\(\hept{\begin{cases}ab+bc+ca-abc=0\\a+b+c-1=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}abc-ab-bc-ca=0\\a+b+c-1=0\end{cases}}\)

\(\Rightarrow\left(abc-ab-bc-ca\right)+\left(a+b+c-1\right)=0\)

\(\Leftrightarrow\left(abc-ab\right)-\left(ac-a\right)-\left(bc-b\right)+\left(c-1\right)=0\)

\(\Leftrightarrow ab\left(c-1\right)-a\left(c-1\right)-b\left(c-1\right)+\left(c-1\right)=0\)

\(\Leftrightarrow\left(a-1\right)\left(b-1\right)\left(c-1\right)=0\)

Vậy..........

\(A=\frac{1}{a}\)\(+\frac{1}{a}\)\(+\frac{1}{a}\)\(+\frac{1}{a}\)\(+\frac{1}{ab}\)\(\ge\frac{25}{4a+ab}\)\(=\frac{25}{a\left(b+4\right)}\)\(\ge\frac{25}{\frac{1}{4}\left(a+b+4\right)^2}\)\(=1\)

\(A_{min=1}\)\(khi\){ a = 5 

                            b = 1

Lần đầu tiên làm toán lớp 8 , có gì sai sót mong bạn chỉ ra hộ mình

( 2x + 1/2 )² - ( 1 - 2x )² = 2/5

⇔ 4x² + 2x + 1/4 - ( 1² - 4x + 4x² ) = 2/5

⇔ 4x² + 2x + 1/4 - 1 + 4x - 4x² = 2/5

⇔ 6x - 3/4 = 2/5

⇔ 6x = 23/20

⇔ x = 23/120

        Bài làm :

Ta có :

( 2x + 1/2 )² - ( 1 - 2x )² = 2/5

<=> 4x² + 2x + 1/4 - 1 + 4x - 4x² = 2/5

<=> 6x - 3/4 = 2/5

<=> 6x = 23/20

<=> x = 23/120

Vậy x=23/120

6x³ - 11x² - x - 2

= 6x³ - 12x² + x² - 2x + x - 2

= ( 6x³ - 12x² ) + ( x² - 2x ) + ( x - 2 )

= 6x²( x - 2 ) + x( x - 2 ) + 1( x - 2 )

= ( x - 2 )( 6x² + x + 1 )

          Bài làm :

Ta có :

6x³ - 11x² - x - 2

= 6x³ - 12x² + x² - 2x + x - 2

= ( 6x³ - 12x² ) + ( x² - 2x ) + ( x - 2 )

= 6x²( x - 2 ) + x( x - 2 ) + 1( x - 2 )

= ( x - 2 )( 6x² + x + 1 )

Nhận xét: f(x) là đơn ánh.

Thật vậy, giả sử f(x₁)=f(x₂) thì: f(f(x₁)+2y)=f(f(x₂)+2y)

=> 4x₁+4y+3=4x₂+4y+3<=>x₁=x₂. Vậy f là đơn ánh

Ta có: f(f(x)+2y)=4x+4y+3=f(f(y)+2x)

Vì f là đơn ánh nên: f(x)+2x=f(y)+2x hay f(x)-2x=f(y)-2y. Với mọi x,y thuộc R

Do đó: f(x)-2x=c, c thuộc R. Thay f(x)=2x+c vào điều kiện ta có c=1

Vậy f(x)=2x+1 (TMĐK)

A = x³ + 3x² + 3x - 899

= (x³ + 3x² + 3x + 1) - 900

= (x + 1)³ - 900

= (29 + 1)³ - 900 = 30³ - 900 = 26100

B = x³ - 6x² + 12x + 10

= (x³ - 6x² + 12x - 8) + 18

= (x - 2)³ + 18

= (12 - 2)³ + 18 = 10³ + 18 = 1000 + 18 = 1018

c) C = 8x³ - 27y³

= (2x)³ - (3y)³

= (2x - 3y)(4x² + 6xy + 9y²)

= (2x - 3y)(4x² - 12xy + 9y²) + (2x - 3y).18xy

= (2x - 3y)(2x - 3y)² + (2x - 3y).18xy

= (2x - 3y)³ + (2x - 3y).18xy

= 5³ + 5.18.4

= 125 - 360

= -235

D = x³ + y³ + 3xy(x² + y²) + 6x²y²(x + y)

= (x + y)(x² - xy + y²) + 3x³y + 3xy³ + 6x²y²

= x² + y² - xy + 3x³y + 3xy³ + 6x²y² 

= (x + y)² - 3xy + 3x³y + 3xy³ + 6x²y² 

= 1 - 3xy(2xy - 1) + 3xy(x² + y²)

= 1 - 3xy(x² + y² + 2xy - 1)

= 1 - 3xy[(x + y)² - 1]

= 1 - 0 = 1

C = ( a + b - c )² - ( a - c )² - 2ab + 2bc

= [ ( a + b ) - c ]² - ( a² - 2ac + c² ) - 2ab + 2bc

= ( a + b )² - 2( a + b )c + c² - a² + 2ac - c² - 2ab + 2bc

= a² + b² + 2ab - 2bc - 2ac - a² + 2ac - 2ab + 2bc

= b²

D = ( a + b + 1 )³ - ( a + b - 1 )³ - 6( a + b )²

= [ ( a + b ) + 1 ]³ - [ ( a + b ) - 1 ]³ - 6( a² + 2ab + b² )

= [ ( a + b )³ + 3( a + b )².1 + 3( a + b ).1² + 1³ ] - [ ( a + b )³ - 3( a + b )².1 + 3( a + b ).1² - 1³ ] - 6a² - 12ab - 6b²

= [ ( a³ + 3a²b + 3ab² + b³ ) + 3( a² + 2ab + b² ) + 3a + 3b + 1 ]  - [ ( a³ + 3a²b + 3ab² + b³ ) - 3( a² + 2ab + b² ) + 3a + 3b - 1 ] - 6a² - 12ab - 6b²

= ( a³ + 3a²b + 3ab² + b³ + 3a² + 6ab + 3b² + 3a + 3b + 1 ) - ( a³ + 3a²b + 3ab² + b³ - 3a² - 6ab - 3b² + 3a + 3b - 1 ) - 6a² - 12ab - 6b²

= a³ + 3a²b + 3ab² + b³ + 3a² + 6ab + 3b² + 3a + 3b + 1 - a³ - 3a²b - 3ab² - b³ + 3a² + 6ab + 3b² - 3a - 3b + 1 - 6a² - 12ab - 6b²

= 2

< D hơi dài nên có thể có sai sót >

Xét tg CID có

\(\widehat{IDC}+\widehat{ICD}=180^o-\widehat{CID}=180^o-50^o=130^o\)

\(\Rightarrow\widehat{D}+\widehat{C}=2\left(\widehat{IDC}+\widehat{ICD}\right)=2.130^o=260^o\)

\(\Rightarrow\widehat{A}+\widehat{B}=360^o-\left(\widehat{C}+\widehat{D}\right)=360^o-260^o=100^o\)

\(\Rightarrow\widehat{A}=\left(100^o+20^o\right):2=60^o\Rightarrow\widehat{B}=100^o-60^o=40^o\)

a) ( A + B + C )²

= [ ( A + B ) + C ]²

= ( A + B )² + 2( A + B )C + C²

= A² + B² + C² + 2AB + 2BC + AC

b) ( A + B - C )²

= [ ( A + B ) - C ]²

= ( A + B )² - 2( A + B )C + C²

= A² + B² + C² + 2AB - 2BC - 2AC

c) ( A - B - C )²

= [ ( A - B ) - C ]²

= ( A - B )² - 2( A - B )C + C²

= A² + B² + C² - 2AB + 2BC - 2AC

        Bài làm :

a) ( A + B + C )²

= [ ( A + B ) + C ]²

= ( A + B )² + 2( A + B )C + C²

= A² + B² + C² + 2AB + 2BC + AC

b) ( A + B - C )²

= [ ( A + B ) - C ]²

= ( A + B )² - 2( A + B )C + C²

= A² + B² + C² + 2AB - 2BC - 2AC

c) ( A - B - C )²

= [ ( A - B ) - C ]²

= ( A - B )² - 2( A - B )C + C²

= A² + B² + C² - 2AB + 2BC - 2AC

a) 127² + 146.127 + 73²

= 127² + 2.73.127 + 73²

= (127 + 73)² = 200² = 40000

b) 9⁸ . 2⁸ - (18⁴ - 1)(18⁴ + 1)

= (9.2)⁸ - 18⁸ + 1

= 18⁸ - 18⁸ + 1 = 1

c) \(\frac{780^2-220^2}{125^2+150.125+75^2}=\frac{\left(780-220\right)\left(780+220\right)}{125^2+2.75.125+75^2}=\frac{560.1000}{\left(125+75\right)^2}=\frac{560000}{200^2}\)

\(=\frac{560000}{40000}=14\)

a) 127² + 146.127 + 73²

= 127² + 2.73.127 + 73²

= ( 127 + 73 )²

= 200² = 40 000

b) 9⁸.2⁸ - ( 18⁴ - 1 )( 18⁴ + 1 ) 

= ( 9.2 )⁸ - [ ( 18⁴ )² - 1² ]

= 18⁸ - 18⁸ + 1

= 1

c) \(\frac{780^2-220^2}{125^2+150\cdot125+75^2}\)

\(=\frac{\left(780-220\right)\left(780+220\right)}{125^2+2\cdot75\cdot125+75^2}\)

\(=\frac{560\cdot1000}{\left(125+75\right)^2}\)

\(=\frac{560000}{200^2}\)

\(=\frac{560000}{40000}=14\)