8x²+4xy-2ax-2y

= 2( x²+2xy-ax-y)

2xy-x²-y²+16

= - (x²-2xy+y²-16)

= - [(x-y)² - 16 ]

=-(x-y-4)(x-y+4)

3a²-6ab+3b²-12c²

=3(a²-2ab+b²-4c²)

=3[(a-b)²-4c²]

=3(a-b-2c)(a-b+2c)

(nếu đề bài là phân tích đa thức thành nhân tử bn nhé)

chúc bn hc tốt

x= 3.x+x

x3.x2=x1.x =x3

x=3++.x3

x=6.3xx=4

a x=5

b m=4.5.

x=4.5-.5.4 +6+

m se co gia tri lon nhat la.4.5.6-7+8

tu di ma tinh tui giai cho roi day neu muon day them goi 0637995421

\(a,\)\(M=\frac{3x+3}{x^3+x^2+x+1}=\frac{3\left(x+1\right)}{x^2\left(x+1\right)+\left(x+1\right)}\)

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

\(b,M\in Z\Leftrightarrow\frac{3}{x^2+1}\in Z\)

\(\Rightarrow3\)\(⋮\)\(x^2+1\)\(\Rightarrow x^2+1\inƯ_3\)

Ta có \(Ư_3=\left\{\pm1;\pm3\right\}\)

Mà \(x^2+1\ge1\)với mọi x 

\(\Rightarrow\orbr{\begin{cases}x^2+1=1\\x^2+1=3\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{2}\end{cases}}}\)

\(c,\)\(M_{max}\Leftrightarrow x^2+1\)nhỏ nhất \(\Rightarrow x^2\)nhỏ nhất \(\Rightarrow x=0\)

\(\Rightarrow M_{max}=3\Leftrightarrow x=0\)

\(3x-7-\left(3-4x\right)\left(2x+1\right)+4x\left(2x-7\right)\)

\(\Rightarrow3x-7-6x-3-8x^2-4x+8x^2-28x=0\)

\(\Rightarrow-35x-10=0\)

\(\Rightarrow-5\cdot\left(7x+2\right)=0\)

Vì -5 < 0

\(\Rightarrow7x+2=0\)

\(\Rightarrow7x=-2\)

\(\Rightarrow x=-\frac{2}{7}\)

\(10x^2-5x\cdot\left(2x+3\right)=15\)

\(\Rightarrow10x^2-10x^2-15x-15=0\)

\(\Rightarrow-15x-15=0\)

\(\Rightarrow-15\cdot\left(x+1\right)=0\)

Vì -15< 0

\(\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

2x³ + 3x² + 2x + 3 = 0 

<=> 2x(x² + 1) + 3(x² + 1) = 0

<=> (x² + 1)(2x + 3) = 0

Vì x² ≥ 0 => x² + 1 ≥ 1

=> 2x + 3 = 0

=> 2x = -3

=> x = -3/2

\(x^2-5x+6\)

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

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

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

\(x^2-5x+6\)

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

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

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

Đặt \(a+\frac{1}{36a}=x\)

pt đã cho trở thành \(9x^2-6x+1=0\)

\(\Leftrightarrow9\left(x^2-\frac{2}{3}x+\frac{1}{9}\right)=0\)

\(\Leftrightarrow9\left(x-\frac{1}{3}\right)^2=0\)

\(\Leftrightarrow x-\frac{1}{3}=0\)

\(\Leftrightarrow x=\frac{1}{3}=a+\frac{1}{36a}=\frac{36a^2+1}{36a}\)

\(\Leftrightarrow12a=36a^2+1\)

\(\Leftrightarrow36a^2-12a+1=0\)

\(\Leftrightarrow\left(6a-1\right)^2=0\)

\(\Leftrightarrow6a-1=0\)

\(\Leftrightarrow a=\frac{1}{6}\Rightarrow a=6\)

Chúc bạn học tốt !!!