a) 2x²-4x-x+2=0

=> 2x(x-2)-(x-2)=0

=> (2x-1)(x-2)=0

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

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

b) 3x²-12x+5x-20=0

=> 3x(x-4)+5.(x-4)=0

=> (x-4)(3x+5)=0

=> \(\left[{}\begin{matrix}x-4=0\\3x+5=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=4\\x=-\dfrac{5}{3}\end{matrix}\right.\)

c)x³+2x²-x²-2x+2x+4=0

=> x²(x+2)-x(x+2)+2(x+2)=0

=>(x²-x+2)(x+2)=0

=> x=-2( vi x²-x+2>0)

d) x³-x²-4x²+4x+4x-4=0

=> x²(x-1)-4x(x-1)+4(x-1)=0

=>(x-1)(x²-4x+4)=0

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

=> \(\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

2x²-5x+2=0

⇔2x²-x-4x+2=0

⇔x(2x-1)-2(2x-1)=0

⇔(x-2)(2x-1)=0

⇔\(\left[{}\begin{matrix}x-2=0\\2x-1=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=2\\2x=1\Leftrightarrow x=\dfrac{1}{2}\end{matrix}\right.\)

sậy S=\(\left\{2;\dfrac{1}{2}\right\}\)

x³+x²+4=0

⇔x³+2x²-x²-2x+2x+4=0

⇔(x³+2x²)-(x²+2x)+(2x+4)=0

⇔x²(x+2)-x(x+2)+2(x+2)=0

⇔(x+2)(x²-x+2)=0

⇔x+2=0 và x²-x+2=0

⇔x=-2 và \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\)(vô lý)

vậy S={-2}

Phân tích đa thức thành nhân tử , ta đươc :

\(x^4+2x^3+5x^2+4x-12=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x_1=-2\\x_2=1\end{array}\right.;x^2+x+6=\left(x+\frac{1}{2}\right)^2+5\frac{3}{4}\ne0\forall x.\)

Vậy pt đã cho các nghiệm : \(x_1=-2;x_2=1.\)

Lời giải:

ĐK: $x\geq 0$

Ta có: \(x^2-5x-2\sqrt{3x}+12=0\)

\(\Leftrightarrow (x^2-6x+9)+(x-2\sqrt{3x}+3)=0\)

\(\Leftrightarrow (x-3)^2+(\sqrt{x}-\sqrt{3})^2=0\)

\(\Leftrightarrow (\sqrt{x}-\sqrt{3})^2(\sqrt{x}+\sqrt{3})^2+(\sqrt{x}-\sqrt{3})^2=0\)

\(\Leftrightarrow (\sqrt{x}-\sqrt{3})^2[(\sqrt{x}+\sqrt{3})^2+1]=0\)

Vì \((\sqrt{x}+\sqrt{3})^2+1\neq 0\Rightarrow (\sqrt{x}-\sqrt{3})^2=0\Rightarrow x=3\) (thỏa mãn)

Vậy..........

ĐKXĐ \(x\ge0\)

\(x^2-5x-2\sqrt{3x}+12=0\)

\(\Rightarrow x^2-6x+9+x-2\sqrt{3x}+3=0\)

\(\Rightarrow\left(x-3\right)^2+\left(\sqrt{x}-\sqrt{3}\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-3\right)^2=0\\\left(\sqrt{x}-\sqrt{3}\right)^2=0\end{cases}}\Leftrightarrow x=3\)

Vậy...

2x³-2x²-3x²+3x=0

<=>2x(x-1)-3x(x-1)=0

<=>(x-1)(2x-3x)=0

<=>-x(x-1)=0

Th₁:-x=0

<=>x=0

Th₂:x-1=0

<=>x=1

Vậy phương trình có tập n_o là S=(0, 1)

\(2x^3-5x^2+3x=0\)

\(\Leftrightarrow2x^3-2x^2-3x^2+3x=0\)

\(\Leftrightarrow2x\left(x-1\right)-3x\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x-3x\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\\x=1\end{matrix}\right.\)

2x3-2x2-3x2+3x=0

<=>2x(x-1)-3x(x-1)=0

<=>(x-1)(2x-3x)=0

<=>-x(x-1)=0

\(\left\{{}\begin{matrix}-x=0\\x-1=0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy phương trình có tập no là S=(0, 1)

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

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

<=>\(\left(x^2+5x+4\right)\left(3x^2-5x\right)=0\)

<=>\(\left(x+1\right)\left(x+4\right)x\left(3x-5\right)=0\)

<=>x+1=0 hoặc x+4=0 hoặc x=0 hoặc 3x-5=0

<=>x=-1 hoặc x=-4 hoặc x=0 hoặc x=5/3

bài này dùng hằng đẳng thức a²-b²= (a-b)(a+b)

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

\(\left(x^2+5x+4\right)\left(3x^2-5x\right)=0\)

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