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

\(\Leftrightarrow\left(4x\right)^2-\left(4x-5\right)^2-15=0\)

\(\Leftrightarrow\left(4x-4x+5\right)\left(4x+4x-5\right)-15=0\)

\(\Leftrightarrow5\left(8x-5\right)-15=0\)

\(\Leftrightarrow40x-25-15=0\)

\(\Leftrightarrow40x-40=0\)

\(\Leftrightarrow x=1\)

Tìm x, biết:

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

\(\left(4x\right)^2-\left(4x-5\right)^2-15=0\)

\(\left(4x-4x+5\right)\left(4x+4x-5\right)-15=0\)

\(5\left(8x-5\right)-15=0\)

\(40x-25-15=0\)

\(40x-40=0\)

\(x=1\)

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

\(\Leftrightarrow\left(4x+4x-5\right)\left(4x-4x+5\right)=15\)

\(\Leftrightarrow5\left(8x-5\right)=15\)

\(\Leftrightarrow8x-5=3\)

\(\Leftrightarrow8x=8\)

\(\Leftrightarrow x=1\)

64x^3 + 1 - 64x^3 + 80x =17

80x                              =16

   x                               =3/10

64x^3 + 1 -64x^3 + 80x = 17

80x                             = 16

   x                              = 3/10

64x^3 + 1 - 64x^3 + 80x = 17

80x                               = 16

   x                                = 3/10

1. \(A=x^{15}+3x^{14}+5=x^{14}\left(x+3\right)+5\)

Thay \(x+3=0\)vào đa thức ta được:\(A=x^{14}.0+5=5\)

2. \(B=\left(x^{2007}+3x^{2006}+1\right)^{2007}=\left[x^{2006}\left(x+3\right)+1\right]^{2007}\)

Thay \(x=-3\)vào đa thức ta được: \(B=\left[x^{2006}\left(-3+3\right)+1\right]^{2017}=\left(x^{2006}.0+1\right)^{2017}=1^{2017}=1\)

3. \(C=21x^4+12x^3-3x^2+24x+15=3x\left(7x^3+4x^2-x+8\right)+15\)

Thay \(7x^3+4x^2-x+8=0\)vào đa thức ta được: \(C=3x.0+15=15\)

4. \(D=-16x^5-28x^4+16x^3-20x^2+32x+2007\)

\(=4x\left(-4x^4-7x^3+4x^2-5x+8\right)+2007\)

Thay \(-4x^4-7x^3+4x^2-5x+8=0\)vào đa thức ta được: \(D=4x.0+2007=2007\)

1. \(A=x^{15}+3x^{14}+5\)

\(A=x^{14}\left(x+3\right)+5\)

\(A=x^{14}+5\)

2. \(B=\left(x^{2007}+3x^{2006}+1\right)^{2007}\)

\(B=\left[x^{2006}\left(x+3\right)+1\right]^{2007}\)

\(B=\left[x^{2006}.\left(-3+3\right)+1\right]^{2007}\)

\(B=1^{2007}=1\)

3. \(C=21x^4+12x^3-3x^2+24x+15\)

\(C=3x\left(7x^2+4x^2-x+8+5\right)\)

\(C=3x\left(0+5\right)\)

\(C=15x\)

4. \(D=-16x^5-28x^4+16x^3-20x^2+32+2007\)

\(D=4x\left(-4x^4-7x^3+4x^2-5x+8\right)+2007\)

\(D=4x.0+2007\)

\(D=2007\)

BÀI 1:

Ta có:   \(VT=\left(7x+1\right)^2-\left(x+7\right)^2\)

                    \(=\left(7x+1+x+7\right)\left(7x+1-x-7\right)\)

                    \(=\left(8x+8\right)\left(6x-6\right)\)

                   \(=8\left(x+1\right).6\left(x-1\right)\)

                  \(=48\left(x^2-1\right)=VP\)  (đpcm)

Bài 2:

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

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

\(\Leftrightarrow\)\(40x=40\)

\(\Leftrightarrow\)\(x=1\)

Vậy...

Bài 3:

\(A=x^2+2x+3=\left(x+1\right)^2+2\ge2\)

Vậy MIN A = 2  khi  x = -1

Bài 1:
\(16x^2-\left(4x-5\right)^2=15\)

\(\Leftrightarrow\left(4x-4x+5\right)\left(4x+4x-5\right)=15\)

\(\Leftrightarrow5\left(8x-5\right)=15\)

\(\Leftrightarrow8x=8\Leftrightarrow x=1\)

Vậy x = 1

Bài 2:

\(VT=\left(7x+1\right)^2-\left(x+7\right)^2\)

\(=\left(7x+1-x-7\right)\left(7x+1+x+7\right)\)

\(=\left(6x-6\right)\left(8x+8\right)\)

\(=48\left(x-1\right)\left(x+1\right)\)

\(=48\left(x^2-1\right)=VP\)

\(\Rightarrowđpcm\)

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

=>4x²+12x+9-4x²+4=49

 =>12x+13=49

=>12x=36

=>x=3

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

=>16x²-16x²+40x-25=15

=>40x-25=15

=>40x=40

=>x=1

c) ( 2x + 1 )^2 - ( x - 1)^2 = 0

=>4x²+4x+1-x²+2x-1=0

=>3x²+6x=0

=>3x(x+2)=0

=>3x=0 hoặc x+2=0

=>x=0 hoặc x=-2

a) \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\\ =>4x^2+12x+9-4x^2+4=49\\=>12x+13=49\\ =>12x=36\\ =>x=3\)

b) \(16x^2-\left(4x-5\right)^2=15\\ =>16x^2-16x^2+40x-25=15\\ =>40x-25=15\\ =>40x=40\\ =>x=1\)

c) \(\left(2x+1\right)^2-\left(x-1\right)^2=0\\ =>4x^2+4x+1-x^2+2x-1=0\\ =>3x^2+6x=0\\ =>3x\left(x+2\right)=0\\ =>\left[\frac{3x=0}{x+2=0}\right]=>\left[\frac{x=0}{x=-2}\right]\)