Bài làm:

Ta có: \(4x^4-12x^2+1\)

\(=4\left(x^4-3x^2+\frac{9}{4}\right)-8\)

\(=4\left(x^2-\frac{3}{2}\right)^2-\left(\sqrt{8}\right)^2\)

\(=\left[2\left(x^2-\frac{3}{2}\right)-2\sqrt{2}\right]\left[2\left(x^2-\frac{3}{2}\right)+2\sqrt{2}\right]\)

\(=4\left(x^2-\frac{3+2\sqrt{2}}{2}\right)\left(x^2-\frac{3-2\sqrt{2}}{2}\right)\)

\(=4\left(x-\sqrt{\frac{3+2\sqrt{2}}{2}}\right)\left(x+\sqrt{\frac{3+2\sqrt{2}}{2}}\right)\left(x-\sqrt{\frac{3-2\sqrt{2}}{2}}\right)\left(x+\sqrt{\frac{3-2\sqrt{2}}{2}}\right)\)

a) 2x^2 + 3 = 2x(x + 4) - 7

<=> 2x^2 + 3 = 2x^2 + 8x - 7

<=> 2x^2 - 2x^2 - 8x = - 7 - 3

<=> -8x = -10

<=> x = -10/-8 = 5/4

b) 4x^2 - 12x + 5 = 0

<=> 4x^2 - 2x - 10x + 5 = 0

<=> 2x(2x - 1) - 5(2x - 1) = 0

<=> (2x - 5)(2x - 1) = 0

<=> 2x - 5 = 0 hoặc 2x - 1 = 0

<=> x = 5/2 hoặc x = 1/2

c) |5 - 2x| = 1 - x
<=> \(\hept{\begin{cases}5-2x\text{ nếu }5-2x\ge0\Leftrightarrow x\ge\frac{5}{2}\\-\left(5-2x\right)\text{ nếu }5-2x< 0\Leftrightarrow x< \frac{5}{2}\end{cases}}\)

+) nếu x >= 5/2, ta có:

5 - 2x = 1 - x

<=> -2x + 1 = 1 - 5

<=> -x = -4

<=> x = 4 (tm)

+) nếu x < 5/2, ta có:

-(5 - 2x) = 1 - x

<=> -5 + 2x = 1 - x

<=> 2x + 1 = 1 + 5

<=> 3x = 6

<=> x = 2 (ktm)

d) \(\frac{2}{x-1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}-\frac{2x+3}{x^2+x+1}\) ; ĐKXĐ: x # 1 

<=> \(\frac{2}{x-1}=\frac{\left(2x-1\right)\left(2x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x+3}{x^2+x+1}\)

<=> \(\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{\left(2x-1\right)\left(2x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{\left(2x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

<=> 2(x^2 + x + 1) = (2x - 1)(2x + 1) - (2x + 3)(x - 1)

<=> 2x^2 + 2x + 2 = 2x^2 - x + 2

<=> 2x^2 - 2x^2 + 2x - x = 2 - 2

<=> x = 0

mạn phép vô đây để kiếm câu trả lời 

\(2x^2+3=2x\left(x+4\right)-7\)

\(< =>2x^2+3=2x.x+4.2x-7\)

\(< =>2x^2+3=2x^2+8x-7\)

\(< =>2x^2+3-2x^2=8x-7\)

\(< =>\left(2x^2-2x^2\right)-8x=-7-3\)

\(< =>-8x=-10< =>8x=10\)

\(< =>x=10:8=\frac{10}{8}=\frac{5}{4}\)

bạn đang học hằng à =))

\(\left(2x-10\right)^2=4x^2-40x+100\)

\(\left(3x-2\right)^2=9x^2-12x+4\)

\(\left(2x-y\right)^2=4x^2-4xy+y^2\)

\(\left(2x-3y\right)^2=4x^2-12xy+9y^2\)

\(\left(x+4\right)^3=x^3+12x^2+48x+64\)

\(\left(x+5\right)^3=x^3+15x^2+75x+125\)

HĐT thì ez mà :))

( 2x - 10 )² = 4x² - 40x + 100

( 3x - 2 )² = 9x² - 12x + 4

( 2x - y )² = 4x² - 4xy + y²

+ ( 2x - 3y )² = 4x² - 12xy + 9y²

+ ( 2x - 3y )³ = 8x³ - 36x²y + 54xy² - 27y³

( x + 4 )³ = x³ + 12x² + 48x + 64

( x + 5 )³ = x³ + 15x² + 75x + 125

\(16x^2-\left(4x-5\right)^2=15\)

\(\Leftrightarrow16x^2-\left(16x^2-40x+25\right)=15\)

\(\Leftrightarrow16x^2-16x^2+40x-25=15\)

\(\Leftrightarrow40x=40\)

\(\Leftrightarrow x=1\)

Vậy \(x=1\)

16x² - ( 4x - 5 )² = 15

<=> 16x² - ( 16x² - 40x + 25 ) = 15

<=> 16x² - 16x² + 40x - 25 = 15

<=> 40x - 25 = 15

<=> 40x = 40

<=> x = 1 

<=> x = 

Bạn vào trang cá nhân của mình đi, có cái này hay lắm

Nếu p>3  mà p là SNT nên p ko chia hết cho 3

Suy ra p^2 chia 3 dư 1

Suy ra p^2+8 chia hết cho 3,mà p^2+8>3 nên p^2+8 là HS(L)

Vậy p nhỏ hơn hoặc bằng 3

Nếu p=2 thì p^2+8 là HS (L)

Khí đó p=3

Suy ra p^3+8p+2=53 là SNT(đpcm)

Bài 1

a, Với \(x=9\)thì \(A=\frac{3+\sqrt{x}}{\sqrt{x}}=\frac{3}{\sqrt{x}}+1=\frac{3}{3}+1=2\)

b, Để \(A=\frac{5}{2}\)thì \(\frac{3+\sqrt{x}}{\sqrt{x}}=\frac{3}{\sqrt{x}}+1=\frac{5}{2}< =>\frac{3}{\sqrt{x}}=\frac{3}{2}< =>x=4\)

Bài 2

a, \(B=\frac{\sqrt{x}-2}{\sqrt{x}}+\frac{4\sqrt{x}+2}{x+\sqrt{x}}\left(đk:x>0\right)\)

\(=1-\frac{2}{\sqrt{x}}+\frac{4\sqrt{x}+2}{x+\sqrt{x}}=\frac{x+5\sqrt{x}+2}{x+\sqrt{x}}-\frac{2}{\sqrt{x}}\)

\(=\frac{x\sqrt{x}+5x+2\sqrt{x}-2x-2\sqrt{x}}{x\sqrt{x}+x}=\frac{x\sqrt{x}+3x}{x\sqrt{x}+x}\)

\(=1+\frac{2x}{x\left(\sqrt{x}+1\right)}=1+\frac{2}{\sqrt{x}+1}=\frac{\sqrt{x}+3}{\sqrt{x}+1}\)

\(A=\frac{3+\sqrt{x}}{\sqrt{x}}\)Thay x = 9 ta có : 

\(VT=\frac{3+\sqrt{9}}{\sqrt{9}}=\frac{3+3}{3}=2\)

Bài ra ta có : \(A=\frac{3+\sqrt{x}}{\sqrt{x}}=\frac{5}{2}\)

\(\Leftrightarrow\frac{3}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}}=\frac{5}{2}\Leftrightarrow\frac{3}{\sqrt{x}}+1=\frac{5}{2}\)

\(\Leftrightarrow\frac{3}{\sqrt{x}}=\frac{3}{2}\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)