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)      Cách 1:

\(6(y - x) - 2(x - y)\)

\( = 6y - 6x - 2x + 2y\)

\( = 8y - 8x\)

Cách 2:

\(6(y - x) - 2(x - y)\\= 6(y-x)+2(y-x)\\=(6+2).(y-x)\\=8.(y-x)\\=8y-8x\)

b)      \(3{x^2} + x - 4x - 5{x^2}\)

\( = (3{x^2} - 5{x^2}) + (x - 4x)\)

\( =  - 2{x^2} - 3x\)

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)(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

\(a,\left(x-5\right)\left(2x+1\right)-2x\left(x-3\right)\\ =x.2x-5.2x+x-5-2x.x-2x.\left(-3\right)\\ =2x^2-10x+x-5-2x^2+6x\\ =2x^2-2x^2-10x+x+6x-5\\ =-3x-5\)

\(b,\left(2+3x\right)\left(2-3x\right)+\left(3x+4\right)^2\\ =\left[2^2-\left(3x\right)^2\right]+\left[\left(3x\right)^2+2.3x.4+4^2\right]\\=4-9x^2+\left(9x^2+24x+16\right)\\ =24x+20\)

a: Ta có: \(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x+3\right)\left(x-3\right)\)

\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)

\(=x^2+10x+25-16x^3-48x^2-36x-2x^3+18x+x^2-9\)

\(=-18x^3-46x^2-8x+16\)

a) 2x(x+3) – 3x²(x+2) + x(3x² + 4x – 6)

= (2x . x + 2x . 3) – (3x² . x + 3x² . 2) + (x . 3x² + x . 4x – x . 6)

= 2x² + 6x – (3x³ + 6x²) + (3x³ + 4x² - 6x)

= 2x² + 6x – 3x³ – 6x² + 3x³ + 4x² - 6x

= (– 3x³ + 3x³ ) + (2x²  - 6x² + 4x² ) + (6x – 6x)

= 0 + 0 + 0

= 0

b) 3x(2x² – x) – 2x²(3x+1) + 5(x² – 1)

= [3x . 2x² + 3x . (-x)] – (2x² . 3x + 2x² . 1) + [5x² + 5 . (-1)]

= 6x³ – 3x² – (6x³ +2x²) + 5x² – 5

= 6x³ – 3x² – 6x³ - 2x² + 5x² – 5

= (6x³ – 6x³ ) + (-3x² – 2x² + 5x²) – 5

= 0 + 0 – 5

= - 5

a: M=A:B

\(=\dfrac{x+\sqrt{x}+10-\sqrt{x}-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{1}=\dfrac{x+7}{\sqrt{x}+3}\)

b: \(M=\dfrac{x-9+16}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{16}{\sqrt{x}+3}\)

=>\(M=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6>=2\sqrt{16}-6=2\)

Dấu = xảy ra khi (căn x+3)^2=16

=>căn x+3=4

=>x=1

a) \(\sqrt{4a^2}=2\left|a\right|=-2a\) ( do a<0)

b) \(\sqrt{4x^2-12x+9}=\sqrt{\left(2x-3\right)^2}=\left|2x-3\right|=3-2x\)(do \(x< \dfrac{3}{2}\Leftrightarrow2x-3< 0\))

\(A=\left|-3\right|+2+\left|5x\right|\)

\(A=3+2+\left(-5x\right)=-5x+5\)