\(8-12x+6x^2-x^3\)

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

\(125x^3-75x^2+15x-1\)

\(=\left(5x-1\right)^3\)

\(x^2-xz-9y^2+3yz\)

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

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

\(x^3-x^2-5x+125\)

\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)

\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)

\(=\left(x+5\right)\left(x^2-6x+25\right)\)

\(x^3+2x^2-6x-27\)

\(=x^3+5x^2+9x-3x^2-15x-27\)

\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)

\(=\left(x-3\right)\left(x^2+5x+9\right)\)

\(12x^3+4x^2-27x-9\)

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

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

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

\(4x^4+4x^3-x^2-x\)

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

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

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

Là 00000000000000

a) \(\dfrac{6}{7}-\dfrac{3}{4}x=-1\dfrac{1}{2}x-3\)

<=> \(\dfrac{6}{7}-\dfrac{3}{4}x=-\dfrac{3}{2}x-3\)

<=> \(-\dfrac{3}{4}x+\dfrac{3}{2}x=-3-\dfrac{6}{7}\)

<=> \(x\left(-\dfrac{3}{4}+\dfrac{3}{2}\right)=-\dfrac{27}{7}\)

<=> \(x=-\dfrac{36}{7}\)

b) (4x-1)² = (4x-1)⁴

<=> (4x-1)² - (4x-1)⁴ = 0

<=> (4x-1)²[1-(4x-1)²] = 0

<=> \(\left\{{}\begin{matrix}4x-1=0\\\left(4x-1\right)^2=1\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\dfrac{1}{4}\\\left\{{}\begin{matrix}4x=2\\4x=0\end{matrix}\right.\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)

c) \(\left(\dfrac{4}{5}-\dfrac{3}{4}:x\right)^3=-\dfrac{1}{27}\)

<=> \(\dfrac{4}{5}-\dfrac{3}{4}:x=-\dfrac{1}{3}\)

<=> \(\dfrac{3}{4}:x=\dfrac{17}{15}\)

<=> \(x=\dfrac{45}{68}\)

d) \(\left|\dfrac{4}{3}-\dfrac{1}{4}x\right|:2\dfrac{1}{3}=0,5\)

<=> \(\left|\dfrac{4}{3}-\dfrac{1}{4}x\right|=\dfrac{7}{6}\)

<=> \(\left\{{}\begin{matrix}\dfrac{4}{3}-\dfrac{1}{4}x=\dfrac{7}{6}\\\dfrac{4}{3}-\dfrac{1}{4}x=-\dfrac{7}{6}\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\dfrac{1}{4}x=\dfrac{1}{6}\\\dfrac{1}{4}x=\dfrac{5}{2}\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)

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

\(\Rightarrow\left(x-3\right)=0\)  ( \(x^2+1>0\forall x\))

\(\Rightarrow x=3\)

đ) \(4.8^2=2^x\)

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

\(2^2.2^6=2^x\)

\(2^8=2^x\)

\(\Rightarrow x=8\)

d) \(\left|x+3\right|=8\)

\(\Rightarrow\orbr{\begin{cases}x+3=8\\x+3=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-11\end{cases}}\)

mấy câu trên dễ rồi tự làm em nhé 

a, 2x + 5 = 7

=> 2x = 2

=> x = 1

vậy____

tìm 2 số nguyên a và b biết :a+b=-1 và a.b=-12.Giup mình nha

P(x) + Q(x)= ( x^5 - 2x^2 + 7x^4 - 9x^3 - 1/4x) + ( 5x^4 - x^5 + 4x^2 - 2x^3 - 1/4)

                 = x^5 - 2x^2 + 7x^4 - 9x^3 - 1/4x + 5x^4 - x^5 + 4x^2 - 2x^3 - 1/4

                 = ( x^5 - x^5 ) - ( 2x^2 + 4x^2) + ( 7x^4 + 5x^4) - ( 9x^3 - 2x^3) - 1/4x - 1/4

                 =                            6x^2 + 12x^4 - 6x^3 - 1/4x - 1/4

P(x) - Q(x)=  ( x^5 - 2x^2 + 7x^4 - 9x^3 -1/4x) - ( 5x^4 - x^5 + 4x^2 - 2x^3 -1/4)

                 = x^5 - 2x^2 + 7x^4 - 9x^3 - 1/4x - 5x^4 + x^5 - 4x^2 + 2x^3 + 1/4

                 = ( x^5 + x^5) - ( 2x^2 - 4x^2) + ( 7x^4 - 5x^4) - ( 9x^3 + 2x^3) - 1/4x + 1/4

                 = 2x^5 - (-2)x^2 + 2x^4 - 11x^3 - 1/4x + 1/4

P(x)=x^5+  7x^4- 9x^3+ 2x^2-1/4x-0

Q(x)=(-x^5+5x^4- 2x^3+ 4x^2+0x-1/4

      =      12x^4-11x^3+ 6x^2-1/4x-1/4

P(x) = x^5 - 2x^2 + 7x^4 - 9x^3 - 1/4x

=x⁵+7x⁴-9x³-2x²-1/4x

Q(x) = 5x^4 - x^5 + 4x^2 - 2x^3 - 1/4         

=-x⁵+5x⁴-2x³+4x²-1/4

P(x)+Q(x)=x⁵+7x⁴-9x³-2x²-1/4x -x⁵+5x⁴-2x³+4x²-1/4

=x⁵-x⁵+7x⁴+5x⁴-9x³-2x³-2x²+4x²-1/4x-1/4

=12x⁴-11x³+2x²-1/4x-1/4

P(x)-Q(x)=x⁵+7x⁴-9x³-2x²-1/4x +x⁵-5x⁴+2x³-4x²+1/4

=x⁵+x⁵+7x⁴-5x⁴-9x³+2x³-2x²-4x²-1/4x-1/4

=2x⁵+2x⁴-7x³-6x²-1/4x-1/4

3: Trường hợp 1: x<-3

Pt sẽ là -x-2-x-3=x

=>-2x-5=x

=>-3x=5

hay x=-5/3(loại)

Trường hợp 2: -3<=x<-2

Pt sẽ là -x-2+x+3=x

=>x=1(loại)

TRường hợp 3: x>=-2

Pt sẽ là x+2+x+3=x

=>2x+5=x

hay x=-5(loại)