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a) x ≠ -5.
b) Ta có P = ( x + 5 ) 2 x + 5 = x + 5
c) Ta có P = 1 Û x = -4 (TMĐK)
d) Ta có P = 0 Û x = -5 (loại). Do vậy x ∈ ∅ .
\(a.\)
\(\dfrac{16x^2-1}{16x^2-8x+1}\\ =\dfrac{\left(4x\right)^2-1}{\left(4x-1\right)^2}\\ =\dfrac{\left(4x-1\right)\left(4x+1\right)}{\left(4x-1\right)^2}\\ =\dfrac{4x+1}{4x-1}\)
\(b.\)
\(\dfrac{4x^2-4xy+y^2}{-\left(4x^2-y^2\right)}\\ =-\dfrac{\left(2x-y\right)^2}{\left(2x-y\right)\left(2x+y\right)}\\ =\dfrac{-\left(2x-y\right)}{2x+y}\\ =\dfrac{y-2x}{y+2x}\)
a) Ta có: \(\dfrac{16x^2-1}{16x^2-8x+1}\)
\(=\dfrac{\left(4x-1\right)\left(4x+1\right)}{\left(4x-1\right)^2}\)
\(=\dfrac{4x+1}{4x-1}\)
b) Ta có: \(\dfrac{4x^2-4xy+y^2}{y^2-4x^2}\)
\(=\dfrac{\left(2x-y\right)^2}{\left(y-2x\right)\left(y+2x\right)}\)
\(=\dfrac{\left(y-2x\right)^2}{\left(y-2x\right)\left(y+2x\right)}\)
\(=\dfrac{y-2x}{y+2x}\)
\(a,\left(2x-1\right)\left(4x^2+2x-1\right)-8x^3+16x\)
\(\Leftrightarrow\)\(8x^3+4x^2-2x-4x^2-2x+1-8x^3+16x\)
\(\Leftrightarrow\)\(4x^2-2x-4x^2-2x+1+16x\)
\(\Leftrightarrow12x+1\)
\(b,\left(x+2\right)^3-x^2\left(x+6\right)-2x-5\)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3-6x^2-2x-5\)
\(\Leftrightarrow6x^2+12x+8-6x^2-2x-5\)
\(\Leftrightarrow10x+3\)
Mình cũng không biết đúng hay sai nữa :V
Bài 1 : Với : \(x>0;x\ne1\)
\(P=\left(1+\frac{1}{\sqrt{x}-1}\right)\frac{1}{x-\sqrt{x}}=\left(\frac{\sqrt{x}}{\sqrt{x}-1}\right).\sqrt{x}\left(\sqrt{x}-1\right)=x\)
Thay vào ta được : \(P=x=25\)
Bài 2 :
a, Với \(x\ge0;x\ne1\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}\)
\(=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)
\(A=\dfrac{16x^2-1}{16x^2-8x+1}\)
\(=\dfrac{\left(4x-1\right)\left(4x+1\right)}{\left(4x-1\right)^2}\)
a) ĐKXĐ:
\(\left(4x-1\right)^2\ne0\Leftrightarrow4x-1\ne0\Leftrightarrow x\ne\dfrac{1}{4}\)
b) \(A=\dfrac{\left(4x+1\right)\left(4x-1\right)}{\left(4x-1\right)^2}=\dfrac{4x+1}{4x-1}\)
a,đkxđ : \(16x^2\ne0\Leftrightarrow x\ne0\)
b, \(\dfrac{16x^2}{1}-\dfrac{1}{16x^2}-\dfrac{8x}{1}+1=\dfrac{256x^4}{16x^2}-\dfrac{1}{16x^2}-\dfrac{128x^3}{16x^2}+\dfrac{16x^2}{16x^2}\)
\(=\dfrac{256x^4-1-128x^3+16x^2}{16x^2}=\dfrac{256x^4-128x^3+16x^2-1}{16x^2}\)
\(=\dfrac{\left(256x^4-128x^3+16^2\right)-1}{16x^2}=\dfrac{16x^2\left(16x^2-8x+1\right)-1}{16x^2}\)
\(=\dfrac{\left(4x\right)^2.\left(\left(4x\right)^2-8x+1\right)-1}{16x^2}=\dfrac{\left(4x\right)^2.\left(4x-1\right)^2-1}{16x^2}\)
\(=\dfrac{\left(16x^2-4x\right)^2-1}{16x^2}=\dfrac{\left(16x^2-4x-1\right)\left(16x^2-4x+1\right)}{16x^2}\)
\(=\dfrac{\left(\left(4x\right)^2-4x-1\right)\left(\left(4x\right)^2-4x+1\right)}{\left(4x\right)^2}\)