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\(A=\left[\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right]:\left[x-2+\frac{10-x^2}{x+2}\right]\) ĐKXĐ : \(x\ne0;x\ne\pm2\)
\(A=\left[\frac{x^2}{x\left(x+2\right)\left(x-2\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right]:\left[\frac{x^2-4}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(A=\left[\frac{3x^2}{3x\left(x+2\right)\left(x-2\right)}-\frac{6x\left(x+2\right)}{3x\left(x+2\right)\left(x-2\right)}+\frac{3x\left(x+2\right)}{3x\left(x+2\right)\left(x-2\right)}\right]:\frac{6}{x+2}\)
\(A=\left[\frac{3x^2-6x^2-12x+3x^2+6x}{3x\left(x+2\right)\left(x-2\right)}\right].\frac{x+2}{6}\)
\(A=\frac{-x}{3x\left(x-2\right)}\)
\(A=\frac{-1}{3x-6}\)
\(M=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)
a) ĐKXĐ : x ≠ -3 , x ≠ 2
\(=\frac{x+2}{x+3}-\frac{5}{x^2-2x+3x-6}-\frac{1}{x-2}\)
\(=\frac{x+2}{x+3}-\frac{5}{x\left(x-2\right)+3\left(x-2\right)}-\frac{1}{x-2}\)
\(=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{x^2-4x+3x-12}{\left(x+3\right)\left(x-2\right)}=\frac{x\left(x-4\right)+3\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}\)
\(=\frac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)
b) Để M = 1/3
=> \(\frac{x-4}{x-2}=\frac{1}{3}\)( x ≠ -3 , x ≠ 2 )
=> 3( x - 4 ) = x - 2
=> 3x - 12 - x + 2 = 0
=> 2x - 10 = 0
=> 2x = 10
=> x = 5 ( tm )
Vậy x = 5 thì M = 1/3
đk: \(x\ne2,x\ne-3\)
a) Ta có: \(M=\frac{-4+x^2}{x^2+x-6}-\frac{5}{x^2+x-6}-\frac{x+3}{x^2+x-6}\)
\(=\frac{x^2-x-12}{x^2+x-6}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\frac{x-4}{x-2}\)
b) \(M=\frac{1}{3}\Rightarrow\frac{x-4}{x-2}=\frac{1}{3}\Leftrightarrow3x-12=x-2\Leftrightarrow x=5\)
\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{x^2-1}\right)\div\frac{x}{x+2019}\)
ĐK : x ≠ ±1 ; x ≠ 0 ; x ≠ -2019
\(=\left(\frac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x+2019}{x}\)
\(=\left(\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x+2019}{x}\)
\(=\left(\frac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{x+2019}{x}\)
\(=\frac{x^2-1}{x^2-1}\times\frac{x+2019}{x}=\frac{x+2019}{x}\)
b. \(A=\frac{x+2019}{x}=1+\frac{2019}{x}\) đạt giá trị lớn nhất
<=> \(\frac{2019}{x}\) đạt giá trị lớn nhất
<=> \(\hept{\begin{cases}x>0\\x\in Z\end{cases}}\) và x đạt giá trị bé nhất
<=> x = 1
Khi đó A = 2020
ĐKXĐ: x \(\ne\)\(\pm\)3; x \(\ne\)-7
a) Ta có: P = \(\left(\frac{x^2+1}{x^2-9}-\frac{x}{x+3}+\frac{5}{3-x}\right):\left(\frac{2x+10}{x+3}-1\right)\)
P = \(\left(\frac{x^2+1}{\left(x-3\right)\left(x+3\right)}-\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\frac{2x+10-x-3}{x+3}\right)\)
P = \(\frac{x^2+1-x^2+3x-5x-15}{\left(x-3\right)\left(x+3\right)}:\frac{x+7}{x+3}\)
P = \(\frac{-2x-14}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{x+7}\)
P = \(\frac{-2\left(x+7\right)}{x-3}\cdot\frac{1}{x+7}=-\frac{2}{x-3}\)
b) Với x \(\ne\)\(\pm\)3 và x \(\ne\)-7
Ta có: x - 1 = 2 <=> x = 3 (ktm)
=> ko tồn tại giá trị P khi x - 1 = 2
c) Với x \(\ne\)\(\pm\)3; và x \(\ne\)-7
Ta có: P = \(\frac{x+5}{6}\)
<=> \(-\frac{2}{x-3}=\frac{x+5}{6}\)
=> (x - 3)(x + 5) = -12
<=> x2 + 2x - 15 = -12
<=> x2 + 2x - 3 = 0
<=> x2 + 3x - x - 3 = 0
<=> (x - 1)(x + 3) = 0
<=> \(\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=1\left(tm\right)\\x=-3\left(ktm\right)\end{cases}}\)
Vậy ...
a) \(P=\left(\frac{x^2+1}{x^2-9}-\frac{x}{x+3}+\frac{5}{3-x}\right):\left(\frac{2x+10}{x+3}-1\right)\left(x\ne\pm3\right)\)
\(=\left(\frac{x^2+1}{\left(x-3\right)\left(x+3\right)}-\frac{x}{x+3}-\frac{5}{x-3}\right):\frac{2x+10-x-3}{x+3}\)
\(=\left(\frac{x^2+1}{\left(x-3\right)\left(x+3\right)}-\frac{x^2-3x}{\left(x-3\right)\left(x+3\right)}-\frac{5x+15}{\left(x-3\right)\left(x+3\right)}\right):\frac{x+7}{x+3}\)
\(=\frac{x^2+1-x^2+3x-5x-15}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{x+7}\)
\(=\frac{\left(-2x-14\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+7\right)}\)
\(=\frac{-2\left(x+7\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+7\right)}=-\frac{2}{x-3}\)
vậy \(P=-\frac{2}{x-3}\left(x\ne\pm3\right)\)
b) ta có \(P=-\frac{2}{x-3}\left(x\ne\pm3\right)\)
có x-1=2
<=> x=3 (không thỏa mãn điều kiện)
vậy không có giá trị P để x-1=2
c) ta có: \(P=-\frac{2}{x-3}\left(x\ne\pm3\right)\)
P=\(\frac{x+5}{6}\)=> \(\frac{-2}{x-3}=\frac{x+5}{6}\)
\(\Leftrightarrow x^2+2x-15=-12\)
\(\Leftrightarrow x^2+2x-3=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=1\end{cases}}}\)
đối chiếu điều kiện ta thấy x=1 thỏa mãn điều kiện
vậy \(P=\frac{x+5}{6}\)đạt được khi x=1