a) 16x^2 - (4x - 5)^2 = 15

<=> 16x^2 - 16x^2 + 40x - 25 = 15

<=> 40x = 40

<=> x = 1

b) (2x + 3)^2 - 4(x - 1)(x + 1) = 49

<=> 4x^2 + 12x + 9 - 4x^2 - 4x + 4x + 4 = 49

<=> 12x + 13 = 49

<=> 12x = 36

<=> x = 3

c) (2x + 1)(1 - 2x) + (1 - 2x)^2 = 18

<=> 1 - 4x^2 + 1 - 4x + 4x^2 = 18

<=> 2 - 4x = 18

<=> -4x = 16

<=> x = -4

d)2(x + 1)^2 - (x - 3)(x + 3) - (x - 4)^2 = 0

<=> 2x^2 + 4x + 2 - x^2 + 3^2 - x^2 + 8x - 16 = 0

<=> 12x - 5 = 0

<=> 12x = 5

<=> x = 5/12

e) (x - 5)^2 - x(x - 4) = 9

<=> x^2 - 10x + 25 - x^2 + 4x = 9

<=> -6x + 25 = 9

<=> -6x = 9 - 25

<=> -6x = -16

<=> x = -16/-6 = 8/3

f) (x - 5)^2 + (x - 4)(1 - x) = 0

<=> x^2 - 10x + 25 + x - x^2 - x - 4 + 4x = 0

<=> -5x + 21 = 0

<=> -5x = -21

<=> x = 21/5

a) A = (x+3)² + (x-3)(x+3) - 2(x+2)(x - 4)

        = (x + 3)(x + 3) + (x - 3)(x + 3) - 2[x(x - 4) + 2(x - 4)]

        = x(x + 3) + 3(x + 3) + x(x + 3) - 3(x + 3) - 2[x² - 4x + 2x - 8]

        = x² + 3x + 3x + 9 + x² + 3x - 3x - 9 - 2(x² - 2x - 8)

        = x² + 3x + 3x + 9 +x² + 3x - 3x - 9 - 2x² + 4x + 16

        = (x² + x² - 2x²) + (3x + 3x + 3x - 3x + 4x) + (9 - 9 + 16) = 10x + 16

Thay x = -1/2 vào biểu thức trên ta có : \(10\cdot\left(-\frac{1}{2}\right)+16=-5+16=11\)

b) \(B=\left(3x+4\right)^2-\left(x-4\right)\left(x+4\right)-10x\)

\(B=9x^2+24x+16-x\left(x+4\right)+4\left(x+4\right)-10x\)

\(B=9x^2+24x+16-x^2-4x+4x+16-10x\)

\(B=\left(9x^2-x^2\right)+\left(24x-4x+4x-10x\right)+\left(16+16\right)\)

\(B=8x^2+14x+32\)

Thay x = -1/10 vào biểu thức trên ta có : \(B=8\cdot\left(-\frac{1}{10}\right)^2+14\cdot\left(-\frac{1}{10}\right)+32=\frac{767}{25}\)

c) \(C=\left(x+1\right)^2-\left(2x-1\right)^2+3\left(x-2\right)\left(x+2\right)\)

\(C=x^2+2x+1-\left(2x-1\right)\left(2x-1\right)+3\left(x^2-4\right)\)

\(C=x^2+2x+1-2x\left(2x-1\right)+1\left(2x-1\right)+3x^2-12\)

\(C=x^2+2x+1-4x^2+2x+2x-1+3x^2-12\)

\(C=\left(x^2-4x^2+3x^2\right)+\left(2x+2x+2x\right)+\left(1-1-12\right)\)

\(C=6x-12\)

Thay x = 1 vào biểu thức ta có : C = 6.1 - 12 = 6 -12 = -6

Còn bài kia làm nốt đi

a) \(9c^2-6c+3\)

\(=\left(9c^2-6c+1\right)+2=\left(3c-1\right)^2+2>0\)

b) \(14m-6m^2-13\)

\(=-6.\left(m^2-\frac{7}{3}m+\frac{13}{6}\right)\)

\(=-6.\left(m^2-2\cdot\frac{7}{6}\cdot m+\frac{49}{36}+\frac{29}{36}\right)\)

\(=-6.\left(m-\frac{7}{6}\right)^2-\frac{29}{6}< 0\)

c) \(a^2-2a+2=\left(a-1\right)^2+1>0\)

d) \(6b-b^2-10=-\left(b^2-6b+9\right)-1=-\left(b-3\right)^2-1< 0\)

a, \(P\left(x\right)=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+...+15\)

\(=x^7-x^7-x^6+x^6+x^5-x^5-x^4+...+15=15\)

Bài 1:(Theo mình câu a nên sửa lại như thế này nhé)

a, a²-5a-14                                                          b,x⁴+x²-2

=a²+2a-7a-14                                                        =x⁴-x³+x³-x²+2x²-2x+2x-2

=(a²+2a)-(7a+14)                                                  =(x⁴-x³)+(x³-x²)+(2x²-2x)+(2x-2)

=a(a+2)-7(a+2)                                                      =x³(x-1)+x²(x-1)+2x(x-1)+2(x-1)

=(a+2)(a-7)                                                            =(x-1)(x³+x²+2x+2)

                                                                              =(x-1)[(x³+x²)+(2x+2)]

                                                                              =(x-1)[x²(x+1)+2(x+1)]

                                                                              =(x-1)(x+1)(x²+2)

Bài 2:

a, x³+x²+x+1=0

<=>(x³+x²)+(x+1)=0

<=>x²(x+1)+(x+1)=0

<=>(x+1)(x²+1)=0

<=>\(\orbr{\begin{cases}x+1=0\\x^2+1=0\left(loại\right)\end{cases}}\)(x² luôn lớn hơn hoặc bằng 0 =>x²+1 luôn lớn hơn hoặc bằng 1 nên x²+1=0 loại nhé)

<=>x= -1

b, x(2x-7)-4x+14=0

<=>x(2x-7)-(4x-14)=0

<=>x(2x-7)-2(2x-7)=0

<=>(2x-7)(x-2)=0

<=>\(\orbr{\begin{cases}2x-7=0\\x-2=0\end{cases}}\)

<=>\(\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)

ĐKXĐ \(2x+1\ge0\Rightarrow x\ge-\frac{1}{2}\)

Khi đó |x² - 1| = 2x + 1

,=> \(\orbr{\begin{cases}x^2-1=2x+1\\x^2-1=-2x-1\end{cases}}\Rightarrow\orbr{\begin{cases}x^2-2x=2\\x^2+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2-2x+1=3\\x^2+2x+1=1\end{cases}}\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=3\\\left(x+1\right)^2=1\end{cases}}\)

Nếu (x - 2)² = 3

=> \(\orbr{\begin{cases}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\sqrt{2}+1\\x=-\sqrt{2}+1\end{cases}}\)(tm)

Nếu (x + 1)² = 1

=> \(\orbr{\begin{cases}x+1=1\\x+1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-2\left(\text{loại}\right)\end{cases}}\Rightarrow x=0\)

a) \(A=\left(x-y\right)^2+\left(x+y\right)^2\)

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

\(=x^2-2xy+y^2+x^2+2xy+y^2\)

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

\(=2x^2+2y^2\)

\(=2.\left(x^2+y^2\right)\)

b) \(B=\left(2a+b\right)^2-\left(2a-b\right)^2\)

\(=\left(4a^2+4ab+b^2\right)-\left(4a^2-4ab+b^2\right)\)

\(=4a^2+4ab+b^2-4a^2+4ab-b^2\)

\(=\left(4a^2-4a^2\right)+\left(4ab+4ab\right)+\left(b^2-b^2\right)\)

\(=8ab\)\

c) \(C=\left(x+y\right)^2-\left(x-y\right)^2\)

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

\(=x^2+2xy+y^2-x^2+2xy-y^2\)

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

\(=4xy\)

d) \(D=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

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

\(=4x^2-4x+1-8x^2+24x-18+4\)

\(=\left(4x^2-8x^2\right)-\left(4x-24x\right)+\left(1-18+4\right)\)

\(=-4x^2+20x-13\)

\(=-4x^2+20x-25+12\)

\(=-\left(4x^2-20x+25\right)-8\)

\(=-\left[\left(2x\right)^2-2.4x.5+5^2\right]-8\)

\(=-\left(2x-5\right)^2-8\)

a) \(5x\left(\frac{1}{5}x-2\right)+3\left(6-\frac{1}{3}x^2\right)=12\)

=> \(x^2-10x+18-x^2=12\)

=> -10x + 18 = 12

=> -10x = -6

=> -5x = -3

=> x = 3/5

b) 7x(x - 2) - 5(x - 1) = 7x² + 3

=> 7x² - 14x - 5x + 5 = 7x² + 3

=> 7x² - 14x - 5x + 5 - 7x² - 3 = 0

=> -19x + 2 = 0

=> -19x = -2

=> x = \(\frac{2}{19}\)

c) 2(5x - 8) - 3(4x - 5) = 4(3x - 4) + 11

=> 10x - 16 - 12x + 15 = 12x - 16 + 11

=> 10x - 16 - 12x + 15 - 12x + 16 - 11 = 0

=> (10x - 12x - 12x) + (-16 + 15 + 16 - 11) = 0

=> -14x + 4 = 0

=> -14x = -4

=> -7x = -2

=> x = 2/7

\(\frac{437^2-363^2}{537^2-463^2}=\frac{\left(437+363\right)\left(437-363\right)}{\left(537+463\right)\left(537-463\right)}=\frac{800.74}{1000.74}=\frac{8}{10}=\frac{4}{5}\)

a) \(x^2+4x+4=\left(x+2\right)^2\)

b) \(x^2-8x+16=\left(x-4\right)^2\)

c) \(\left(x+5\right)\left(x-5\right)=x^2-25\)

d) \(x^2+2x+1=\left(x+1\right)^2\)

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

f) \(\left(2x+3y\right)^2+2\left(2x+3y\right)+1=\left(2x+3y+1\right)^2\)

g) \(\left(2+bx^2\right)\left(bx^2-2\right)=\left(bx^2+2\right)\left(bx^2-2\right)=\left(bx^2\right)^2-4=b^2x^4-4\)