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a)(2x-3)2=(x+5)2
=>4x2-12x+9=x2+10x+25
=>3x2-22x-16=0
=>3x2+2x-24x-16=0
=>x(3x+2)-8(3x+2)=0
=>(x-8)(3x+2)=0
=>x=8 hoặc x=-2/3
b)X2.(x-1)-4x2+8x-4=0
=>x2(x-1)-4x2+4x+4x-4=0
=>x2(x-1)-4x(x-1)-4(x-1)=0
=>x2(x-1)-(4x-4)(x-1)=0
=>(x2-4x+4)(x-1)=0
=>(x-2)2(x-1)=0
=>x=2 hoặc x=1
c) 4x2- 25 - (2x- 5) . ( 2x+7)=0
=>4x2-25-(4x2+14x-10x-35)=0
=>4x2-25-4x2-14x+10x+35=0
=>-4x+10=0
=>-4x=-10 <=>x=5/2
d) x3+27+(x+3).(x-9)=0
=>x3+33+(x+3)(x-9)=0
=>(x+3)(x2-3x+9)+(x+3)(x-9)=0
=>(x2-3x+9+x-9)(x+3)=0
=>(x2-2x)(x+3)=0
=>x(x-2)(x+3)=0
=>x=0 hoặc x=2 hoặc x=-3
e) (x-2).(x+5)- x2+4=0
=>(x-2)(x+5)-(x-2)(x+2)=0
=>(x-2)(x+5-x-2)=0
=>3(x-2)=0 <=>x=2
Sau khi khai triển hằng đẳng thức và thực hiện chuyển vế bạn sẽ đk kết quả như này!(\(\left(2x-3\right)^2=\left(x+5\right)^2=3x^2-22x-14\)
b) \(3x\left(x+5\right)-2x-10=0\)
\(\Leftrightarrow3x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-5\end{cases}}\)
c) \(x^3-9x=0\)
\(\Leftrightarrow x\left(x^2-9\right)=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)
TH1: \(x=0\)
TH2: \(x-3=0\Rightarrow x=3\)
\(x+3=0\Rightarrow x=-3\)
Vậy:..
d) \(\left(5+2x\right)\left(2x-7\right)=4x^2-25\)
\(\Leftrightarrow\left(5+2x\right)\left(2x-7\right)=\left(2x-5\right)\left(2x+5\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(2x-7-2x+5\right)=0\)
\(\Leftrightarrow-2\left(2x+5\right)=0\)
\(\Leftrightarrow2x+5=0\)
\(\Leftrightarrow x=-\frac{5}{2}\)
e) \(x^2-11x+30=0\)
\(\Leftrightarrow x^2-5x-6x+30=0\)
\(\Leftrightarrow x\left(x-5\right)-6\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=5\end{cases}}\)
\(x\left(2x-7\right)-4x+14=0\Leftrightarrow\left(x-2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{2}\end{matrix}\right.\)
\(x^2\left(x-1\right)-4\left(x-1\right)=\left(x^2-4\right)\left(x-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=1\end{matrix}\right.\)
\(x^4-x^3-x^2+x=x\left(x^3+1\right)-x^2\left(x+1\right)=x\left(x+1\right)\left(x^2-x+1-x^2\right)=x\left(x+1\right)\left(1-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
a) \(x\left(2x-7\right)-4x+14-0\Leftrightarrow2x^2-11x+14=0\Leftrightarrow2x^2-4x-7x+14=0\Leftrightarrow2x\left(x-2\right)-7\left(x-2\right)=0\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=2\end{matrix}\right.\)
b) \(x^2\left(x-1\right)-4x+4=0\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)
c) \(x+x^2-x^3-x^4=0\Leftrightarrow x\left(x^3+x^2-x-1\right)=0\Leftrightarrow x\left[x\left(x^2-1\right)+\left(x^2-1\right)\right]=0\Leftrightarrow x\left(x+1\right)\left(x^2-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
d) \(2x^3+3x^2+2x+3=0\Leftrightarrow x^2\left(2x+3\right)+2x+3=0\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\Leftrightarrow x=-1,5\left(x^2+1>0\forall x\right)\)
e) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\Leftrightarrow2x-5=0\Leftrightarrow x=2,5\)
g) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)
a ) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\)
\(\Leftrightarrow-2\left(2x-5\right)=0\)
\(\Leftrightarrow2x-5=0\Leftrightarrow x=\dfrac{5}{2}.\)
Vậy .........
b) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+4x\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-4\end{matrix}\right.\)
Vậy .........
c ) \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x^2=-1\left(loại\right)\end{matrix}\right.\)
Vậy .........
\(a,x^3=x\)
\(\Rightarrow x^3-x=0\)
\(\Rightarrow x\left(x^2-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-1=0\end{cases}}\)
\(\Rightarrow x^2=\left(-1\right)^2=1\)
\(KL:x=0;x=1\)
c) \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}\right)\left(\frac{2x^3+3x^2+2x+3}{x+\frac{3}{2}}\right)=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}\right)\left(2x^2+2\right)=0\) (bạn tự thực hiện phép chia đa thức giúp mình)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{3}{2}=0\\2x^2+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\2\left(x^2+1\right)=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x^2+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\left(C\right)\\x^2=-1\left(L\right)\end{cases}}\)
Vậy đa thức có nghiệm duy nhất \(x=-\frac{3}{2}\)
a) \(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+2x+7\right)+2\left(x-2\right)\left(x+2\right)-5\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left[x^2+2x+7+2\left(x+2\right)-5\right]=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+4x+6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x^2+4x+6=0\end{matrix}\right.\)
Ta có:
\(x^2+4x+6\)
\(=x^2+2.x.2+4+2\)
\(=\left(x+2\right)^2+2\)
Vì \(\left(x+2\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x+2\right)^2+2\ge2\) với mọi x
\(\Rightarrow x^2+4x+6\) vô nghiệm
\(\Rightarrow x-2=0\)
\(\Rightarrow x=2\)
b) \(3x\left(x-1\right)+\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(3x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
c) \(2\left(x+3\right)x^2-3x=0\)
\(\Rightarrow x\left[2\left(x+3\right)x-3\right]=0\)
\(\Rightarrow x\left(2x^2+6x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x^2+6x-3=0\end{matrix}\right.\)
Ta có:
\(2x^2+6x-3\)
\(=2\left(x^2+3x-\dfrac{3}{2}\right)\)
\(=2\left(x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{9}{4}-\dfrac{3}{2}\right)\)
\(=2\left(x+\dfrac{3}{2}\right)^2-\dfrac{15}{2}\)
Vì \(2\left(x+\dfrac{3}{2}\right)^2\ge0\) với mọi x
\(\Rightarrow2\left(x+\dfrac{3}{2}\right)^2-\dfrac{15}{2}\ge-\dfrac{15}{2}\) với mọi x
\(\Rightarrow2x^2+6x-3\) vô nghiệm
\(\Rightarrow x=0\)
a: \(\Leftrightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\)
=>(x-2)(x^2+4x+6)=0
=>x-2=0
=>x=2
b: =>(2x-5)(2x+5)-(2x-5)(2x+7)=0
=>(2x-5)(2x+5-2x-7)=0
=>2x-5=0
=>x=5/2
c: =>(x+3)(x^2-3x+9+x-9)=0
=>(x+3)(x^2-2x)=0
=>\(x\in\left\{0;2;-3\right\}\)
a) ( 5 - 2x )( 2x + 7 ) - 4x2 + 25 = 0
<=> ( 5 - 2x )( 2x + 7 ) + ( 5 - 2x )( 5 + 2x ) = 0
<=> ( 5 - 2x )( 2x + 7 + 5 + 2x ) = 0
<=> ( 5 - 2x )( 4x + 12 ) = 0
<=> \(\orbr{\begin{cases}5-2x=0\\4x+12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-3\end{cases}}\)
b) ( 5x2 + 3x - 2 )2 - ( 4x2 - x - 5 )2 = 0 ( như này chứ nhỉ ? )
<=> [ ( 5x2 + 3x - 2 ) - ( 4x2 - x - 5 ) ][ ( 5x2 + 3x - 2 ) + ( 4x2 - x - 5 ) ] = 0
<=> ( 5x2 + 3x - 2 - 4x2 + x + 5 )( 5x2 + 3x - 2 + 4x2 - x - 5 ) = 0
<=> ( x2 + 4x + 3 )( 9x2 + 2x - 7 ) = 0
<=> ( x2 + x + 3x + 3 )( 9x2 + 9x - 7x - 7 ) = 0
<=> [ x( x + 1 ) + 3( x + 1 ) ][ 9x( x + 1 ) - 7( x + 1 ) ] = 0
<=> ( x + 1 )( x + 3 )( x + 1 )( 9x - 7 ) = 0
<=> ( x + 1 )2( x + 3 )( 9x - 7 ) = 0
<=> x + 1 = 0 hoặc x + 3 = 0 hoặc 9x - 7 = 0
<=> x = -1 hoặc x = -3 hoặc x = 7/9
c) 15x4 - 8x3 - 14x2 - 8x + 15 = 0
<=> 15x4 + 22x3 - 30x3 + 15x2 + 15x2 - 44x2 - 30x + 22x + 15 = 0
<=> ( 15x4 + 22x3 + 15x2 ) - ( 30x3 + 44x2 + 30x ) + ( 15x2 + 22x + 15 ) = 0
<=> x2( 15x2 + 22x + 15 ) - 2x( 15x2 + 22x + 15 ) + ( 15x2 + 22x + 15 ) = 0
<=> ( 15x2 + 22x + 15 )( x2 - 2x + 1 ) = 0
<=> ( 15x2 + 22x + 15 )( x - 1 )2 = 0
<=> \(\orbr{\begin{cases}15x^2+22x+15=0\\\left(x-1\right)^2=0\end{cases}}\)
+) ( x - 1 )2 = 0 <=> x = 1
+) 15x2 + 22x + 15 = 15( x2 + 22/15x + 121/225 ) + 104/15 = 15( x + 11/25 )2 + 104/15 ≥ 104/15 > 0 ∀ x
Vậy phương trình có nghiệm duy nhất là x = 1
Ta có : 4x2 - 25 - (2x - 5)(2x + 7) = 0
<=> (2x)2 - 52 - (2x - 5)(2x + 7) = 0
=> (2x - 5)(2x + 5) - (2x - 5)(2x + 7) = 0
=> (2x - 5)(2x + 5 - 2x - 7) = 0
=> (2x - 5)(-2) = 0
=> 2x - 5 = 0
=> 2x = 5
=> x = 5/2
b) ta có: x^3 +27+(x+3)(x-9)=0
<=>x^3 +27 +x^2 -6x-27=0
<=>x^3 +x^2-6x=0
<=>(x^3 -2x^2) +(3.x^2 -6x)=0
<=>x^2(x-2)+3x(x-2)=0
<=>(x^2 +3x)(x-2)=0
<=>x(x+3)(x-2)=0=> x=0 hoặc x+3=0 hoặc x-2=0=>x=0 hoặc x=-3 hoặc x=2