a, Đặt A = \(x^2-10x+26\)

= \(x^2-10x+25+1\)

=\(x^2-2.x.5+5^2+1\)

=\(\left(x-5\right)^2+1\)

tại x = 105 thay vào A ta được :

A = \(\left(105-5\right)^2+1\)= 10000+1=10001

b Đặt B = \(x^2+0,2x+0,01\)

= \(x^2+2.x.0,1+0,1^2\)

= \(\left(x+0,1\right)^2\)

Tại x = 0,9 thay vào B ta được :

B = \(\left(0,9+0,1\right)^2=1^2=1\)

Ket qua la - 3

 12,04 - 11 - 0,04 =

= ( 12,04 - 0,04 ) - 11

= 12 - 11

= 1

52884

Câu 1 :

\(A=\frac{\sqrt{x}+1}{\sqrt{x-1}}\) khi x = 9

tại x = 9 thay vào A ta được : \(\frac{\sqrt{9}+1}{\sqrt{9}-1}\) = \(\frac{3+1}{3-1}=\frac{4}{2}=2\)

Câu 2 :

a, Ta có : P = \(\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

= \(\left(\frac{x-2}{\sqrt{x}.\sqrt{x}+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\left(\frac{x-2}{\sqrt{x}\left(\sqrt{x+2}\right)}+\frac{1}{\sqrt{x}+2}\right)\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

= \(\left(\frac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

= \(\left(\frac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

=\(\frac{x-2+2\sqrt{x}-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)+2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}}\) => đpcm

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