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a) x vô nghiệm
b)<=>(x2-3x+3)(x2-2x+3)-2x2=(x-3)(x-1)(x2-x+3)
=>(x-3)(x-1)(x2-x+3)=0
TH1:x-3=0
=>X=3
TH2:x-1=0
=>x=1
TH3:x2-x+3=0
<=>(-1)2-4(1.3)=-11
vì -11<0
=>x=1 hoặc 3
bạn tự tiếp làm đi dễ mà
\(4x^2-25+\left(2x+7\right)\left(5-2x\right)\)
\(=\left(2x-5\right)\left(2x+5\right)+\left(2x+7\right)\left(5-2x\right)\)
\(=\left(2x-5\right)\left(2x+5\right)-\left(2x-7\right)\left(2x-5\right)\)
\(=\left(2x-5\right)\left(2x+5-2x+7\right)\)
\(=\left(2x-5\right).12\)
Những câu khác làm tương tự
Bài 2: a) \(3x^3-3x=0\Leftrightarrow3x\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
b) \(x^2-x+\frac{1}{4}=0\Leftrightarrow x^2-2.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
a/ \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\)
<=> \(x^3-3x^2+3x-1+\left(2-x\right)\left(x+2\right)^2+3x^2+6x=17\)
<=> \(x^3+9x-1+2\left(x+2\right)^2-x\left(x+2\right)^2=17\)
<=> \(x^3+9x-1+2\left(x^2+2x+1\right)-x\left(x^2+2x+1\right)=17\)
<=> \(x^3+9x-1+2x^2+4x+2-x^3-2x^2-x=17\)
<=> \(12x+1=17\)
<=> \(12x=16\)
<=> \(x=\frac{4}{3}\)
b/ \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)
<=> \(\left(x+2\right)\left(x-2\right)^2-x\left(x^2-2\right)=15\)
<=> \(x\left(x-2\right)^2-x\left(x^2-2\right)+2\left(x-2\right)^2=15\)
<=> \(x\left(x^2-2x+1\right)-x\left(x^2-2\right)+2\left(x-2\right)^2=15\)
<=> \(x\left[x^2-2x+1-\left(x^2-2\right)\right]+2\left(x-2\right)^2=15\)
<=> \(x\left(x^2-2x+1-x^2+2\right)+2\left(x-2\right)^2=15\)
<=> \(x\left(3-2x\right)+2\left(x^2-2x+1\right)=15\)
<=> \(3x-2x^2+2x^2-4x+2=15\)
<=> \(2-x=15\)
<=> \(x=-13\)
2)
a) \(3x^3-3x=0\)
\(\Leftrightarrow3x\left(x^2-1\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy x=0 ; x=-1 ; x=1
b) \(x^2-x+\dfrac{1}{4}=0\)
\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
1)
a) \(\left(x-2\right)\left(x^2+3x+4\right)\)
\(\Leftrightarrow x^3+3x^2+4x-2x^2-6x-8\)
\(\Leftrightarrow x^3+x^2-2x-8\)
b) \(\left(x-2\right)\left(x-x^2+4\right)\)
\(=x^2-x^3+4x-2x+2x^2-8\)
\(=3x^2-x^3+2x-8\)
c) \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
d) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=18x^2+12x-9x-6-6x^3-4x^2+3x^2+2x\)
\(=17x^2+5x-6-6x^3\)
a) \(3x^2-3y^2-2\left(x-y\right)^2=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=\left(x-y\right)\left[3\left(x+y\right)-2\left(x-y\right)\right]\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
b) \(x^3-4x^2-9x+36\)
\(=x^2\left(x-4\right)-9\left(x-4\right)\)
\(=\left(x-4\right)\left(x^2-9\right)=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)
c) \(x^2-y^2-2x-2y=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)=\left(x+y\right)\left(x-y-2\right)\)
d) \(\left(x-3\right)\left(x-1\right)-3=x^2-4x+3-3=x\left(x-4\right)\)
a) (x-3)(x+3)-(x-1)^2=0
=> (x^2-9)-(x^2-2x+1)=0
=>x^2-9-x^2+2x-1=0
=>(x^2-x^2)-9-1+2x=0
=>-10+2x=0
=>-2.(-5-x)=0
=>-5-x=0
=>-x=0+5
=>x=-5
vậy x=-5
b) x^3-3x^2+3x-1=0
=>(x-1)^3=0
=>x-1=0
=>x=0+1
=>x=1
vậy x=1
c) 4x^2-28x=0
=>4x.(x-7)=0
=> 2 TH
* 4x=0=>x=0
*x-7=0=>x=0+7=>x=7
vậy x=0 hoặc x=7
`(x-2)^3 - (x-3).(x^2+3x+9)+6.(x+1)^2 = 15`
`x^3 -6x^2 +12x-8 - x^3-27 + 6x^2 +12x+6=15`
`24x-29 = 15`
`24x=15+29`
`24x=44`
`x=44:24`
`x=11/6`