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b) \(27x^3-54x^2+36x=8\)
\(\Rightarrow27x^3-54x^2+36x-8=0\)
\(\Rightarrow\left(3x\right)^3-3.\left(3x\right)^2.2+3.3x.2^2-2^3=0\)
\(\Rightarrow\left(3x-2\right)^3=0\)
\(\Rightarrow3x-2=0\)
\(\Rightarrow3x=2\)
\(\Rightarrow x=\dfrac{2}{3}\)
(2x-5)^2-(5+2x)^2=0
<=>(2x-5-5-2x)(2x-5+5+2x)=0
<=>(-10).(4x)=0
<=>(-40x)=0
<=>x =0
27x^3-54x^2+36x=8
<=>27x^3-54x^2+36x-8=0
<=>(3x-2)^3=0
<=>3x-2=0
<=>3x=2
<=>x=2/3
Bài 2:
a: \(=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\)
b: \(=2xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(2xy-1\right)\)
Bài 3:
=>x^2=5
hay \(x=\pm\sqrt{5}\)
phân tích đa thức sau thành nhân tử:
a) x2+2x-y2+1
=x\(^2\)+2x+1-y\(^2\)
=(x+1)\(^2\)-y\(^2\)
=(x+1-y)(x+1+y)
b) x2+3x-y2+3y
=x\(^2\)-y\(^2\)+3x+3y
=(x-y)(x+y)+3(x+y)
=(x+y)(x-y+3)
c) 3(x+3)-x2+9
=3(x+3)-(x\(^2\)-3\(^2\))
=3(x+3)-(x-3)(x+3)
=(x+3)[3-(x-3)]
=(x+3)(3-x+3)
a)
\(2x+3=(2x+3)^2\)
\(\Leftrightarrow (2x+3)^2-(2x+3)=0\)
\(\Leftrightarrow (2x+3)(2x+3-1)=0\)
\(\Leftrightarrow (2x+3)(2x+2)=0\Rightarrow \left[\begin{matrix} 2x+3=0\\ 2x+2=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-\frac{3}{2}\\ x=-1\end{matrix}\right.\)
b) \((x-5)^2=5-x\)
\(\Leftrightarrow (x-5)^2+(x-5)=0\)
\(\Leftrightarrow (x-5)(x-5+1)=0\)
\(\Leftrightarrow (x-5)(x-4)=0\)
\(\Rightarrow \left[\begin{matrix} x-5=0\\ x-4=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=5\\ x=4\end{matrix}\right.\)
c) \((x+2)^3=x+2\)
\(\Leftrightarrow (x+2)^3-(x+2)=0\)
\(\Leftrightarrow (x+2)[(x+2)^2-1]=0\)
\(\Leftrightarrow (x+2)(x+2-1)(x+2+1)=0\)
\(\Leftrightarrow (x+2)(x+1)(x+3)=0\)
\(\Rightarrow \left[\begin{matrix} x+2=0\\ x+1=0\\ x+3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-2\\ x=-1\\ x=-3\end{matrix}\right.\)
d)
\(|3x-1|=(1-3x)^2\)
\(\Leftrightarrow |3x-1|=|3x-1|^2\)
\(\Leftrightarrow |3x-1|^2-|3x-1|=0\)
\(\Leftrightarrow |3x-1|(|3x-1|-1)=0\)
\(\Rightarrow \left[\begin{matrix} |3x-1|=0\\ |3x-1|-1=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} |3x-1|=0\\ |3x-1|=1\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} 3x-1=0\\ 3x-1=\pm 1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{1}{3}\\ x=\frac{2}{3}\\ x=0\end{matrix}\right.\)
e)
\(2x+(x+3)(3-x)+(x+1)(x-1)=7\)
\(\Leftrightarrow 2x+(3^2-x^2)+(x^2-1^2)=7\)
\(\Leftrightarrow 2x=-1\Rightarrow x=-\frac{1}{2}\)
a>16-x/4=2x+1/3
<=>3[16-x)=4(2x+1)
<=>48-3x=8x+8
<=>-3x-8x=8-48
<=>-5x=-40
<=>x=8
a) \(\dfrac{x+5}{3\left(x-1\right)}+1=\dfrac{3x+7}{5\left(x-1\right)}\) ( đk: \(x\ne1\))
\(\Leftrightarrow\dfrac{5\left(x+5\right)}{15\left(x-1\right)}+\dfrac{15\left(x-1\right)}{15\left(x-1\right)}=\dfrac{3\left(3x+7\right)}{15\left(x-1\right)}\)
\(\Rightarrow5\left(x+5\right)+15\left(x-1\right)=3\left(3x+7\right)\)
\(\Leftrightarrow5x+25+15x-15=9x+21\)
\(\Leftrightarrow5x+15x-9x=21-25+15\)
\(\Leftrightarrow11x=11\Leftrightarrow x=1\) (loại)
Vậy tập nghiệm: \(S=\varnothing\)
b) \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}-\dfrac{8}{x^2+2x-3}=1\) (đk: \(x\ne1,x\ne-3\))
\(\Leftrightarrow\dfrac{\left(3x-1\right)\left(x+3\right)}{x^2+2x-3}-\dfrac{\left(2x+5\right)\left(x-1\right)}{x^2+2x-3}-\dfrac{8}{x^2+2x-3}=\dfrac{x^2+2x-3}{x^2+2x-3}\)
\(\Rightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)-8=x^2+2x-3\)
\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5-8=x^2+2x-3\)
\(\Leftrightarrow3x=3\Leftrightarrow x=1\) (loại)
Vậy tập nghiệm: \(S=\varnothing\)
a.
3x – 2 = 2x – 3
⇔ 3x – 2x = -3 + 2
⇔ x = -1.
Vậy phương trình có nghiệm x = -1.
b.
2(x-3)+5x(x-1)=5x^2
<=>2x-6+5x^2-5x=5x^2
<=>2x+5x^2-5x-5x^2=6
<=>-3x=6
<=>x=-2
Vậy nghiệm của pt là x=-2
\(\left(1-2x\right)^2=\left(3x-2\right)^2\)
\(=\left(1-2x\right)^2-\left(3x-2\right)^2=0\)
\(\left(1-2x-3x+2\right)\left(1-2x+3x-2\right)=0\)
\(\left(3-5x\right)\left(x-1\right)=0\)
\(\Rightarrow3-5x=0\) \(x-1=0\)
\(\Rightarrow x=\frac{3}{5}\) or \(x=1\)
b)\(\left(x-2\right)^3+\left(5-2x\right)^3\)
=\(\left(x-2+5-2x\right)\left(\left(x-2\right)^2-\left(x-2\right)\left(5-2x\right)+\left(5-2x\right)^2\right)\)
\(\left(3-x\right)\left(x^2-4x+4-5x+2x^2+10-4x+25-20x+4x^2\right)\)
(\(\left(3-x\right)\left(7x^2-33x+39\right)\)
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