Ta có: \(x^4+x^3+x+1=0\)

\(\Leftrightarrow\left(x^4+x^3\right)+\left(x+1\right)=0\)

\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2\left(x^2+x+1\right)=0\)

Mà \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\left(\forall x\right)\)

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

a) Nối A với C

Xét tam giác ABC có : AB< BC+AC (qh giữa các cạnh trong tam giác)(1)

Xét tam giác ADC có: AC<AD+DC( ---------------------------------------)(2)

Cộng vế 1 và 2 vào ta sẽ có:

AB+AC< BC+AC+AD+CD=> AB+BC< CD +AD

b) Xét tam giác ABC , ta có: AC< AB+BC

Xét tam giác ADC , ta có: AC< AD+DC

=> 2AC< a+b+c+d nên AC<( AB+BC+CD+AD):2 (1)

tương tự như vậy BD<(AB+BC+CD+AD):2 (2)

Từ 1 và 2 suy ra AC+BD<AB+BC+DC+AD

a) Ta có: \(x^4y^4+x^2y^2+1\)

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

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

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

c) \(4x^4+1\)

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

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

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

a) x² - y² - 4x + 4y

= (x² - 4x + 4) - (y² - 4y + 4)

= (x - 2)² - (y - 2)²

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

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

b) (xy + 4)² - 4(x + y)²

= (xy + 4)² - [2(x + y)]²

= (xy + 4)² - (2x + 2y)²

= (xy + 4 - 2x - 2y)(xy + 4 + 2x + 2y)

c) 25 - x² + 2xy - y²

= 25 - (x² - 2xy + y²)

= 5² - (x - y)²

=> (5 - x + y)(5 + x - y)

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

\(=\left(x-y\right)\left(x+y-4\right)\)

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

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

Ta có: \(\left(a^2+b^2\right)\left(c^2+d^2\right)\)

\(=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)

\(=\left(a^2c^2+2abcd+b^2d^2\right)+\left(a^2d^2-2abcd+b^2c^2\right)\)

\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)

=> đpcm

a) \(3\left(x-3\right)-5\left(-x+1\right)=x+6\)

\(\Leftrightarrow3x-9+5x-5-x-6=0\)

\(\Leftrightarrow7x=20\)

\(\Rightarrow x=\frac{20}{7}\)

b) \(\left|4x-2\right|=8\Leftrightarrow\orbr{\begin{cases}4x-2=8\\4x-2=-8\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=10\\4x=-6\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)

c) \(-3\left|6x+1\right|=-12\)

\(\Leftrightarrow\left|6x+1\right|=4\Leftrightarrow\orbr{\begin{cases}6x+1=4\\6x+1=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}6x=3\\6x=-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{5}{6}\end{cases}}\)

                                                   Bài giải

a, \(3\left(x-3\right)-5\left(-x+1\right)=x+6\)

\(3x-9+5x-5-x-6=0\)

\(7x-20=0\)

\(7x=20\)

\(x=\frac{20}{7}\)

b, \(\left|4x-2\right|=8\)

\(4x-2=\pm8\)

\(\Rightarrow\orbr{\begin{cases}4x-2=-8\\4x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}4x=-6\\4x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)

Vậy \(x\in\left\{-3\text{ ; }2\right\}\)

c, \(-3\left|6x+1\right|=-12\)

\(\left|6x+1\right|=4\)

\(6x+1=\pm4\)

\(\Rightarrow\orbr{\begin{cases}6x+1=-4\\6x+1=4\end{cases}}\Rightarrow\orbr{\begin{cases}6x=-5\\6x=3\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{5}{6}\\x=\frac{1}{2}\end{cases}}\)

\(\Rightarrow\text{ }x\in\left\{-\frac{5}{6}\text{ ; }\frac{1}{2}\right\}\)

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

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

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

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

\(=4x^2+4x+1-\left(x^2+4x+4\right)-3x^2-6x\) 

\(=4x^2+4x+1-x^2-4x-4-3x^2-6x\) 

\(=-6x-3\) 

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

Bài làm :

\(\left(9x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\)

\(=9x^3-45x^2+9x-2x^2+10x-2-x^3-11x\)

\(=\left(9x^3-x^3\right)+\left(-45x^2-2x^2\right)+\left(9x+10x-11x\right)-2\)

\(=8x^3-47x^2+8x-2\)

Học tốt nhé

( 9x - 2 )( x² - 5x + 1 ) - x( x² + 11 )

= 9x³ - 45x² + 9x - 2x² + 10x - 2 - x³ - 11x

= 8x³ - 47x² + 8x - 2