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+ Rút gọn:
(Quy đồng với MTC = 5x(x + 5))
(Rút gọn nhân tử chung x(x + 5))
+ Tại x = -4 giá trị của phân thức rút gọn bằng:
Ta được ngày 1 tháng 5. Đó là ngày Quốc tế Lao động.
ĐKXĐ: \(x\notin\left\{0;-5\right\}\)
\(A=\dfrac{x^2}{5x+25}+\dfrac{2\left(x-5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)
\(=\dfrac{x^2}{5\left(x+5\right)}+\dfrac{2\left(x-5\right)}{x}+\dfrac{5x+50}{x\left(x+5\right)}\)
\(=\dfrac{x^3+2\cdot5\left(x-5\right)\left(x+5\right)+5\left(5x+50\right)}{5x\left(x+5\right)}\)
\(=\dfrac{x^3+10x^2-250+25x+250}{5x\left(x+5\right)}\)
\(=\dfrac{x^3+10x^2+25x}{5x\left(x+5\right)}=\dfrac{x\left(x^2+10x+25\right)}{5x\left(x+5\right)}\)
\(=\dfrac{\left(x+5\right)^2}{5\left(x+5\right)}=\dfrac{x+5}{5}\)
\(A=\dfrac{x^2}{5x+25}+\dfrac{2\left(x-5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\left(ĐKXĐ:x\ne0;x\ne-5\right)\)
\(A=\dfrac{x^2}{5\left(x+5\right)}+\dfrac{2\left(x-5\right)}{x}+\dfrac{50+5x}{x\left(x+5\right)}\)
\(A=\dfrac{x^2.x}{5x\left(x+5\right)}+\dfrac{2.5\left(x+5\right)\left(x-5\right)}{5x\left(x+5\right)}+\dfrac{5\left(50+5x\right)}{5x\left(x+5\right)}\)
\(A=\dfrac{x^3}{5x\left(x+5\right)}+\dfrac{10.\left(x^2-25\right)}{5x\left(x+5\right)}+\dfrac{250+25x}{5x\left(x+5\right)}\)
\(A=\dfrac{x^3}{5x\left(x+5\right)}+\dfrac{10x^2-250}{5x\left(x+5\right)}+\dfrac{250+25x}{5x\left(x+5\right)}\)
\(A=\dfrac{x^3+10x^2-250+250+25x}{5x\left(x+5\right)}\)
\(A=\dfrac{x^3+10x^2+25x}{5x\left(x+5\right)}\)
\(A=\dfrac{x\left(x^2+10x+25\right)}{5x\left(x+5\right)}\)
\(A=\dfrac{\left(x+5\right)^2}{5\left(x+5\right)}\)
\(A=\dfrac{x+5}{5}\)
a: \(P=\dfrac{x\left(x+2\right)}{2\left(x+5\right)}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-50-5x+50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)
a) ĐKXĐ: \(x\ne-10;x\ne0;x\ne-5\)
b) \(P=\dfrac{x^2+2x}{2x+20}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+2x}{2\left(x+10\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x^2+2x\right)\left(x+5\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{2\left(x-5\right)\left(x+10\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{\left(50-5x\right)\left(x+10\right)}{2x\left(x+5\right)\left(x+10\right)}\)
\(=\dfrac{x^4+7x^3+10x^2+2x^2+10x-100+500-5x^2}{2x\left(x+10\right)\left(x+5\right)}\)
\(=\dfrac{x^4+7x^3+7x^2+10x+400}{2x\left(x+10\right)\left(x+5\right)}\)
c) \(P=0\Rightarrow x^4+7x^3+7x^2+10x+400=0\Leftrightarrow...\)
Số xấu thì câu c, d làm cũng như không. Bạn xem lại đề.
a) đK: \(x\ne0;2\)
B = \(\dfrac{3x-4}{x\left(x-2\right)}.\dfrac{x\left(x-2\right)}{x^2-4-x^2}=\dfrac{3x-4}{-4}=\dfrac{4-3x}{4}\) \(\dfrac{x-4+2x}{x\left(x-2\right)}:\dfrac{\left(x-2\right)\left(x+2\right)-x^2}{x\left(x-2\right)}\)
= \(\dfrac{3x-4}{x\left(x-2\right)}.\dfrac{x\left(x-2\right)}{x^2-4-x^2}=\dfrac{4-3x}{4}\)
b) Thay x = -2 (TMDK) vào B, ta có:
\(B=\dfrac{4-3.\left(-2\right)}{4}=\dfrac{4+6}{4}=\dfrac{5}{2}\)
c) Để \(\left|B\right|-2x=5\)
<=> \(\left|\dfrac{4-3x}{4}\right|-2x=5\)
TH1: \(x\le\dfrac{4}{3}\)
<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{4-3x}{4}\)
PT <=> \(\dfrac{4-3x}{4}-2x=5\)
<=> \(\dfrac{4-3x-8x}{4}=5\)
<=> \(4-11x=20\)
<=> x = \(\dfrac{-16}{11}\) (Tm)
TH2: \(x>\dfrac{4}{3}\)
<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{3x-4}{4}\)
PT <=> \(\dfrac{3x-4}{4}-2x=5\)
<=> \(\dfrac{3x-4-8x}{4}=5\)
<=> \(-5x-4=20\)
<=> \(x=\dfrac{-24}{5}\left(l\right)\)
d) Xét (2-x)B = \(\dfrac{\left(2-x\right)\left(4-3x\right)}{4}\) = \(\dfrac{3x^2-10x+8}{4}\)
= \(\dfrac{3\left(x-\dfrac{5}{3}\right)^2-\dfrac{1}{3}}{4}\)
Mà \(3\left(x-\dfrac{5}{3}\right)^2\ge\) 0
=> (2-x)B \(\ge\dfrac{\dfrac{-1}{3}}{4}=\dfrac{-1}{12}\)
Dấu "=" <=> x = \(\dfrac{5}{3}\left(tm\right)\)
e) Số nguyên âm lớn nhất là -1
Để B = -1
<=> \(\dfrac{4-3x}{4}=-1\)
<=> 4 - 3x = -4
<=> \(x=\dfrac{8}{3}\left(tm\right)\)
g)
TH1: \(x\le\dfrac{4}{3}\)
<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{4-3x}{4}\)
BDT <=> \(\dfrac{4-3x}{4}< 2x-4\)
<=> \(4-3x< 8x-16\)
<=> \(x>\dfrac{20}{11}\left(l\right)\)
TH2: \(x>\dfrac{4}{3}\)
<=> \(\left|\dfrac{4-3x}{4}\right|=\dfrac{3x-4}{4}\)
BDT <=> \(\dfrac{3x-4}{4}< 2x-4\)
<=> \(3x-4< 8x-16\)
<=> x > \(\dfrac{12}{5}\)
KHDK: \(x>\dfrac{12}{5}\)
a,ĐK: \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)
c, Với x = 4 thỏa mãn ĐKXĐ thì
\(A=\frac{-3}{4-3}=-3\)
d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)
\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)
Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)
1) \(\dfrac{15-5x}{5x^2-15x}=\dfrac{5\left(3-x\right)}{5x\left(x-3\right)}=-\dfrac{5\left(x-3\right)}{5x\left(x-3\right)}=-\dfrac{1}{x}\)
Chọn A
2) \(\dfrac{x\left(x-5\right)}{x^2+25}=\dfrac{x\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{x}{x+5}\)
\(A=0\Leftrightarrow\dfrac{x}{x+5}=0\Leftrightarrow x=0\)
Chọn B
3) \(\dfrac{2x-5}{5-2x}=-\dfrac{5-2x}{5-2x}=-1\)
Chọn D