)\left(a^2+1\right)\)
\(=a\left(a-1\right)\left(a+1\right)\left(a^2-4+5\right)=a\left(a-1\right)\left(a+1\right)\left(a^2-4\right)+5a\left(a-1\right)\left(a+1\right)\)
\(=a\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)+5a\left(a-1\right)\left(a+1\right)\)
Do a(a-1)(a+1)(a-2)(a+2) là tích của 5 số hạng liên tiếp nên chia hết cho 2,3 và 5
Lại có a(a-1)(a+1) là tích của 3 số hạng liên tiếp nên chia hết cho 2,3 suy ra 5a(a-1)(a+1) chia hết cho 2,3,5
Từ đó:a(a-1)(a+1)(a-1)(a+2)+5a(a-1)(a+1) chia hết cho 2,3,5 hay a(a-1)(a+1)(a-2)(a+2)+5a(a-1)(a+1) chia hết cho 30 \(\Leftrightarrow a^5-a\) chia hết cho 30
Tương tự ta có\(b^5-b\) chia hết cho 30, \(c^5-c\) chia hết cho 30
Do đó:\(a^5-a+b^5-b+c^5-c⋮30\)
\(\Leftrightarrow a^5+b^5+c^5-\left(a+b+c\right)⋮30\)
Mà a+b+c=0 nên;
\(a^5+b^5+c^5⋮30\left(ĐCCM\right)\)