a: ĐKXĐ: x<>0; x<>5; x<>5/2; x<>-5

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

\(=\dfrac{x^2-x^2+10x-25}{x\left(x-5\right)\left(x+5\right)}\cdot\dfrac{x\left(x+5\right)}{2x-5}=\dfrac{1}{x-5}\)

a) ĐKXĐ: \(\begin{cases}x\ne0\\x+5\ne0\end{cases}\Leftrightarrow\begin{cases}x\ne0\\x\ne-5\end{cases}\)

b)\(A=\frac{x^2+2x}{2x+10}+\frac{x+5}{x}-\frac{50-5x}{2x\left(x+5\right)}=\frac{x^2+2x}{2.\left(x+5\right)}+\frac{x+5}{x}-\frac{50-5x}{2x\left(x+5\right)}\)

\(=\frac{x^2+2x}{2x.\left(x+5\right)}+\frac{2\left(x+5\right)^2}{2x\left(x+5\right)}-\frac{50-5x}{2x\left(x+5\right)}\)

\(=\frac{x^2+2x+2x^2+20x+50-50+5x}{2x\left(x+5\right)}=\frac{3x^2+27x}{2x\left(x+5\right)}=\frac{3x.\left(x+9\right)}{2x\left(x+5\right)}=\frac{3x+27}{2x+10}\)

c)Để A=1 thì: \(\frac{3x+27}{2x+10}=1\Rightarrow3x+27=2x+10\Leftrightarrow x=-17\)(nhận)

Vậy x=-17 thì A=1

\(a.ĐKXĐ:\hept{\begin{cases}1-3x\ne0\\3x+1\ne0\\x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{3}\\...\\x\ge0\end{cases}}}\)

\(b,M=\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right):\frac{6x^2+10}{1-6x+9x^2}\)

\(=\left(\frac{3x\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\frac{2x\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}\right).\frac{\left(1-3x\right)^2}{6x^2+10}\)

\(=\left(\frac{3x+9x^2+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}\right).\frac{\left(1-3x\right)^2}{6x^2+10}\)

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

==>Sai đề không mem

\(P=\left(\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}\right):\frac{2x-5}{x^2+5x}+\frac{x}{5-x}\)

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

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

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

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

\(=\frac{5}{x-5}-\frac{x}{x-5}\)

\(=\frac{5-x}{x-5}\)

\(=\frac{-\left(x-5\right)}{x-5}\)

\(=-1\)

=> biểu thức  P k phụ thuộc vào x

a)Đk:\(\begin{cases}x^2-25\ne0\\x^2+5x\ne0\end{cases}\)\(\Leftrightarrow\begin{cases}\left(x-5\right)\left(x+5\right)\ne0\\x\left(x+5\right)\ne0\end{cases}\)\(\Leftrightarrow\begin{cases}x\ne5\\x\ne-5\\x\ne0\end{cases}\)

\(A=\frac{x}{x^2-25}-\frac{x-5}{x^2+5x}=\frac{x}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{x\left(x+5\right)}\)

\(=\frac{x^2}{x\left(x-5\right)\left(x+5\right)}-\frac{\left(x+5\right)^2}{x\left(x-5\right)\left(x+5\right)}=\frac{x^2-\left(x^2+10x+25\right)}{x\left(x-5\right)\left(x+5\right)}\)\(=\frac{10x-25}{x^3-25x}\)

a) ĐKXĐ : \(x\ne0;-5\)

b) \(A=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)

\(A=\frac{x\left(x^2+2x\right)}{2x\left(x+5\right)}+\frac{2\left(x+5\right)\left(x-5\right)}{2x\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}\)

\(A=\frac{x^3+2x^2+x^2-50+20-5x}{2x\left(x+5\right)}\)

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

\(A=\frac{x-1}{2}\)

c) \(A=1\Leftrightarrow\frac{x-1}{2}=1\Leftrightarrow x=3\)( thỏa )

\(A=-3\Leftrightarrow\frac{x-1}{2}=-3\Leftrightarrow x=-5\)( loại )

a, hông biết

b,

\(A=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x.\left(x+5\right)}\)

\(A=\frac{x^2+2x}{2x\left(x+5\right)}+\frac{x-5}{x}+\frac{50-5x}{2x+\left(x+5\right)}\)

\(A=\frac{x\left(x^2+2x\right)}{2x\left(x+5\right)}+\frac{2\left(x+5\right).\left(x-5\right)}{2x\left(x+5\right)}+\frac{50-5x}{2x.\left(x+5\right)}\)

\(A=\frac{x\left(x^2+2x\right)+2\left(x+5\right).\left(x-5\right)+50-5x}{2x.\left(x+5\right)}\)

\(A=\frac{x^3+2x+2\left(x^2-25\right)+50-5x}{2x\left(x+5\right)}\)

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

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

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

\(A=\frac{x.\left(x+5\right)-\left(x+5\right)}{2\left(x+5\right)}\)

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

\(A=\frac{x-1}{2}\)

c,

\(\left[{}\begin{matrix}\frac{x-1}{2}=1\\\frac{x-1}{2}=-3\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x-1=1.2=2\\x-1=-3.2=-6\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=2+1=3\\x=-6+1=-5\end{matrix}\right.\)

Vậy \(x\in\left\{3;-5\right\}\)

YNguyễn Thị Diễm Quỳnhlê thị hương giangBHLightning FarronoNguyễn Huy Túàng Tử HàonkiBastkVõ Đông Anh TuấnoongNguyễn Trần NhãVũ Huy HoàngPhan Huy Hoàng Asoyeon_Tiểubàng giảiMysterious PerPhương AnsonNguRibi Nkok Ngokyễn TNguyễn Thanh Hằngrần ThànDương Bá Gia Bảoh Nguyễn Thị Linh ChiĐạtnh

a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)

b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)

\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)

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

ĐKXĐ : \(\hept{\begin{cases}2x+10\ne0\\x\ne0\\2x\left(x+5\right)\ne0\end{cases}\Rightarrow x\ne0;x\ne-2\left(1\right)}\)

Ta có P = \(\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50+5x}{2x\left(x+5\right)}\)

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

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

\(=\frac{x^3+2x^2+2x^2-50+50+5x}{2x\left(x+5\right)}=\frac{x^3+4x^2+5x}{2x\left(x+5\right)}=\frac{x\left(x^2+4x+5\right)}{2x\left(x+5\right)}\)

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

c) P = 1

<=> \(\frac{x^2+4x+5}{2\left(x+5\right)}=1\Rightarrow x^2+4x+5=2\left(x+5\right)\)

=> x² + 4x + 5 - 2x - 10 = 0

=> x² + 2x - 5 = 0

=> x² + 2x + 1 - 6 = 0

=> (x + 1)² = 6

=> \(\orbr{\begin{cases}x+1=\sqrt{6}\\x+1=-\sqrt{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\sqrt{6}-1\\x=-\sqrt{6}-1\end{cases}}\)(tm (1))

d) P = -1/2

<=> \(\frac{x^2+4x+5}{2\left(x+5\right)}=-\frac{1}{2}\)

=> 2(x² + 4x + 5) = -2(x + 5)

=> 2x² + 8x + 10 = -2x - 10

=> 2x² + 8x + 10 + 2x + 10 = 0

=> 2x² + 10x + 20 = 0

=> 2(x² + 5x + 10) = 0

=> x² + 5x + 10 = 0

=> \(x^2+2.\frac{5}{2}x+\frac{25}{4}+\frac{15}{4}=0\)

=> \(\left(x+\frac{5}{2}\right)^2+\frac{15}{4}=0\)

=> \(x\in\varnothing\left(\text{Vì }\left(x+\frac{5}{2}\right)^2+\frac{15}{4}>0\forall x\right)\)

Vậy không tồn tại x để P = -1/2

\(P=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50+5x}{2x\left(x+5\right)}\)

a) ĐK : x ≠ 0 ; x ≠ -5

b) \(P=\frac{x\left(x+2\right)}{2\left(x+5\right)}+\frac{x-5}{x}+\frac{50+5x}{2x\left(x+5\right)}\)

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

\(=\frac{x^3+2x^2}{2x\left(x+5\right)}+\frac{2\left(x^2-25\right)}{2x\left(x+5\right)}+\frac{50+5x}{2x\left(x+5\right)}\)

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

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

\(=\frac{x^2+4x+5}{2x+10}\)

c) Để P = 1

thì \(\frac{x^2+4x+5}{2x+10}=1\)

=> x² + 4x + 5 = 2x + 10

=> x² + 4x + 5 - 2x - 10 = 0

=> x² - 2x - 5 = 0

=> ( x² - 2x + 1 ) - 6 = 0

=> ( x - 1 )² - ( √6 )² = 0

=> ( x - 1 - √6 )( x - 1 + √6 ) = 0

=> x = 1 + √6 hoặc x = 1 - √6

Cả hai giá trị đều thỏa x ≠ 0 ; x ≠ -5

Vậy x = 1 + √6 hoặc x = 1 - √6

d) Để P = -1/2

thì \(\frac{x^2+4x+5}{2x+10}=\frac{-1}{2}\)

=> 2( x² + 4x + 5 ) = -2x - 10

=> 2x² + 8x + 10 + 2x + 10 = 0

=> 2x² + 10x + 20 = 0

=> 2( x² + 5x + 10 ) = 0

=> x² + 5x + 10 = 0 (*)

Ta có : x² + 5x + 10 = ( x² + 5x + 25/4 ) + 15/4 = ( x + 5/2 )² + 15/4 ≥ 15/4 > 0 ∀ x

tức (*) không xảy ra

Vậy không có giá trị của x để P = -1/2

a,\(ĐKXĐ:\hept{\begin{cases}2x+10\ne0\\x\ne0\\2x\left(x+5\right)\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne-5\\x\ne0\\x\ne-5\end{cases}\Rightarrow}x\ne0;x\ne-5}\)

b.\(P=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50+5x}{2x\left(x+5\right)}\)

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

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

\(=\frac{x\left(x^2+2x\right)+2\left(x^2-25\right)+50+5x}{2x\left(x+5\right)}\)

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

\(=\frac{x^3+4x^2+5x}{2x\left(x+5\right)}=\frac{x\left(x^2+4x+5\right)}{2x\left(x+5\right)}=\frac{x^2+4x+5}{2x+10}\)

c,Để \(P=1\Leftrightarrow\frac{x^2+4x+5}{2x+10}=1\)

\(\Leftrightarrow x^2+4x+5=2x+10\)

\(\Leftrightarrow x^2+4x+5-2x-10=0\)

\(\Leftrightarrow x^2-2x-5=0\)

\(\Leftrightarrow\left(x^2-2x+1\right)-6=0\)

\(\Leftrightarrow\left(x-1\right)^2-\left(\sqrt{6}\right)^2=0\)

\(\Leftrightarrow\left(x-1-\sqrt{6}\right)\left(x-1+\sqrt{6}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1-\sqrt{6}\\x=1+\sqrt{6}\end{cases}}\)

d,Để \(P=-\frac{1}{2}\Leftrightarrow\frac{x^2+4x+5}{2x+10}=-\frac{1}{2}\)

\(\Leftrightarrow2\left(x^2+4x+5\right)=-2x-10\)

\(\Leftrightarrow2x^2+8x+10=-2x-10\)

\(\Leftrightarrow2x^2+8x+10+2x+10=0\)

\(\Leftrightarrow2x^2+10x+20=0\)

\(\Leftrightarrow2\left(x^2+5x+10\right)=0\)

\(\Leftrightarrow x^2+5x+10=0\)

\(\Leftrightarrow x^2+2x\frac{5}{2}+\frac{25}{5}-\frac{25}{4}+10=0\)

\(\Leftrightarrow\left(x+\frac{5}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\forall x\)

=> x² +5x+10 =0 không xảy ra 

Vậy không có giá trị nào tm để P=-1/2