Bài 4: Những hằng đẳng thức đáng nhớ (Tiếp)

`A=x^2 -x +1 = x^2 -2*x*1/2 +(1/2)^2 +1 -(1/2)^2`

`A=(x-1/2)^2 + 3/4 >0 AAx `

Vậy `A=x^2 -x +1 >0`

`B=x^2 +x+1 = x^2 +2*x*1/2 +(1/2)^2 +1 -(1/2)^2`

`B= (x+1/2)^2 +3/4 > 0 AAx`

Vậy `B=x^2 +x +1 >0`

`C=x^2 +2x+2 = x^2 +2*x*1 +1^2 +2-1^2`

`C=(x+1)^2 +1 >0`

Vậy `C=x^2 +2x+2>0`

a: \(=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

b: \(=x^2+x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

c: \(=x^2+2x+1+1=\left(x+1\right)^2+1>0\)

\(A=-3\left(x^2-4x+4\right)+12=-3\left(x-2\right)^2+12\le12\)

\(A_{max}=12\) khi \(x=4\)

\(B=-\left(x^2-2x+1\right)-\left(x^2+2xy+y^2\right)+1=-\left(x-1\right)^2-\left(x+y\right)^2+1\le1\)

\(B_{max}=1\) khi \(\left(x;y\right)=\left(1;-1\right)\)

\(C=-\left(4x^2+12x+9\right)+9=-\left(2x+3\right)^2+9\le9\)

\(C_{max}=9\) khi \(x=-\dfrac{3}{2}\)

A = - 3 (x^2 - 4x + 4) + 12 

= - 3 (x - 2)^2 + 12 ≤ 12

= max => 12 khi x = 4

B = - (x^2 - 2x + 1) - (x^2 + 2xy + y^2) + 1

= - (x - 1)^2 - (x + y ) + 1 ≤ 1 

= max => 1 khi (x,y) khi (1,1)

a) A = 153² - 53²

=> A = (153+53)(153-53)

=> A = 206.100

=> A = 20600

b) B = 2020² - 2019² + 2018² - 2017² +...+2² - 1²

=>B = (2020+2019)(2020-2019)+(2018 + 2017).(2018-2017) + ... +(2-1)(2+1)

=> B = 4039 + 4035 +...+3

=> B = (3+4039). 1010 : 2 

=> B = 2041210

c) C = x³ - 3x² + 3x - 1

=> C = (x³ -1) - (3x² - 3x)

=> C = (x-1)(x² +x+1)- 3x(x-1)

=>C = (x-1)(x² +x+1 - 3x)

Thay x= 101 vào C

=> C = (101-1)(101² +101+1-3.101)

=> C = 1000000

d) D = x³ + 6x² +12x + 8

=> D = (x³ + 2³) + (6x² +12x)

=> D = (x +2)(x² - 2x + 4)+ 6x(x+2)

=> D = (x+2)(x² +4x+4)

=> D = (x+2)(x+2)²

=> D = (x+2)³

=> D = (8+2)³

=> D = 1000

e) Ta có x = 9 => 10 = (x+1)

=> E = x²⁰¹³ - (x+1).x²⁰¹² + (x+1).x²⁰¹¹ +...-(x+1).x² + (x+1).x-(x+1)

=> E = x²⁰¹³ - x²⁰¹³ - x²⁰¹² + x²⁰¹² + x²⁰¹¹ +..- x³ - x² + x² + x - x - 1

=> E = -1 

a) \(A=153^2-53^2=\left(153-53\right)\left(153+53\right)=100.206=20600\)

b) \(B=2020^2-2019^2+2018^2-2017^2+...+2^2-1^2\)

\(=\left(2020-2019\right)\left(2020+2019\right)+\left(2018-2017\right)\left(2018+2017\right)+...+\left(2-1\right)\left(2+1\right)\)

\(=\left(2020+2019\right)+\left(2018+2017\right)+...+\left(2+1\right)\)

\(=2020+2019+2018+2017+...+2+1\) (có \(\left(2020-1\right)+1=2020\) (số hạng))

\(=\left(2020+1\right)+\left(2019+2\right)+...+\left(1011+1010\right)\) (có \(2020:2=1010\) (cặp))

\(=2021+2021+...+2021\) (có 1010 (số hạng))

\(=1010.2021\)

\(=2041210\)

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

Thay \(x=101\) vào biểu thức C, ta được:

\(C=\left(101-1\right)^3=100^3=1000000\)

d) \(D=x^3+6x^2+12x+8=\left(x+2\right)^3\)

Thay \(x=8\) vào biểu thức D, ta được:

\(D=\left(8+2\right)^3=10^3=1000\)

e) Vì \(x=9\Rightarrow x+1=10,\) thay vào biểu thức E ta được:

\(E=x^{2013}-\left(x+1\right)x^{2012}+\left(x+1\right)x^{2011}+...-\left(x+1\right)x^2+\left(x+1\right)x-\left(x+1\right)\)

\(=x^{2013}-x^{2013}-x^{2012}+x^{2012}+x^{2011}+...-x^3-x^2+x^2+x-x-1\)

\(=-1\)

`a)(5x-1)(5x+1)=15`

`<=>25x^2-1=15`

`<=>25x^2=16`

`<=>x^2=16/25`

`<=>x=+-4/5`

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`b)x^2-8x+16-(5x+2)^2=0`

`<=>(x-4)^2-(5x+2)^2=0`

`<=>(x-4-5x-2)(x-4+5x+2)=0`

`<=>(-4x-6)(6x-2)=0`

`<=>[(x=-3/2),(x=1/3):}`

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`c)x^2-10x+9=0`

`<=>x^2-9x-x+9=0`

`<=>(x-9)(x-1)=0`

`<=>[(x=9),(x=1):}`

a, \(\left(5x-1\right)\left(5x+1\right)=15\)

\(\Leftrightarrow\left(5x\right)^2-1^2=15\)

\(\Leftrightarrow25x^2-1=15\)

\(\Leftrightarrow25x^2=16\)

\(\Leftrightarrow x^2=\dfrac{16}{25}=\left(\pm\dfrac{4}{5}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

b, \(x^2-8x+16-\left(5x+2\right)^2=0\)

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

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

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

\(\Leftrightarrow\left(-4x-6\right)\left(6x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-4x-6=0\\6x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-4x=6\\6x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

c, \(x^2-10x+9=0\)

\(\Leftrightarrow\left(x^2-9x\right)-\left(x-9\right)=0\)

\(\Leftrightarrow x\left(x-9\right)-\left(x-9\right)=0\)

\(\Leftrightarrow\left(x-9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-9=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)

a: \(x^2+12x+36=\left(x+6\right)^2\)

b: \(x^2+81+18x=\left(x+9\right)^2\)

c: \(x^2-16x+64=\left(x-8\right)^2\)

d:\(16-2x+\dfrac{1}{16}x^2=\left(4-\dfrac{1}{4}x\right)^2\)

e: \(9x^2-6xy+y^2=\left(3x-y\right)^2\)

f: \(16a^2-9b^2=\left(4a-3b\right)\left(4a+3b\right)\)

\(a,=-\left(x^3-3x^2+3x-1\right)\\ =-\left(x-1\right)^3\\ b,=x^3+3.x^2\dfrac{1}{3}+3.\dfrac{1}{9}x+\left(\dfrac{1}{3}\right)^3\\ =\left(x+\dfrac{1}{3}\right)^3\)

\(a,-x^3+3x^2-3x+1=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
\(b,x^3+x^2+\dfrac{1}{3}x+\dfrac{1}{27}=\left(x-\dfrac{1}{3}\right)^2\)
- Học tút-

`(x+y/x)^3`

`=x^3+3x^2 . y/x+3x. (y/x)^2+(y/x)^3`

`=x^3+3xy+3[y^2]/x+[y^3]/[x^3]`

\(x^3-3.4.x^2+3.x.16-64-8=0\\ \left(x-4\right)^3-8=0\\ \left(x-4-2\right)\left[\left(x-4\right)^2+2\left(x-4\right)+4\right]=0\\ \left(x-6\right)\left(x^2-8x+16+2x-8+4\right)=0\\ \left(x-6\right)\left(x^2-6x+12\right)=0\\ \left[{}\begin{matrix}x-6=0\\x^2-6x+12=0\left(vono\right)\end{matrix}\right.=>x=6\)

\(\Leftrightarrow6x^3+18x^2+18x+6+2x^3-2-2x^3-6x^2-6x-2=32\)

\(\Leftrightarrow6x^3+12x^2+12x+2=32\)

\(\Leftrightarrow6x^3+12x^2+12x-30=0\)

hay x=1

\(=3\left(x^2-2x+674\right)\\ =3\left(x^2-2x+1\right)+673\\ =3\left(x-1\right)^2+673\ge673\forall x\\ \)

Dấu = xảy ra khi

\(x-1=0\\ x=1\)

Vậy GTNN : \(673khix=1\)

\(=3x^2-6x+3+2019=3\left(x-1\right)^2+2019>=2019\)

Dấu '=' xảy ra khi x=1