`a)(2x-1)^2+(x+3)^2-5(x-7)(x+7)`

`=4x^2-4x+1+x^2+6x+9-5(x^2-49)`

`=5x^2-5x^2-4x+6x+1+9+245`

`=2x+255`

`b)(x-2)(x^2+2x+4)-(25+x^3)`

`=x^3-8-x^3-25=-33`

Lời giải:

a. 

$(2x-1)^2+(x+3)^2-5(x-7)(x+7)$

$=4x^2-4x+1+(x^2+6x+9)-5(x^2-49)$

$=5x^2+2x+10-(5x^2-245)=2x+255$

b.

$(x-2)(x^2+2x+4)-(25+x^3)=(x^3-2^3)-(25+x^3)$

$=-8-25=-33$

a: \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)\)

\(=4x^2-4x+1+x^2+6x+9-5x^2+245\)

\(=2x+255\)

b: \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x^3+25\right)\)

\(=x^3-8-x^3-25\)

=-33

a: A=(-1)+(-1)+...+(-1)+2023

=2023-1011=1012

a)

`4*(2y+3x)-3(x-3y)`

`=8y+12x-3x+9y`

`=8y+9y+12x-3x`

`=17y+9x`

b)

`x^2 +2x-x(7x-3)`

`=x^2 +2x-7x^2 +3x`

`=x^2 -7x^2 +2x+6x`

`= -6x^2 +8x`

\(a,A=7\sqrt{5}+6\sqrt{5}-5\sqrt{5}-6\sqrt{5}=2\sqrt{5}\\ b,B=12-5\cdot2=2\\ c,C=\left[2-\dfrac{\sqrt{7}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}\right]\left[2+\dfrac{\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}+1}\right]\\ C=\left(2-\sqrt{7}\right)\left(2+\sqrt{7}\right)=4-7=-3\)

cíuuu

a: =căn 3+căn 5-căn 3=căn5
b: \(=\sqrt{x-2-2\sqrt{x-2}+1}=\sqrt{\left(\sqrt{x-2}-1\right)^2}\)

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

A=(-1)+(-1)+...+(-1)+2023

=2023-1011

=1012

1012

a.\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(\sqrt{3}+2\right)^2}=\left|\sqrt{3}+2\right|=\sqrt{3}+2\)

b.\(\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}-2\right|=\sqrt{5}-2\)

c.\(\sqrt{14+6\sqrt{5}}=\sqrt{\left(\sqrt{5}+3\right)^2}=\left|\sqrt{5}+3\right|=\sqrt{5}+3\)

d.\(\sqrt{17-12\sqrt{2}}=\sqrt{\left(2\sqrt{2}-3\right)^2}=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)