\(\left(x-1\right)\left(x+1\right)\left(x+2\right)=0\)

\(TH1:x-1=0\Leftrightarrow x=1\)

\(TH2:x+1=0\Leftrightarrow x=-1\)

\(TH3:x+2=0\Leftrightarrow x=-2\)

nhân đa thức vs đa thức , ko phải tìm x đâu bạn ạ! dù sao cững cảm ơn nh!

1.

=3/5x(3/7+4/7)+2/5x(13/9-4/9)

=3/5x1+2/5x1

=3/5+2/5

=1

2.Xx(3/4+4/5)=7/10

Xx31/20=7/10

X         =7/10:31/20

X         =14/31

\(\frac{3}{5}\cdot\frac{3}{7}+\frac{3}{5}\cdot\frac{4}{7}+\frac{2}{5}\cdot\frac{13}{9}-\frac{2}{5}\cdot\frac{4}{9}\)

\(=\frac{3}{5}\cdot\left(\frac{3}{7}+\frac{4}{7}\right)+\frac{2}{5}\cdot\left(\frac{13}{9}-\frac{4}{9}\right)\)

\(=\frac{3}{5}\cdot1+\frac{2}{5}\cdot1\)\(=\frac{3}{5}+\frac{2}{5}=1\)

_________________________________________

\(\frac{3}{4}\cdot x+\frac{4}{5}\cdot x=\frac{7}{10}\)

\(\left(\frac{3}{4}+\frac{4}{5}\right)\cdot x=\frac{7}{10}\)

\(\frac{31}{20}\cdot x=\frac{7}{10}\)

\(x=\frac{7}{10}:\frac{31}{20}\)

\(x=\frac{14}{31}\)

4x3-2=10

{ [ ( 1 + 2 ) x 3 - 4 ] : 5 + 6 }x7^8

= { [ 3 x 3 - 4 ] : 5 + 6 }

= { [ 9 - 4 ] : 5 + 6 }

= { 5 : 5 + 6 }

= { 1 + 6 }

= 7

KẾT BẠN NHA

a​) S=1 + 2 + 2^2 + 2^3 +...+ 2^63

   2S=2 + 2^2 + 2^3 + 2^4 +...+ 2^64

   S=2S-S=(2 + 2^2 + 2^3 + 6^4 +...+ 2^64)-(1 + 2 + 2^2 + 2^3 +...+ 2^63)

   S=2 + 2^2 + 2^3 + 2^4 +...+ 2^64 - 1 - 2 - 2^2 - 2^3 -...- 2^63

   S=2^64 - 1

Lời giải:
$x^2+4x+n=(x^2-2x)+(6x-12)+12+n=x(x-2)+6(x-2)+12+n$
$=(x-2)(x+6)+12+n$

Vậy $x^2+4x+n$ chia $x-2$ được thương là $x+6$ và dư $12+n$

1.

a) \(\left(-2x^3\right)\)\(\left(x^2+5x-\frac{1}{2}\right)\) = \(-2x^5\)\(-10x^4\) \(+x^3\)

b) (\(6x^3-7x^2\)\(-x+2\))\(:\left(2x+1\right)\)=\(3x^2-5x+2\)

2.

a) 9x(3x-y) + 3y (y-3x)=9x(3x-y)-3y(3x-y)

= (9x-3y)(3x-y)

= 3(3x-y)(3x-y)

= 3(3x-y)^2

b) \(x^3-3x^2\)\(-9x+27\)= \(\left(x^3-3x^2\right)\)\(-\left(9x-27\right)\)

= \(x^2\left(x-3\right)\)\(-9\left(x-3\right)\)

= \(\left(x^2-9\right)\left(x-3\right)\)

= \(\left(x+3\right)\left(x-3\right)\left(x-3\right)\)

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

Bài 1 ) a ) \(\left(-2x^3\right)\left(x^2+5x-\frac{1}{2}\right)\)

\(=-2x^5-10x^4+x^3\)

b ) \(\left(6x^3-7x^2+x+2\right):\left(2x+1\right)\)

\(=3x^2-5x+2\)

2 ) a ) \(9x\left(3x-y\right)+3y\left(y-3x\right)\)

\(=9x\left(3x-y\right)-3y\left(3x-y\right)\)

\(=\left(3x-y\right)\left(9x-3y\right)\)

\(=3\left(3x-y\right)\left(x-y\right)\)

b ) \(x^3-3x^2-9x+27\)

\(=\left(x^3-3x^2\right)-\left(9x-27\right)\)

\(=x^2\left(x-3\right)-9\left(x-3\right)\)

\(=\left(x^2-9\right)\left(x-3\right)\)

\(=\left(x-3\right)\left(x+3\right)\left(x-3\right)\)

C1

a) -7x(3x-2)=-21x^2+14x

b) 87^2+26.87+13^2=87^2+2.13.87+13^2=(87+13)^2=100^2

C2

a) (x-5)(x+5)

b)3x(x+5)-2(x+5)=(3x-2)(x+5)=0

\(\Rightarrow\left[\begin{array}{nghiempt}3x-2=0\\x+5=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{2}{3}\\x=-5\end{array}\right.\)

Vậy S={-5;2/3}

C3:

a)3x^3-2x^2+2=(x+1)(3x^2-5x-5)-3

b) Để A chia hết cho B=> x+1\(\inƯ\left(-3\right)\)

\(\Rightarrow\begin{cases}x+1=3\\x+1=-3\\x+1=1\\x+1=-1\end{cases}\)\(\Rightarrow\begin{cases}x=2\\x=-4\\x=0\\x=-2\end{cases}\)