d) \(\sqrt{17+12\sqrt{2}}=\sqrt{8+12\sqrt{2}+9}=\sqrt{\left(2\sqrt{2}+3\right)^2}=2\sqrt{2}+3\)

a) \(=\sqrt{16-8\sqrt{7}+7}\) \(=\sqrt{\left(4-\sqrt{7}\right)^2}\) \(=\left|4-\sqrt{7}\right|\) \(=4-\sqrt{7}\)

b) \(=\sqrt{16-8\sqrt{2}+2}\) \(=\sqrt{\left(4-\sqrt{2}\right)^2}\) \(=\left|4-\sqrt{2}\right|\) \(=4-\sqrt{2}\)

c) \(=\sqrt{9+6\sqrt{3}+3}\) \(=\sqrt{\left(3+\sqrt{3}\right)^2}\) \(=\left|3+\sqrt{3}\right|\) \(=3+\sqrt{3}\)

d) \(=\sqrt{17+6\sqrt{8}}\) \(=\sqrt{9+6\sqrt{8}+8}\) \(=\sqrt{\left(3+\sqrt{8}\right)^2}\) \(=\left|3+\sqrt{8}\right|\) \(=3+\sqrt{8}\) \(=3+2\sqrt{2}\)

1,        84/98

2,        20/24

làm lời giải ra hộ mình nha 

minh sẽ k cho

bài 1: 84/98

bài 2: 20/24

Bài 1 : 

\(\left(x-2\right)^2-\left(x-3^2\right)=\left(x-2\right)^2-\left(x-9\right)\)

\(=x^2-4x+4-x+9=x^2-5x+13\)

Bài 2 : 

a, \(P=\frac{1-4x^2}{4x^2-4x+1}=\frac{\left(1-2x\right)\left(2x+1\right)}{\left(2x-1\right)^2}\)

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

b, Thay x = -4 ta được : 

\(\frac{-2.\left(-4\right)-1}{2.\left(-4\right)-1}=\frac{8-1}{-8-1}=-\frac{7}{9}\)

\(\dfrac{7x^2-14x+7}{3x^2-3x}=\dfrac{7\left(x^2-2x+1\right)}{3x\left(x-1\right)}=\dfrac{7\left(x-1\right)^2}{3x\left(x-1\right)}=\dfrac{7\left(x-1\right)}{3x}\)

\(=\dfrac{7\left(x^2-2x+1\right)}{3x\left(x-1\right)}=\dfrac{7\left(x-1\right)^2}{3x\left(x-1\right)}=\dfrac{7\left(x-1\right)}{3x}\)

\(\dfrac{7x^2-14x+7}{3x^2-3x}=\dfrac{7\left(x^2-2x+1\right)}{3x\left(x-1\right)}=\dfrac{7\left(x-1\right)^2}{3x\left(x-1\right)}=\dfrac{7\left(x-1\right)}{3x}\)

1:

\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)

2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)

\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)

\(=\dfrac{20-6}{2}=7\)