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\(a,ĐKXĐ:x-1\ne0;1-x\ne0;1+x\ne0\)
\(\Rightarrow\left[{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
\(b,C=\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}\)
\(C=\dfrac{x^2+x}{2\left(x-1\right)\left(x+1\right)}+\dfrac{-x^2-1}{2\left(x+1\right)\left(x-1\right)}=\dfrac{1}{2x+2}\)
\(c,C=-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2x+2}=-\dfrac{1}{2}\Leftrightarrow-2x-2=2\Leftrightarrow x=0\)
điều kiện của x để gtrị của biểu thức đc xác định
=>\(2x+10\ne0;x\ne0:2x\left(x+5\right)\ne0\)
\(2x+5\ne0;x\ne0\)
=>\(x\ne-5;x\ne0\)
vậy đkxđ là \(x\ne-5;x\ne0\)
rút gon giống với bạn nguyen thuy hoa đến \(\dfrac{x-1}{2}\)
b,để bt =1=>\(\dfrac{x-1}{2}=1\)
=>x-1=2
=>x=3 thỏa mãn đkxđ
c,d giống như trên
a,
ĐKXĐ: 2x - 2 \(\ne\)0 <=> 2x \(\ne\)2 <=> x \(\ne\)1
2 - 2x2 \(\ne\)0 <=> 2( 1 - x2) \(\ne\)0 <=> (1 - x)(1 + x) \(\ne\)0
<=> x \(\ne\)-1
b,
C = \(\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}\)
= \(\dfrac{x}{2x-2}-\dfrac{x^2+1}{2\left(x^2-1\right)}\)
= \(\dfrac{x}{2\left(x-1\right)}-\dfrac{x^2+1}{2\left(x+1\right)\left(x-1\right)}\)
= \(\dfrac{x\left(x+1\right)-x^2-1}{2\left(x-1\right)\left(x+1\right)}\)
= \(\dfrac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}\)
= \(\dfrac{x-1}{2\left(x-1\right)\left(x+1\right)}\)
= \(\dfrac{1}{2\left(x+1\right)}\)
c,
Để C = \(-\dfrac{1}{2}\)
<=> \(\dfrac{1}{2\left(x+1\right)}=\dfrac{-1}{2}\)
<=> \(\dfrac{1}{x+1}=-1\)
<=> x + 1 = -1
<=> x = -2
d,
Để C > 0
<=> \(\dfrac{1}{2\left(x+1\right)}\)> 0
<=> 2(x + 1) > 0
<=> x + 1 > 0
<=> x > -1
\(ĐKXĐ:x\ne-3;2\)
\(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}-\frac{1}{x-2}=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x+2\right)}-\frac{1}{x-2}\)
\(=\frac{x^2+4x+4}{\left(x+3\right)\left(x+2\right)}-\frac{5}{\left(x+3\right)\left(x+2\right)}-\frac{x+3}{\left(x+2\right)\left(x+3\right)}\)
\(=\frac{x^2+4x+4-5-x-3}{\left(x+2\right)\left(x+3\right)}=\frac{x^2+3x-4}{\left(x+3\right)\left(x+2\right)}=\frac{\left(x+4\right)\left(x-1\right)}{\left(x+3\right)\left(x+2\right)}\)
\(x^2-9=0\Leftrightarrow x=3\left(vì:x\ne-3\right)\)
\(\Rightarrow P=\frac{7}{15}\)
\(P\inℤ\Leftrightarrow x^2+3x-4⋮x^2+5x+6\Leftrightarrow2x+10⋮x^2+5x+6\Leftrightarrow12⋮x^2+5xx+6\)
\(................\left(dễ\right)\)
P/s: shitbo sai rồi nha bạn!Nếu không tin thì thay x = 3 vào P ban đầu và giá trị P sau khi rút gọn sẽ thấy sự khác biệt =)
ĐK: \(x\ne-3;x\ne2\)
a) \(P=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}-\frac{1}{x-2}\)
\(=\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)
b) \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow x=\pm3\)
Thay vào điều kiện,tìm loại x = -3 .Tìm được x =3
Ta có: \(P=\frac{x-4}{x-2}=\frac{3-4}{3-2}=-1\)
c)Ta có: \(P=\frac{x-4}{x-2}=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\)
Để P có giá trị nguyên thì \(\frac{2}{x-2}\) nguyên hay \(x-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Suy ra \(x=\left\{0;1;3;4\right\}\)
a) P xác định \(\Leftrightarrow\hept{\begin{cases}2x+10\ne0\\x\ne0\\2x\left(x+5\right)\ne0\end{cases}\Leftrightarrow x\ne\left\{-5;0\right\}}\)
b) \(P=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^2\left(x+2\right)}{2x\left(x+5\right)}+\frac{2\left(x-5\right)\left(x+5\right)}{2x\left(x+5\right)}+\frac{5\left(10-x\right)}{2x\left(x+5\right)}\)
\(P=\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^3+5x^2-x^2-5x}{2x\left(x+5\right)}\)
\(P=\frac{x^2\left(x+5\right)-x\left(x+5\right)}{2x\left(x+5\right)}\)
\(P=\frac{\left(x+5\right)\left(x^2-x\right)}{2x\left(x+5\right)}\)
\(P=\frac{x\left(x-1\right)}{2x}\)
\(P=\frac{x-1}{2}\)
c) Để P = 0 thì \(x-1=0\Leftrightarrow x=1\)( thỏa mãn ĐKXĐ )
Để P = 1/4 thì \(\frac{x-1}{2}=\frac{1}{4}\)
\(\Leftrightarrow4\left(x-1\right)=2\)
\(\Leftrightarrow4x-4=2\)
\(\Leftrightarrow4x=6\)
\(\Leftrightarrow x=\frac{3}{2}\)( thỏa mãn ĐKXĐ )
d) Để P > 0 thì \(\frac{x-1}{2}>0\)
Mà 2 > 0, do đó để P > 0 thì \(x-1>0\Leftrightarrow x>1\)
Để P < 0 thì \(\frac{x-1}{2}< 0\)
Mà 2 > 0, do đó để P < 0 thì \(x-1< 0\Leftrightarrow x< 1\)
\(\text{a) }A=\dfrac{x}{2x+2}+\dfrac{x^2+1}{2-2x^2}\\ A=\dfrac{x}{2\left(x+1\right)}+\dfrac{x^2+1}{2\left(1-x^2\right)}\\ A=\dfrac{x}{2\left(x+1\right)}+\dfrac{x^2+1}{2\left(1-x\right)\left(1+x\right)}\\ A=\dfrac{x\left(1-x\right)}{2\left(x+1\right)\left(1-x\right)}+\dfrac{x^2+1}{2\left(x+1\right)\left(1-x\right)}\\ A=\dfrac{x-x^2+x^2+1}{2\left(x+1\right)\left(1-x\right)}\)
\(\Rightarrow\) Để \(A\) có nghĩa
\(\text{thì }\Rightarrow2\left(x+1\right)\left(1-x\right)\ne0\\ \Rightarrow\left\{{}\begin{matrix}x+1\ne0\\1-x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne1\end{matrix}\right.\)
\(\text{b) }A=\dfrac{x-x^2+x^2+1}{2\left(x+1\right)\left(1-x\right)}\\ A=\dfrac{x+1}{2\left(x+1\right)\left(1-x\right)}\\ A=\dfrac{1}{2\left(1-x\right)}\\ A=\dfrac{1}{2-2x}\)
c) Để \(A=\dfrac{1}{2}\)
\(\text{thì }\Rightarrow\dfrac{1}{2-2x}=\dfrac{1}{2}\\ \Leftrightarrow2-2x=2\\ \Leftrightarrow2x=0\\ \Leftrightarrow x=0\)
Vậy......................
a) Để M có nghĩa thì \(x\ne1;x\ne-1\)
b) \(M=\dfrac{x}{2x-2}+\dfrac{x^2-1}{2-2x^2}\)
\(=\dfrac{x^2+x}{2\left(x-1\right)\left(x+1\right)}-\dfrac{x^2-1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x+1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{2x-2}\)
c) Để \(M=\dfrac{1}{2}\) thì \(\dfrac{1}{2x-2}=\dfrac{1}{2}\)
\(\Leftrightarrow2x-2=2\)
\(\Leftrightarrow x=2\)