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a) \(A=\left(2x+1\right)^2-\left(x+2\right)\left(x-2\right)-2x\left(x+1\right)\)
\(A=4x^2+4x+1-x^2+4-2x^2-2x\)
\(A=x^2+2x+5\)
b) Để A = 4
=> \(x^2+2x+5=4\)
\(\Leftrightarrow x^2+2x+1=0\)
\(\Leftrightarrow\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
c) Ta có A = x2 + 2x + 5
A = ( x + 1 )2 + 4
=> \(A\ge4>0\left(đpcm\right)\)
a,\(A=\left(2x+1\right)^2-\left(x+2\right)\left(x-2\right)-2x\left(x+1\right)\)
\(=4x^2+4x+1-x^2+4-2x^2-2x\)
\(=x^2+2x+5\)
b,\(A=x^2+2x+5=4\)
\(\Rightarrow x^2+2x+5-4=0\)
\(x^2+2x+1=0\)
\(\left(x+1\right)^2=0\)
\(x+1=0\)
\(x=-1\)
c, Ta có: \(A=x^2+2x+5=\left(x^2+2x+1\right)+4=\left(x+1\right)^2+4\ge4>0\)
Hay: A > 0 => đpcm
=.= hok tốt!!
a) -4x2+2x
b) -4x2+2x=0
x(-4x+2)=0
=> x=0 hoặc -4x+2=0
-4x = -2
x=1/2(đpcm)
c) Thay x=-1/4 vào -4x2+2x ta có : -4 (-1/4)2 +2(-1/4) = ... (tự tính )
a) A = (x - 3)(x + 1) - (2x - 1)^2 - (x + 2)(x - 2)
A = x^2 - 2x - 3 - 4x^2 + 4x - 1 - x^2 + 4
A = -4x^2 + 2x
b) 4x^2 - 2x = 0
<=> 2x(2x - 1) = 0
<=> 2x = 0 hoặc 2x - 1 = 0
<=> x = 0 hoặc x = 1/2
c) với x = -1/4, ta có:
4(-1/4)^2 - 2(-1/4) = 3/4
\(1.a,Q=\frac{x+3}{2x+1}-\frac{x-7}{2x+1}=\frac{x+3}{2x+1}+\frac{7-x}{2x+1}\)
\(=\frac{x+3+7-x}{2x+1}=\frac{10}{2x+1}\)
\(b,\) Vì \(x\inℤ\Rightarrow\left(2x+1\right)\inℤ\)
Q nhận giá trị nguyên \(\Leftrightarrow\frac{10}{2x+1}\) nhận giá trị nguyên
\(\Leftrightarrow10⋮2x+1\)
\(\Leftrightarrow2x+1\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
Mà \(\left(2x+1\right):2\) dư 1 nên \(2x+1=\pm1;\pm5\)
\(\Rightarrow x=-1;0;-3;2\)
Vậy.......................
\(2;A=\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right):\left(\frac{1-x}{x+2}\right)\)
\(ĐKXĐ:\hept{\begin{cases}x^2-4\ne0\\1-x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne\pm2\\x\ne1\end{cases}}\)
\(a,A=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right).\frac{x+2}{1-x}\)
\(A=\left(\frac{x+x-2-2x-4}{\left(x+2\right)\left(x-2\right)}\right).\frac{x+2}{1-x}\)
\(A=\frac{-6}{\left(x+2\right)\left(x-2\right)}.\frac{x+2}{1-x}=\frac{-6}{\left(x-2\right)\left(1-x\right)}\)
b, Khi x = -4
\(A=\frac{-6}{\left(-4-2\right)\left(1+4\right)}=\frac{-6}{-6.5}=\frac{1}{5}\)
BÀI 1:
a) \(ĐKXĐ:\) \(\hept{\begin{cases}x-2\ne0\\x+2\ne0\end{cases}}\) \(\Leftrightarrow\)\(\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}\)
b) \(A=\left(\frac{2}{x-2}-\frac{2}{x+2}\right).\frac{x^2+4x+4}{8}\)
\(=\left(\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right).\frac{\left(x+2\right)^2}{8}\)
\(=\frac{2x+4-2x+4}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x+2\right)^2}{8}\)
\(=\frac{x+2}{x-2}\)
c) \(A=0\) \(\Rightarrow\)\(\frac{x+2}{x-2}=0\)
\(\Leftrightarrow\) \(x+2=0\)
\(\Leftrightarrow\)\(x=-2\) (loại vì ko thỏa mãn ĐKXĐ)
Vậy ko tìm đc x để A = 0
p/s: bn đăng từng bài ra đc ko, mk lm cho
Bài 3 : a, ĐK : \(x\ne0;-2\)
b, \(A=\frac{x^2}{x^2+2x}+\frac{2}{x+2}+\frac{2}{x}=\frac{x^2}{x\left(x+2\right)}+\frac{2x}{x\left(x+2\right)}+\frac{2\left(x+2\right)}{x\left(x+2\right)}\)
\(=\frac{x^2+4x+4}{x\left(x+2\right)}=\frac{\left(x+2\right)^2}{x\left(x+2\right)}=\frac{x+2}{x}\)
c, \(A=\frac{-\frac{3}{2}+2}{-\frac{3}{2}}=\frac{\frac{1}{2}}{-\frac{3}{2}}=-\frac{1}{12}\)
d) \(A>0\Leftrightarrow\frac{-1}{x-2}>0\)
\(\Leftrightarrow x-2< 0\) ( vì \(-1< 0\))
\(\Leftrightarrow x< 2\)
\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(A=\)\(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(A=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
\(A=\frac{-1}{x-2}\)
a, A = 4x^2+4x+1-5x^2+5x+2x^2-2x-4 = x^2+7x-3
b, Với x=-1/2 thì A = 1/4 - 7/2 - 3 = -25/4
c, Để A =-3 thì x^2+7x-3 = -3
<=> x^2+7x-3+3 = 0
<=> x^2+7x = 0
<=> x.(x+7) = 0
<=> x=0 hoặc x+7 = 0
<=> x=0 hoặc x=-7
k mk nha
sai