\(3\left(x-1\right)^2-3x\left(2-5\right)=21\)

\(\Leftrightarrow3x^2-6x+3+9x-21=0\)

\(\Leftrightarrow3x^2+3x-18=0\)

\(\Leftrightarrow3\left(x^2+x-6\right)=0\)

\(\Leftrightarrow3\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

Vậy \(S=\left\{2;-3\right\}\)

mik cần gấp nhé😱😱😱

điiii mừ , giúp mik đyyy

\(\left(1+3^2\right)\left(x^2+y^2\right)\ge\left(x+3y\right)^2=10^2\)

\(\Leftrightarrow x^2+y^2\ge10\)

Dấu \(=\)khi \(\hept{\begin{cases}x+3y=10\\\frac{x}{1}=\frac{y}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}\).

Áp dụng bđt Cauchy-Schwarz dạng Engel : \(x^2+y^2=\frac{x^2}{1}+\frac{9y^2}{9}\ge\frac{\left(x+9y\right)^2}{1+9}=\frac{10^2}{10}=10\)

Dấu "=" xảy ra \(\Leftrightarrow\frac{x}{1}=\frac{3y}{9}=\frac{x+3y}{1+9}=\frac{10}{10}=1\Rightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}\)

a) \(3x^2-6x=x^2-4x+4\)

\(\Leftrightarrow2x^2-2x-4=0\)

\(\Leftrightarrow2\left(x^2-x-2\right)=0\)

\(\Leftrightarrow2\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

Vậy \(S=\left\{2;-1\right\}\)

b) \(x^3-7x^2+6x=0\)

\(\Leftrightarrow x\left(x^2-7x+6\right)=0\)

\(\Leftrightarrow x\left(x-6\right)\left(x-1\right)=0\)

\(\Leftrightarrow x=0;x-6=0;x-1=0\)

\(\Leftrightarrow x=0;x=6;x=1\)

Vậy \(S=\left\{0;6;1\right\}\)

c) \(x^4+4x^3+4x^2=25\)

\(\Leftrightarrow x^2\left(x^2+4x+4\right)=25\)

\(\Leftrightarrow x^2\left(x+2\right)^2-25=0\)

\(\Leftrightarrow\left[x\left(x+2\right)\right]^2-5^2=0\)

\(\Leftrightarrow\left(x^2+2x-5\right)\left(x^2+2x+5\right)=0\)

\(\Leftrightarrow\left(x+1-\sqrt{6}\right)\left(x+1+\sqrt{6}\right)=0\) (vì \(x^2+2x+5>0\) )

\(\Leftrightarrow\orbr{\begin{cases}x+1-\sqrt{6}=0\\x+1+\sqrt{6}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{6}-1\\x=-\sqrt{6}-1\end{cases}}\)

Vậy \(S=\left\{\sqrt{6}-1;-\sqrt{6}-1\right\}\)

a,\(3x^2-6x=\left(x^2-4x+4\right)\)
\(3x^2-6x-x^2+4x-4=0\)

\(3x^2-x^2-6x+4x-4=0\)
\(2x^2-2x-4=0\)
\(2x^2+2x-4x-4=0\)
\(2x\left(x+1\right)-4\left(x+1\right)=0\)
\(\left(2x-4\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
b, \(x^3-7x^2+6x=0\)
\(x^3-x^2-6x^2+6x=0\)
\(x^2\left(x-1\right)-6x\left(x-1\right)=0\)
\(\left(x-1\right)\left(x^2-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\x=\pm6\end{cases}}\)
 

Để x⁴ + ax² + b chia hết cho x² + x + 1 thì x⁴ + ax² + b khi phân tích phải có nhân tử là x² + x + 1

Sau khi phân tích thì x⁴ + ax² + b có dạng ( x² + x + 1 )( x² + cx + d )

=> x⁴ + ax² + b = ( x² + x + 1 )( x² + cx + d )

<=> x⁴ + ax² + b = x⁴ + cx³ + dx² + x³ + cx² + dx + x² + cx + d

<=> x⁴ + ax² + b = x⁴ + ( c + 1 )x³ + ( c + d + 1 )x² + ( c + d )x + d

Đồng nhất hệ số ta có : \(\hept{\begin{cases}c+1=0\\c+d+1=a\\c+d=0\end{cases}};d=b\Rightarrow\hept{\begin{cases}a=b=d=1\\c=-1\end{cases}}\)

Vậy a = b = 1

x^4+ax^2+1
= x^4+2x^2+1+ax^2-2x^2
=(x^2+1)^2-x^2+x^2(a-1)
= (x^2+x+1)(x^2-x+1)+x^2(a-1)
= (x^2+x+1)(x^2-x+1)+(a-1)(x^2+x+1) -(a-1)(x-1). 
để x^4+ax^2+1 chia hết cho x^2+x+1 
thì số dư =0 
<=> (a-1)(x-1) =0 
<=> a=1

Mình làm bừa thôi, sai thông cảm :>

a) 16 + 2x³y³

= 2( 8 + x³y³)

= 2[ 2³ + (xy)³]

= 2 (2+xy)(4 - 2xy + x²y²)

b) 100a² - (a² + 25)²

= (10a)² - (a² +25)²

= (10a - a² - 25)(10a + a² +25)

= -(a² - 2a.5 + 5²)(a² + 2a.5 + 5²)

=-(a-5)² (a+5)²

a) \(16+2x^3y^3\)

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

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

b) \(100a^2-\left(a^2+25\right)^2\)

\(=\left(10a-a^2-25\right)\left(10a+a^2+25\right)\)

\(=-\left(a-5\right)^2\left(a+5\right)^2\)

giúp e giải 3 phần cuối bài 1 và bài 3 nhé e đang cần gấp mong mn giúp

Bài 1 : 

d, \(3x^2+6xy-48z^2+3y^2=3\left[\left(x^2+2xy+y^2\right)-16z^2\right]\)

\(=3\left[\left(x+y\right)^2-\left(4z\right)^2\right]=3\left(x+y-4z\right)\left(x+y+4z\right)\)

e, \(x^2-z^2+4y^2-4t^2-4xy+4zt=x^2-4xy+4y^2-\left(z^2-4zt+4t^2\right)\)

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

f, \(x^3+2x^2y+xy^2-16x=x\left(x^2+2xy+y^2\right)-16x\)

\(=x\left[\left(x+y\right)^2-16\right]=x\left(x+y-4\right)\left(x+y+4\right)\)