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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 của A : \(\hept{\begin{cases}2x-3\ne0\\2x+3\ne0\\9-4x^2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}2x\ne3\\2x\ne-3\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne\frac{3}{2}\\x\ne-\frac{3}{2}\end{cases}}}\)
=> Giá trị của biểu thức A được xác định khi x khác 3/2 và x khác -3/2
\(A=\frac{5}{2x-3}+\frac{2}{2x+3}-\frac{2x+5}{9-4x^2}\)
\(=\frac{5}{2x-3}+\frac{2}{2x+3}+\frac{2x+5}{\left(2x+3\right)\left(2x-3\right)}\)
\(=\frac{5.\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}+\frac{2.\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}+\frac{2x+5}{\left(2x+3\right)\left(2x-3\right)}\)
\(=\frac{10x+15+4x-6+2x+5}{\left(2x+3\right)\left(2x-3\right)}\)
..... chắc tôi làm sai oy !
a, ĐKXĐ: \(\hept{\begin{cases}5x+25\ne0\\x\ne0\\x^2+5x\ne0\end{cases}\Rightarrow\hept{\begin{cases}5\left(x+5\right)\ne0\\x\ne0\\x\left(x+5\right)\ne0\end{cases}\Rightarrow}}\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)
b, \(P=\frac{x^2}{5x+25}+\frac{2x-10}{x}+\frac{50+5x}{x^2+5x}\)
\(=\frac{x^3}{5x\left(x+5\right)}+\frac{5\left(2x-10\right)\left(x+5\right)}{5x\left(x+5\right)}+\frac{\left(50+5x\right).5}{5x\left(x+5\right)}\)
\(=\frac{x^3+10\left(x-5\right)\left(x+5\right)+250+25x}{5x\left(x+5\right)}\)
\(=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}=\frac{x+5}{5}\)
c, \(P=-4\Rightarrow\frac{x+5}{5}=-4\Rightarrow x+5=-20\Rightarrow x=-25\)
d, \(\frac{1}{P}\in Z\Rightarrow\frac{5}{x+5}\in Z\Rightarrow5⋮\left(x+5\right)\Rightarrow x+5\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\Rightarrow x\in\left\{-10;-6;-4;0\right\}\)
Mà x khác 0 (ĐKXĐ của P) nên \(x\in\left\{-10;-6;-4\right\}\)
a) \(ĐKXĐ:\hept{\begin{cases}5x+25\ne0\\x\ne0\\x^2+5x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)
b) \(P=\frac{x^2}{5x+25}+\frac{2x-10}{x}+\frac{50+5x}{x^2+5x}\)
\(P=\frac{x^3}{5x\left(x+5\right)}+\frac{10x^2-250}{5x\left(x+5\right)}+\frac{250+25x}{5x\left(x+5\right)}\)
\(P=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}=\frac{x+5}{5}\)
c) \(P=4\Leftrightarrow\frac{x+5}{5}=4\Leftrightarrow x+5=20\Leftrightarrow x=15\)
d) \(\frac{1}{P}=\frac{5}{x+5}\in Z\Leftrightarrow5⋮x+5\)
\(\Leftrightarrow x+5\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Lập bảng nhé
e) \(Q=P+\frac{x+25}{x+5}=\frac{x+30}{x+5}=1+\frac{25}{x+5}\)
\(Q_{min}\Leftrightarrow\frac{25}{x+5}_{min}\)
a) ĐK: \(x\ne\left\{0;\pm7;49\right\}\)
b) \(\left(\frac{x}{x^2-49}-\frac{x-7}{x^2-7x}\right):\frac{2x-7}{x^2+7x}-\frac{x}{x-7}\)
Xét \(\frac{x}{x^2-49}-\frac{x-7}{x^2-7x}\)
= \(\frac{x^2}{x\left(x-7\right)\left(x+7\right)}-\frac{\left(x-7\right)\left(x+7\right)}{x\left(x-7\right)\left(x+7\right)}\)
=\(\frac{x^2-\left(x-7\right)\left(x+7\right)}{x\left(x-7\right)\left(x+7\right)}\)
=\(\frac{x^2-\left(x^2-49\right)}{x.\left(x-7\right)\left(x+7\right)}\)
=\(\frac{49}{x\left(x+7\right)\left(x-7\right)}\) (ĐK: x\(\ne\) { 0; 7;-7;49}
sau đó: chia trước trừ sau
Đoạn sau dễ chắc bạn tự làm được
Làm bài tốt
a) xác định khi x khác +-1
b)
\(A=\left(\frac{\left(2x+1\right).\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\right).\frac{\left(x-1\right)}{\left(x+1\right)}\)
\(A=\left(\frac{\left(2x^2+3x+1\right)+8-\left(x^2-2x+1\right)}{\left(x-1\right)\left(x+1\right)}\right).\frac{\left(x-1\right)}{\left(x+1\right)}=\frac{x^2+5x+8}{\left(x-1\right)\left(x+1\right)}.\frac{x-1}{x+1}\)
\(A=\frac{x^2+5x+8}{\left(x+1\right)^2}=1+\frac{3\left(x+1\right)+4}{\left(x+1\right)^2}\)
c)
GTNN \(B=\frac{3y+4}{y^2}\ge-\frac{9}{16}\)
GTNN \(A=\frac{7}{16}\)
a. ĐK \(\hept{\begin{cases}x\ne0\\x+5\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}}\)
b. \(A=\frac{x^2+2x}{2x\left(x+5\right)}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}=\frac{x\left(x^2+2x\right)+2\left(x-5\right)\left(x+5\right)+50-5x}{2x\left(x+5\right)}\)
\(=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(=\frac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\frac{x-1}{2}\)
Để \(A=1\Rightarrow\frac{x-1}{2}=1\Rightarrow x=3\)
Để \(A=-3\Rightarrow\frac{x-1}{2}=-3\Rightarrow x=-5\)
Vậy với x=3 thì A=1 ; với x=-5 thì A=-3
\(B=\frac{x^2-2}{x^2+1}=\frac{x^2+1-3}{x^2+1}=1-\frac{3}{x^2+1}\)
\(B_{min}\Rightarrow\left(\frac{3}{x^2+1}\right)_{max}\Rightarrow\left(x^2+1\right)_{min}\)
\(x^2+1\ge1\). 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\(\left(x^2+2x-2\right)_{min}\Leftrightarrow x=-1\)
Dài quá trôi hết đề khỏi màn hình: nhìn thấy câu nào giải cấu ấy
Bài 4:
\(A=\frac{\left(x-1\right)+\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{2}{\left(x+1\right)\left(x-1\right)}=\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
a) DK x khác +-1
b) \(dk\left(a\right)\Rightarrow A=\frac{2}{\left(x+1\right)}\)
c) x+1 phải thuộc Ước của 2=> x=(-3,-2,0))
1. a) Biểu thức a có nghĩa \(\Leftrightarrow\hept{\begin{cases}x+2\ne0\\x^2-4\ne0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+2\ne0\\x-2\ne0\\x+2\ne0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}\)
Vậy vs \(x\ne2,x\ne-2\) thì bt a có nghĩa
b) \(A=\frac{x}{x+2}+\frac{4-2x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{4-2x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-2x+4-2x}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-4x+4}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x-2}{x+2}\)
c) \(A=0\Leftrightarrow\frac{x-2}{x+2}=0\)
\(\Leftrightarrow x-2=\left(x+2\right).0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)(ko thỏa mãn điều kiện )
=> ko có gía trị nào của x để A=0
\(A=\frac{2x}{x^2-25}+\frac{5}{5-x}-\frac{1}{x+5}\left(ĐKXĐ:x\ne\pm5\right)\)
\(=\frac{2x}{\left(x-5\right)\left(x+5\right)}-\frac{5}{x-5}-\frac{1}{x+5}\)
\(=\frac{2x}{\left(x-5\right)\left(x+5\right)}-\frac{5\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{2x-5x-25-x+5}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)
\(=\frac{-4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}\)
\(=-\frac{4}{x-5}\)