Bài 3: Rút gọn phân thức

Ta có:

\(\dfrac{x^{24}+x^{20}+x^{16}+x^{12}+...+x^4+1}{x^{26}+x^{24}+x^{22}+x^{20}+...+x^2+1}\)

Xét \(M=x^{24}+x^{20}+x^{16}+x^{12}+...+x^4+1\)

\(\Rightarrow x^4M=x^{28}+x^{24}+x^{20}+x^{16}+...+x^8+x^4\)

\(\Rightarrow x^4M-M=\left(x^{28}+x^{24}+x^{20}+...+x^8+x^4\right)-\left(x^{24}+x^{20}+x^{16}+...+x^4+1\right)\)

\(\Rightarrow\left(x^4-1\right)M=x^{28}-1\)

\(\Rightarrow M=\dfrac{x^{28}-1}{x^4-1}\)

Xét \(N=x^{26}+x^{24}+x^{22}+x^{20}+...+x^2+1\)

\(\Rightarrow x^2N=x^{28}+x^{26}+x^{24}+x^{20}+...+x^4+x^2\)

\(\Rightarrow x^2N-N=\left(x^{28}+x^{26}+x^{24}+...+x^4+x^2\right)-\left(x^{26}+x^{24}+x^{22}+...+x^2+1_{ }\right)\)

\(\Rightarrow\left(x^2-1\right)N=x^{28}-1\)

\(\Rightarrow N=\dfrac{x^{28}-1}{x^2-1}\)

Ta có:

\(\dfrac{x^{24}+x^{20}+x^{16}+x^{12}+...+x^4+1}{x^{26}+x^{24}+x^{22}+x^{20}+...+x^2+1}\)

\(=\dfrac{M}{N}=\dfrac{\dfrac{x^{28}-1}{x^4-1}}{\dfrac{x^{28}-1}{x^2-1}}\)

\(=\dfrac{x^{28}-1}{x^4-1}.\dfrac{x^2-1}{x^{28}-1}=\dfrac{x^2-1}{x^4-1}\)

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

Chúc bạn học tốt!

Bài 1: 

8: \(=\dfrac{x+3}{x\left(x-3\right)}\)

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

10: \(=1:\dfrac{a-1}{a}=\dfrac{a}{a-1}\)

12: \(=\dfrac{6\left(x+1\right)}{3x\left(x+1\right)}=\dfrac{2}{x}\)

13: \(\dfrac{3}{x+3}-\dfrac{x-6}{x\left(x+3\right)}\)

\(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2x+6}{x\left(x+3\right)}=\dfrac{2}{x}\)

a, \(\sqrt{b}\) tồn tại \(\Leftrightarrow b>0\)

\(\left\{{}\begin{matrix}\sqrt{b}+1\ne0\\\sqrt{b}-1\ne0\\b-1\ne0\end{matrix}\right.\Leftrightarrow b\ne1\)

Vậy B có nghĩa khi \(\left\{{}\begin{matrix}b>0\\b\ne1\end{matrix}\right.\)

b,

\(B=\dfrac{\sqrt{b}}{\sqrt{b}+1}-\dfrac{\sqrt{b}}{\sqrt{b}-1}-\dfrac{2}{b-1}\)

\(=\dfrac{\sqrt{b}}{\sqrt{b}+1}-\dfrac{\sqrt{b}}{\sqrt{b}-1}-\dfrac{2}{\left(\sqrt{b}-1\right)\left(\sqrt{b}+1\right)}\)

\(=\dfrac{\sqrt{b}\left(\sqrt{b}-1\right)-\sqrt{b}\left(\sqrt{b}+1\right)-2}{\left(\sqrt{b}-1\right)\left(\sqrt{b}+1\right)}=\dfrac{b-\sqrt{b}-b-\sqrt{b}-2}{\left(\sqrt{b}-1\right)\left(\sqrt{b}+1\right)}=\dfrac{-2\left(\sqrt{b}+1\right)}{\left(\sqrt{b}-1\right)\left(\sqrt{b}+1\right)}=\dfrac{-2}{\sqrt{b}-1}=\dfrac{2}{1-\sqrt{b}}\)

c,

\(B>1\Leftrightarrow2>1-\sqrt{b}\)

\(\Leftrightarrow2-\left(1-\sqrt{b}\right)=1+\sqrt{b}>0\) (luôn đúng với mọi b)

=> Với mọi b có ĐKXĐ là b khác 0 và b > 1 thì B > 1

a )

ĐKXĐ : \(y\ne0\) , \(y\ne-1\)

\(P=\left(\dfrac{2y^2+1}{y^3+1}-\dfrac{y}{y+y^2}\right):\left(1-\dfrac{y^2-2y-1}{y^2-y+1}\right)\)

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

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

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

\(=\dfrac{y}{y+2}\)

Câu b :

\(\left|2y+5\right|=3\)

\(\Leftrightarrow\left[{}\begin{matrix}2y+5=3\\-2y-5=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=-1\\y=-4\end{matrix}\right.\)

Thay \(y=-1\) vào P ta được : \(P=\dfrac{-1}{-1+2}=-1\)

Thay \(y=-4\) vào P ta được : \(P=\dfrac{-4}{-4+2}=2\)

Câu c :

T chỉ biết lập luận thôi :

Để P chia hết cho 4 thì \(\dfrac{y}{y+2}\) chia hết cho 4 hay \(\dfrac{y}{y+2}\) phải là bội của 4.

Do \(y< y+2\) nên \(\dfrac{y}{y+2}\) không thể là các số 4 ; 8 ;12 ;.........

Nên \(\dfrac{y}{y+2}=0\) thì sẽ chia hết cho 4 . \(\Leftrightarrow y=0\) ( Loại )

Nên không có giá trị y nào hết .

Câu d :

\(P=3-m>2\)

\(\Leftrightarrow-m>-1\)

\(\Leftrightarrow m< 1\)

1/ đkxđ: x≠\(\pm\)1; x≠1/2

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

\(=\left(\dfrac{x+1}{\left(1-x\right)\left(1+x\right)}+\dfrac{2\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

\(=\dfrac{x+1+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)

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

b/ A nguyên <=> 1 - 2x ∈ Ư(2)

<=> 1 - 2x = {-2;-1;1;2}

<=> -2x = {-3; -2; 0;1}

<=> x = {3/2; 1; 0; -1/2}

mà x nguyên => x = {1;0}

c/ \(\left|A\right|=A\Leftrightarrow\left|\dfrac{2}{1-2x}\right|=\dfrac{2}{1-2x}\)

+) Với x > 1/2 có:

\(\dfrac{2}{1-2x}=\dfrac{2}{1-2x}\Leftrightarrow\dfrac{2}{1-2x}-\dfrac{2}{1-2x}=0\Leftrightarrow0x=0\)

=> x>1/2 thỏa mãn là nghiệm

+) Với x < 1/2 có:

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

\(\Leftrightarrow\dfrac{2}{1-2x}-\dfrac{2}{2x-1}=0\Leftrightarrow\dfrac{2}{1-2x}+\dfrac{2}{1-2x}=0\)

\(\Leftrightarrow\dfrac{4}{1-2x}=0\) mà 1 - 2x ≠ 0 => vô nghiệm

Vậy x>1/2

Ta có:

\(ax+by+cz=0\Rightarrow\left(ax+by+cz\right)^2=0\)

\(\Rightarrow a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz=0\)

\(\Rightarrow a^2x^2+b^2y^2+c^2z^2=-2axby-2bycz-2axcz\)

Ta có:

\(bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2=bc\left(y^2-2yz+z^2\right)+ca\left(z^2-2xz+x^2\right)+ab\left(x^2-2xy+y^2\right)\)

\(=bcy^2-2bcyz+bcz^2+acz^2-2acxz+acx^2+abx^2-2abxy+aby^2\)

\(=bcy^2+bcz^2+acz^2+acx^2+abx^2+aby^2-2axby-2bycz-2axcz\)

\(=bcy^2+bcz^2+acz^2+acx^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2\)

\(=\left(abx^2+a^2x^2+acx^2\right)+\left(bcy^2+aby^2+b^2y^2\right)+\left(bcz^2+acz^2+c^2z^2\right)\)

\(=ax^2\left(b+a+c\right)+by^2\left(c+a+b\right)+cz^2\left(b+a+c\right)\)

\(=\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)\)

Thay vào A ta được:

\(A=\dfrac{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}{ax^2+by^2+cz^2}=a+b+c\)

Bài2:

a: \(Q=\left(\dfrac{2x+1}{x\left(x-5\right)}-\dfrac{2x}{x\left(x+5\right)}\right)\cdot\dfrac{x\left(x-5\right)\left(x+5\right)}{21x-2}\)

\(=\dfrac{2x^2+11x+5-2x^2+10x}{x\left(x-5\right)\left(x+5\right)}\cdot\dfrac{x\left(x-5\right)\left(x+5\right)}{21x-2}\)

\(=\dfrac{21x+5}{21x-2}\)

b: Để Q là số nguyên thì \(21x-2+7⋮21x-2\)

\(\Leftrightarrow21x-2\in\left\{1;-1;7;-7\right\}\)

hay \(x\in\left\{\dfrac{1}{7};\dfrac{1}{21};\dfrac{3}{7};-\dfrac{5}{21}\right\}\)

B1:

a) A = \(\dfrac{1}{x+2}+\dfrac{x^2-x-2}{x^2-7x+10}-\dfrac{2x-4}{x-5}\)

= \(\dfrac{1}{x+2}+\dfrac{\left(x^2-2x\right)+\left(x-2\right)}{\left(x^2-2x\right)-\left(5x-10\right)}-\dfrac{2\left(x-2\right)}{x-5}\)

= \(\dfrac{1}{x+2}+\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{2\left(x-2\right)}{x-5}\) [ĐKXĐ: x ≠ -2; x ≠ 5]

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

= \(\dfrac{-x^2+4x+5}{\left(x+2\right)\left(x-5\right)}\)

= \(\dfrac{-x\left(x-5\right)-\left(x-5\right)}{\left(x+2\right)\left(x-5\right)}\)

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

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

b) Thay x = 3 vào A, ta có:

A = \(-\dfrac{3+1}{3+2}=-\dfrac{4}{5}\)

c) A = 1

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

<=> -2x = 3 <=> x = \(\dfrac{-3}{2}\)

d) A = \(\dfrac{-\left(x+1\right)}{x+2}\)= \(\dfrac{-\left(x+2\right)+1}{x+2}\)= -1 + \(\dfrac{1}{x+2}\)

A đạt giá trị nguyên khi 1 chia hết cho x + 2 hay x + 2 ∈ Ư(1) = {1;-1}

* x + 2 = 1 <=> x = -1

* x + 2 = -1 <=> x = -3

B2: M = \(\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)

= \(\dfrac{x\left(x+2\right)}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{5\left(10-x\right)}{2x\left(x+5\right)}\)[ĐKXĐ: x ≠ 0; x ≠ -5

= \(\dfrac{x^2\left(x+2\right)+2\left(x+5\right)\left(x-5\right)+5\left(10-x\right)}{2x\left(x+5\right)}\)

= \(\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)

= \(\dfrac{x^2+4x-5}{2\left(x+5\right)}\)

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

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

b) Thay x = 3 vào M, ta có:

M = \(\dfrac{3-1}{2}=1\)

Thay x = 5 vào M, ta có:

M = \(\dfrac{5-1}{2}=2\)

Bài 2: 

a: \(M=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2+2\left(x-5\right)\left(x+5\right)+50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2+50-5x+2x^2-50}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)

b: Khi x=3 thì \(M=\dfrac{3-1}{2}=\dfrac{2}{2}=1\)

Khi x=5 thì \(M=\dfrac{5-1}{2}=\dfrac{4}{2}=2\)