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Ta có:
$10x^2-x-11=0$
$\Delta=(-1)^2-4\cdot10\cdot(-11)=441=21^2$
$x=\dfrac{1\pm21}{20}$
$x=1$ hoặc $x=-\dfrac{11}{10}$
Vậy $x\in\left{1,-\dfrac{11}{10}\right}$.
b)$2x^2-3x-2=0$
$\Delta=(-3)^2-4\cdot2\cdot(-2)=25=5^2$
$x=\dfrac{3\pm5}{4}$
$x=2$ hoặc $x=-\dfrac12$
Vậy $x\in\left{2,-\dfrac12\right}$.
$2x^2-8=0$
$2(x^2-4)=0$
$(x-2)(x+2)=0$
$x=2$ hoặc $x=-2$
Vậy $x=\pm2$.
d)$3x^2-5x=0$
$x(3x-5)=0$
$x=0$ hoặc $x=\dfrac53$
Vậy $x\in\left{0,\dfrac53\right}$.
Câu 1:
\(a^2+2ab+b^2-ac-bc\)
\(=\left(a+b\right)^2-c\left(a+b\right)\)
\(=\left(a+b\right)\left(a+b-c\right)\)
Câu 2:
\(5x^2-5y^2-10x+10y\)
\(=5\left(x-y\right)\left(x+y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)\left(5x+5y-10\right)\)
\(=5\left(x-y\right)\left(x+y-2\right)\)
Câu 3:
\(3x^2-6xy+3y^2-12z^2\)
\(=3\left[\left(x-y\right)^2-4z^2\right]\)
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
Câu 4:
\(x^4+x^3+x^2-1\)
\(=x^3\left(x+1\right)+\left(x-1\right)\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+x-1\right)\)
Câu 5:
\(x^3-3x^2+3x-1-y^3\)
\(=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right)y+y^2\right]\)
\(=\left(x-y-1\right)\left(x^2-2x+1+xy-y+y^2\right)\)
Câu 6:
\(x^4-x^2+2x-1\)
\(=x^4-\left(x-1\right)^2\)
\(=\left(x^2-x+1\right)\left(x^2+x-1\right)\)
Câu 7:
\(\left(x+y\right)^3-x^3-y^3\)
\(=\left(x+y\right)^3-\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]\)
\(=3xy\left(x+y\right)\)
h)\(=x^2+2x-3x-6=x\left(x+2\right)-3\left(x+2\right)=\left(x-3\right)\left(x+2\right)\)
\(e=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left[\left(x+y\right)-1\right]\)
k=\(x^4+5x^2-x^2-5=x^2\left(x^2+5\right)-\left(x^2+5\right)=\left(x^2-1\right)\left(x^2+5\right)\)
\(x^2-4+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2+x-2\right)\)
\(=2x\left(x-2\right)\)
\(x+\sqrt{3}=2\Rightarrow\sqrt{3}=2-x\Rightarrow3=\left(2-x\right)^2\Rightarrow x^2-4x+1=0\)
Ta có:
\(B=x^5-4x^4+x^4-4x^3+x^3+5x^2+x^2-20x+5+2013\)
\(\Rightarrow B=x^5-4x^4+x^3+x^4-4x^3+x^2+5x^2-20x+5+2013\)
\(\Rightarrow B=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2013\)
\(\Rightarrow B=x^3.0+x^2.0+5.0+2013=2013\)