bạn đăng tách ra nhé

a, \(\left(2x+1\right)\left(x-4\right)=\left(2x+1\right)^2\)

\(\Leftrightarrow2x^2-7x-4=4x^2+4x+1\Leftrightarrow2x^2+11x+5=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)=0\Leftrightarrow x=-5;x=-\frac{1}{2}\)

b, sửa đề :  \(\left(x-4\right)\left(x^2+4x+16\right)-\left(x^2-6\right)=2\)

\(\Leftrightarrow x^3-64-x^2+6=2\Leftrightarrow x^3-x^2-60=0\Leftrightarrow x=4,27...\)

c, \(\left(2x-1\right)^2-\left(3x+4\right)^2=0\Leftrightarrow\left(2x-1+3x+4\right)\left(2x-1-3x-4\right)=0\)

\(\Leftrightarrow\left(5x+3\right)\left(-x-5\right)=0\Leftrightarrow x=-\frac{3}{5};x=-5\)

d, \(\left(9x+2\right)\left(x-1\right)-\left(3x-1\right)^2=0\)

\(\Leftrightarrow9x^2-7x-2-9x^2+6x-1=0\Leftrightarrow-x-3=0\Leftrightarrow x=-3\)

e, \(\left(2x+3\right)^2-4\left(x-1\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow4x^2+12x+9-4\left(x-1\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow4x^2+12x+9-4\left(x^3-x-x^2+1\right)=0\)

\(\Leftrightarrow4x^2+12x+9-4x^3+4x+4x^2-4=0\)

\(\Leftrightarrow-4x^3+8x^2+16x+5=0\Leftrightarrow x=-0,9...;x=-0,41...;x=3,31...\)

f, \(15x\left(x+4-6x-24\right)=0\Leftrightarrow15\left(-5x-20\right)=0\)

\(\Leftrightarrow-75x-300=0\Leftrightarrow x=-4\)

g, \(\left(4x-10\right)\left(2-3x\right)-30^2=0\)

\(\Leftrightarrow8x-12x^2-20+30x-900=0\Leftrightarrow-12x^2+38x-920=0\)

vô nghiệm 

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\(M=x^3+18x^2+108x+16\)

\(M=x^3+18x^2+108x+216-200\)

\(M=\left(x+6\right)^3-200\)

\(M=\left(-20\right)^3-200\)

\(M=-8000-200\)

\(M=-8200\)

Ta có a - b = 1

=> (a - b)² = 1

<=> a² + b² - 2ab = 1

<=> a² + b² = 25

=> (a² + b²)² = 625

<=> a⁴ + b⁴ + 2(ab)² = 625

<=> a⁴ + b⁴ = 625 - 2.12² = 337

(20^2+18^2+16^2+.........+4^2+2^2)-(19^2+17^2+15^2+.........+3^2+1^2)

=(20+18+16+..+4+2)^2-(19+17+15+...+3+1)^2

=100^2-100^2

=0

https://h7.net/hoi-dap/toan-8/tinh-nhanh-20-2-18-2-16-2-4-2-2-2-19-2-17-2--faq432725.html

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\(Q=2x^2+8x+9\)

\(Q=\left(2x^2+8x+2\sqrt{2}^2\right)+1\)

\(Q=\left(\sqrt{2}x+2\sqrt{2}\right)^2+1\ge1>0\)

vậy bt luôn dương với mọi biến x

ta có: 4x²+16x+18=4x²+16x+16+2=(2x+4)²+2 khác 0 suy ra 2.(2x^2+8x+9)khác 0 suy ra 2x^2+8x+9 khác 0 (1) 

ta có: \(\left(\sqrt{2}x+3\right)^2=2x^2+6.\sqrt{2}x+9\)>=0 với mọi x

nếu x>0 thì ta có: 2x^2+8x+9>0 (2)

nếu x<0 :

\(6\sqrt{2}>8\Rightarrow6\sqrt{2}x< 8\left(x< 0\right)\Rightarrow2x^2+6.\sqrt{2}x+9< 2x^2+8x+9\)

mà \(2x^2+6.\sqrt{2}x+9>=0\)suy ra \(2x^2+8x+9>=0\)(3)

từ (1,2,3) suy ra 2x^2+8x+9 có gt dương với mọi biến