7x-3=6x+7

7x-3=6x+7

7x-6x=7+3

x=10

1) x² - 4 = 0

=> x² = 4

=> x = \(\pm\)2

2) 2x² - 8 = 0

=> 2x² = 8

=> x² = 4

=> x = \(\pm2\)

3) (x + 3)² = 4 => (x + 3)² = 2²

=> \(\orbr{\begin{cases}x+3=2\\x+3=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=-5\end{cases}}\)

4) (x - 7)² = 36

=> (x - 7)² = 6²

=> \(\orbr{\begin{cases}x-7=6\\x-7=-6\end{cases}}\Rightarrow\orbr{\begin{cases}x=13\\x=1\end{cases}}\)

5) x² - 14x = -49

=> x² - 14x + 49 = 0

=> x² - 7x - 7x + 49 = 0

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

=> (x - 7)² = 0

=> x = 7

6) x² + 6x + 5 = 0

=> x² + x + 5x + 5 = 0

=> x(x + 1) + 5(x + 1) = 0

=> (x + 1)(x + 5) = 0

=> \(\orbr{\begin{cases}x=-1\\x=-5\end{cases}}\)

7) x² - 14x + 13 = 0

=> x² - x - 13x + 13 = 0

=> x(x - 1) - 13(x - 1) = 0

=> (x - 1)(x - 13) = 0

=> \(\orbr{\begin{cases}x=1\\x=13\end{cases}}\)

8) x² + 10x +16 = 0

=> x² + 2x + 8x + 16 = 0

=> x(x + 2) + 8(x + 2) = 0

=> (x + 2)(x + 8) = 0

=> \(\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)

Bài 1.

2n²( n + 1 ) - 2n( n² + n - 3 )

= 2n³ + 2n² - 2n³ - 2nⁿ + 6n

= 6n \(⋮6\forall n\inℤ\)( đpcm )

Bài 2.

P = ( m² - 2m + 4 )( m + 2 ) - m³ + ( m + 3 )( m - 3 ) - m² - 18

P = m³ + 8 - m³ + m² - 9 - m² - 18

P = 8 - 9 - 18 = -19

=> P không phụ thuộc vào biến M ( đpcm )

( 3x + 1 )² + ( x + 1 )² = 10( x - 1 )( x + 1 )

<=> 9x² + 6x + 1 + x² + 2x + 1 = 10( x² - 1 ) 

<=> 10x² + 8x + 2 = 10x² - 10

<=> 10x² + 8x - 10x² = -10 - 2

<=> 8x = -12

<=> x = -12/8 = -3/2

c. => 9x² + 6x + 1 + x² + 2x + 1 = 10 . ( x² - 1 )

=> 10x² + 8x + 2 = 10x² - 10

=> 10x² + 8x + 2 - 10x² + 10 = 0

=> 8x + 12 = 0

=> 8x = 12

=> x = 3/2

TA CÓ ; F=1-X-X^2=-(X^2+2.X.1/2+1/4)+5/4=-(X-1/2)^2+5/4=<5/4

Vậy GTLN của F là 5/4 xảy ra khi x=1/2

\(F=1-x-x^2\)

\(F=\left(-x^2-x-\frac{1}{4}\right)+\frac{5}{4}\)

\(F=-\left[x^2+2\cdot\frac{1}{2}x+\left(\frac{1}{2}\right)^2\right]+\frac{5}{4}\)

\(F=-\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\)

\(-\left(x+\frac{1}{2}\right)^2\le0\forall x\Rightarrow-\left(x+\frac{1}{2}\right)^2+\frac{5}{4}\le\frac{5}{4}\)

Dấu " = " xảy ra <=> x + 1/2 = 0 => x = -1/2

=> MaxF = 5/4 <=> x = -1/2

a)

(x-2).(x+2)-(x+2)^2=4

<=>(x^2-2^2)-(x^2+4x+4)=4

<=> x^2-4-x^2-4x-4=4

<=> -4x=12

<=> x=-3

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

<=> x² - 4 - ( x² + 4x + 4 ) = 4

<=> x² - 4 - x² - 4x - 4 = 4

<=> -4x - 8 = 4

<=> -4x = 12

<=> x = -3

b) 4( x + 1 )² + ( 2x - 1 )² - 8( x - 1 )( x + 1 ) = 11

<=> 4( x² + 2x + 1 ) + 4x² - 4x + 1 - 8( x² - 1 )

<=> 4x² + 8x + 4 + 4x² - 4x + 1 - 8x² + 8 = 11

<=> 4x + 13 = 11

<=> 4x = -2

<=> x = -2/4 = -1/2

a. \(2x^3+3x^2+2x+3=2x\left(x^2+1\right)+3\left(x^2+1\right)=\left(2x+3\right)\left(x^2+1\right)\)

b. \(a^2-ab+a-b=a\left(a+1\right)-b\left(a+1\right)=\left(a-b\right)\left(a+1\right)\)

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

d. \(x^4-2x^3+10x^2-20x=x\left(x^3-2x^2+10x-20\right)\)

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

e. \(x^3+2x^2+x=x^2\left(x+1\right)+x\left(x+1\right)=\left(x^2+x\right)\left(x+1\right)\)

f. \(xy+y^2-x-y=x\left(y-1\right)+y\left(y-1\right)=\left(x+y\right)\left(y-1\right)\)

a) 2x³ + 3x² + 2x + 3

= ( 2x³ + 2x ) + ( 3x² + 3 )

= 2x( x² + 1 ) + 3( x² + 1 )

= ( x² + 1 )( 2x + 3 )

b) a² - ab + a - b

= ( a² + a ) - ( ab + b )

= a( a + 1 ) - b( a + 1 )

= ( a - b )( a + 1 )

c) 2x² + 4x + 2 - 2y²

= ( 2x² - 2y² ) + ( 4x + 2 )

= 2( x² - y² ) + 2( 2x + 1 )

= 2( x² - y² + 2x + 1 )

= 2[ ( x² + 2x + 1 ) - y² ]

= 2[ ( x + 1 )² - y² ]

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

d) x⁴ - 2x³ + 10x² - 20x

= x( x³ - 2x² + 10x - 20 )

= x[ ( x³ - 2x² ) + ( 10x - 20 ) ]

= x[ x²( x - 2 ) + 10( x - 2 ) ]

= x( x - 2 )( x² + 10 )

e) x³ + 2x² + x = x( x² + 2x + 1 ) = x( x + 1 )²

f) xy + y² - x - y

= ( xy - x ) + ( y² - y )

= x( y - 1 ) + y( y - 1 )

= ( x + y )( y - 1 )

@hoàng đây là tính hay gì bạn . Nếu tính thì :

a) (2x + 5y)²

= (2x + 5y)(2x + 5y)

= 2x(2x + 5y) + 5y(2x + 5y)

= 4x² + 10xy + 10xy + 25y²

= 4x² + 20xy + 25y²

b) Bạn sửa đề lại nhé

c) (4x - 7y)² = (4x - 7y)(4x - 7y)

= 4x(4x - 7y) - 7y(4x - 7y)

= 16x² - 28xy - 28xy + 49y²

= 16x² - 56xy + 49y²

d) (3x³ - 2y²)²

= (3x³ - 2y²)(3x³ - 2y²)

= 3x³(3x³ - 2y²) - 2y²(3x³ - 2y²)

= 9x⁶ - 6x³y² - 6x³y² + 4y⁴

= 9x⁶ - 12x³y² + 4y⁴

@huanhoahong bạn không biết làm thì đừng có vô đây để trả lời và nói xấu bạn

a)

\(=x^2\left(2x+3\right)+\left(2x+3\right)\)

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

b)

\(=a\left(a-b\right)+a-b\)

\(=\left(a+1\right)\left(a-b\right)\)

c)

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

\(=2\left(x+1-y\right)\left(x+1+y\right)\)

d)

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

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

e)

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

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

f)

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

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

a,2x³+3x²+2x+3

=(2x³+2x)+(3x²+3)

=2x(x²+1)+3(x²+1)

=(x²+1)(2x+3)

b,a²-ab+a-b

=(a²-ab)+(a-b)

=a(a-b)+(a-b)

=(a-b)(a+1)

c,2x²+4x+2-2y²

=2(x²+2x+1-y²)

=2[(x²+2x+1)-y² ]

=2[(x+1)²-y² ]

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

d,x⁴-2x³+10x²-20x

=(x⁴-2x³)+(10x²-20x)

=x³(x-2)+10x(x-2)

=(x-2)(x³+10x)

=(x-2)[x(x²+10)]

e,x³+2x²+x

=x(x²+2x+1)

=x(x+1)²

f,xy+y²-x-y

=(xy+y²)-(x-y)

=y(x+y)-(x+y)

=(x+y)(y-1)

1)\(8x^6-\frac{1}{125}y^3=\left(2x^2\right)^3-\left(\frac{1}{5}y\right)^3\)

Bạn tự lm tiếp.AD HĐT số (7)

2)\(\left(x+4\right)^3-64=\left(x+4\right)^3-4^3\)

AD HĐT số (7).Tự lm tiếp

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

AD HĐT số (7).Tự lm tiếp

4)\(x^9+1=\left(x^3\right)^3+1\)

AD HĐT số (7).Tự lm tiếp

5,\(x^{12}-y^4=\left(x^6\right)^2-\left(y^2\right)^2\)

AD HĐT số (3).Tự lm tiếp

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

AD HĐT số (4)

7)\(x^3-15x^2+75x-125=\left(x-5\right)^3\)

AD HĐT số (5)

8)\(27a^3-54a^2b+36ab^2-8b^3\)

\(=\left(3a\right)^3-3.\left(3a\right)^2.2b+3.3a.\left(2b\right)^2-\left(2b\right)^3\)

\(=\left(3a-2b\right)^3\)

AD HĐT số (5)