Bài 3: Những hằng đẳng thức đáng nhớ

a: \(50.5^2-50.4^2=\left(50.5-50.4\right)\cdot\left(50.5+50.4\right)\)

\(=100.9\cdot0.1=10.09\)

b; \(202\cdot198=\left(200+2\right)\left(200-2\right)\)

\(=200^2-2^2=40000-4=39996\)

c: \(10.2^2=\left(10+0.2\right)^2\)

\(=10^2+2\cdot10\cdot0.2+0.2^2\)

\(=100+0.04+4=104.04\)

d: \(101^2-202\cdot71+71^2\)

\(=101^2-2\cdot101\cdot71+71^2\)

\(=\left(101-71\right)^2=50^2=2500\)

a: \(\left(1-4x\right)\left(1+4x\right)=1-\left(4x\right)^2=1-16x^2\)

b: \(\left(-2x-5y\right)\left(2x-5y\right)\)

\(=-\left(2x+5y\right)\left(2x-5y\right)\)

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

c: \(\left(x^3-3x\right)\left(x^3+3x\right)=\left(x^3\right)^2-\left(3x\right)^2=x^6-9x^2\)

d: \(\left(1+x+x^2\right)\left(1+x-x^2\right)\)

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

a: \(\left(x^2+4y^2\right)\left(x+2y\right)\left(x-2y\right)\)

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

\(=x^4-16y^4\)

b: \(\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\)

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

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

\(=x^8-1\)

a) \(\left(4x-5\right)^2\)

\(=\left(4x\right)^2-2\cdot4x\cdot5+5^2\)

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

b) \(\left(-x+0,3\right)^2\)

\(=\left(-x\right)^2+2\cdot\left(-x\right)\cdot0,3+\left(0,3\right)^2\)

\(=x^2-0,6x+0,09\)

c) \(\left(-x-10y\right)^2\)

\(=\left(-x\right)^2+2\cdot\left(-x\right)\cdot\left(-10y\right)+\left(-10y\right)^2\)

\(=x+20xy+100y^2\)

d) \(\left(a^3-3a\right)^2\)

\(=\left(a^3\right)^2-2\cdot a^3\cdot3a+\left(3a\right)^2\)

\(=a^6-6a^4+9a^2\)

a: \(\left(4x-5\right)^2=\left(4x\right)^2-2\cdot4x\cdot5+5^2=16x^2-40x+25\)

c: \(\left(-x+0.3\right)^2=\left(-x\right)^2+2\cdot\left(-x\right)\cdot0.3+0.3^2\)

\(=x^2-0.6x+0.09\)

d: \(\left(-x-10y\right)^2=\left(x+10y\right)^2\)

\(=x^2+2\cdot x\cdot10y+\left(10y\right)^2\)

\(=x^2+20xy+100y^2\)

e: \(\left(a^3-3a\right)^2\)

\(=\left(a^3\right)^2-2\cdot a^3\cdot3a+\left(3a\right)^2\)

\(=a^6-6a^4+9a^2\)

`(x+y)^2 -2(x+y)(x-y)+(x-y)^2`

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

a) \(107^2-93^2\)

\(=\left(107-93\right)\left(107+93\right)\)

\(=14\cdot200\)

\(=2800\)

b) \(106\cdot94\)

\(=\left(100+6\right)\left(100-6\right)\)

\(=100^2-6^2\)

\(=10000-36\)

\(=9964\)

c) \(102^2\)

\(=\left(100+2\right)^2\)

\(=100^2+2\cdot100\cdot2+2^2\)

\(=10000+400+4\)

\(=10404\)

d) \(199^2\)

\(=\left(200-1\right)^2\)

\(=200^2-2\cdot200\cdot1+1^2\)

\(=40000-400+1\)

\(=39601\)

e) \(234^2-233\cdot235\)

\(=234^2-\left(234-1\right)\left(234+1\right)\)

\(=234^2-\left(234^2-1\right)\)

\(=234^2-234^2+1\)

\(=1\)

f) \(788\cdot790-789^2\)

\(=\left(789-1\right)\left(789+1\right)-789^2\)

\(=\left(789^2-1\right)-789^2\)

\(=789^2-1-789^2\)

\(=-1\)

g) \(234^2+132\cdot234+66^2\)

\(=234^2+2\cdot66\cdot234+66^2\)

\(=\left(234+66\right)^2\)

\(=300^2\)

\(=90000\)

h) \(368^2-68\cdot736+68^2\)

\(=368^2-2\cdot368\cdot68+68^2\)

\(=\left(368-68\right)^2\)

\(=300^2\)

\(=90000\)

\(f,F=x^2+9y^2-8x+4y+27\) (sửa đề)

\(=\left(x^2-8x+16\right)+\left(9y^2+4y+\dfrac{4}{9}\right)+\dfrac{95}{9}\)

\(=\left(x^2-2\cdot x\cdot4+4^2\right)+\left[\left(3y\right)^2+2\cdot3y\cdot\dfrac{2}{3}+\left(\dfrac{2}{3}\right)^2\right]+\dfrac{95}{9}\)

\(=\left(x-4\right)^2+\left(3y+\dfrac{2}{3}\right)^2+\dfrac{95}{9}\)

Ta thấy: \(\left(x-4\right)^2\ge0\forall x\)

             \(\left(3y+\dfrac{2}{3}\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-4\right)^2+\left(3y+\dfrac{2}{3}\right)^2\ge0\forall x;y\)

\(\Rightarrow\left(x-4\right)^2+\left(3y+\dfrac{2}{3}\right)^2+\dfrac{95}{9}\ge\dfrac{95}{9}>0\forall x;y\)

hay \(F\) luôn dương với mọi \(x;y\).

\(Toru\)

Viết liền thế ai mà đọc được

1:

a: \(=4x^2+4x+1-x^2-4x+5\)

\(=3x^2+6\)

b: \(=9x^2-24x+16+\left(2x-3\right)^2\)

\(=9x^2-24x+16+4x^2-12x+9\)

\(=13x^2-36x+25\)

c: \(=2\left(x^2-2x+1\right)-3\left(x^2+6x+9\right)+4x^2-1\)

\(=2x^2-4x+2-3x^2-18x-27+4x^2-1\)

\(=3x^2-22x-26\)

`#3107.101107`

`D = x^3 - y^3 - 3xy` biết `x - y - 1 = 0`

Ta có:

`x - y - 1 = 0`

`=> x - y = 1`

`D = x^3 - y^3 - 3xy`

`= (x - y)(x^2 + xy + y^2) - 3xy`

`= 1 * (x^2 + xy + y^2) - 3xy`

`= x^2+ xy + y^2 - 3xy`

`= x^2 - 2xy + y^2`

`= x^2 - 2*x*y + y^2`

`= (x - y)^2`

`= 1^2 = 1`

Vậy, với `x - y = 1` thì `D = 1`

________

`E = x^3 + y^3` với `x + y = 5; x^2 + y^2 = 17`

`x + y = 5`

`=> (x + y)^2 = 25`

`=> x^2 + 2xy + y^2 = 25`

`=> 2xy = 25 - (x^2 + y^2)`

`=> 2xy = 25 - 17`

`=> 2xy = 8`

`=> xy = 4`

Ta có:

`E = x^3 + y^3`

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

`= 5 * [ (x^2 + y^2) - xy]`

`= 5 * (17 - 4)`

`= 5 * 13`

`= 65`

Vậy, với `x + y = 5; x^2 + y^2 = 17` thì `E = 65`

________

`F = x^3 - y^3` với `x - y = 4; x^2 + y^2 = 26`

Ta có:

`x - y = 4`

`=> (x - y)^2 = 16`

`=> x^2 - 2xy + y^2 = 16`

`=> (x^2 + y^2) - 2xy = 16`

`=> 2xy = (x^2 + y^2) - 16`

`=> 2xy = 26 - 16`

`=> 2xy = 10`

`=> xy = 5`

Ta có:

`F = x^3 - y^3`

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

`= 4 * [ (x^2 + y^2) + xy]`

`= 4 * (26 + 5)`

`= 4*31`

`= 124`

Vậy, với `x - y = 4; x^2 + y^2 = 26` thì `F = 124.`

\(9^n-2\cdot3^n+x^2+5+4x=0\)

\(\Leftrightarrow\left(9^n-2\cdot3^n+1\right)+\left(x^2+4x+4\right)=0\)

\(\Leftrightarrow\left[\left(3^n\right)^2-2\cdot3^n\cdot1+1^2\right]+\left(x^2+2\cdot x\cdot2+2^2\right)=0\)

\(\Leftrightarrow\left(3^n-1\right)^2+\left(x+2\right)^2=0\)

Ta thấy: \(\left(3^n-1\right)^2\ge0\forall n\)

              \(\left(x+2\right)^2\ge0\forall x\)

\(\Rightarrow\left(3^n-1\right)^2+\left(x+2\right)^2\ge0\forall x;n\)

Mặt khác: \(\left(3^n-1\right)^2+\left(x+2\right)^2=0\)

nên ta có: \(\left\{{}\begin{matrix}3^n-1=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3^n=1\\x=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}n=0\\x=-2\end{matrix}\right.\left(tm\right)\)

Vậy \(n=0;x=-2\).

#\(Toru\)