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a, \(x^2-25-\left(x+5\right)=0\)
\(\Rightarrow x^2-5^2-\left(x+5\right)=0\)
\(\Rightarrow\left(x-5\right)\times\left(x+5\right)-\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\times\left(x-5-1\right)=0\)
\(\Rightarrow\left(x+5\right)\times\left(x-6\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+5=0\\x-6=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=0-5=\left(-5\right)\\x=0+6=6\end{cases}}\)
b, \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\Rightarrow\left(2x-1\right)^2-\left(\left(2x\right)^2-1^2\right)=0\)
\(\Rightarrow\left(2x-1\right)^2-\left(2x-1\right)\times\left(2x+1\right)=0\)
\(\Rightarrow\left(2x-1\right)\times\left(2x-1-\left(2x+1\right)\right)=0\)
\(\Rightarrow\left(2x-1\right)\times\left(2x-1-2x-1\right)=0\)
\(\Rightarrow\left(2x-1\right)\times\left(-2\right)=0\)\(\Rightarrow\left(-4x\right)+2=0\)
\(\Rightarrow\left(-4x\right)=0-2=-2\)
\(\Rightarrow x=\frac{-2}{-4}=\frac{1}{2}\)
c, \(x^2\times\left(x^2+4\right)-x^2-4=0\)
\(\Rightarrow x^2\times\left(x^2+4\right)-\left(x^2+4\right)=0\)
\(\Rightarrow\left(x^2-1\right)\times\left(x^2+4\right)=0\)
\(\Rightarrow\hept{\begin{cases}x^2-1=0\\x^2+4=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x^2=1\\x^2=\left(-4\right)\end{cases}}\)
\(\Rightarrow x=1\)
a) \(=x^2-2x+3-2x-x^2-2+4x=1\)
b)\(=6x+10x^2-6x+2x=10x^2+2x=2x\left(5x+1\right)\)
c)\(=3x^{n-2}.x^{n+2}-3x^{n-2}.y^{n+2}+y^{n+2}.3x^{n-2}-y^{n+2}.y^{n-2}\)
\(=3x^{2n}-y^{2n}\)
\(2x\left(x-3\right)-x+3=0\)
<=> \(2x\left(x-3\right)-\left(x-3\right)=0\)
<=> \(\left(x-3\right)\left(2x-1\right)=0\)
<=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}\)
Vậy...
x3 - 2x2 - 8x = 0
⇔ x( x2 - 2x - 8 ) = 0
⇔ x( x2 - 4x + 2x - 8 ) = 0
⇔ x[ x( x - 4 ) + 2( x - 4 ) ] = 0
⇔ x( x - 4 )( x + 2 ) = 0
⇔ x = 0 hoặc x - 4 = 0 hoặc x + 2 = 0
⇔ x = 0 hoặc x = 4 hoặc x = -2
x( x - 1 ) - x2 + 2x = 5
⇔ x2 - x - x2 + 2x = 5
⇔ x = 5
4x3 - 36x = 0
⇔ 4x( x2 - 9 ) = 0
⇔ 4x( x - 3 )( x + 3 ) = 0
⇔ 4x = 0 hoặc x - 3 = 0 hoặc x + 3 = 0
⇔ x = 0 hoặc x = 3 hoặc x = -3
2x2 - 2x = ( x - 1 )2
⇔ 2x( x - 1 ) - ( x - 1 )2 = 0
⇔ ( x - 1 )( 2x - x + 1 ) = 0
⇔ ( x - 1 )( x + 1 ) = 0
⇔ x - 1 = 0 hoặc x + 1 = 0
⇔ x = 1 hoặc x = -1
( x - 7 )( x2 - 9x + 20 )( x - 2 ) = 72
⇔ [ ( x - 7 )( x - 2 ) ]( x2 - 9x + 20 ) - 72 = 0
⇔ ( x2 - 9x + 14 )( x2 - 9x + 20 ) - 72 = 0
Đặt t = x2 - 9x + 17
⇔ ( t - 3 )( t + 3 ) - 72 = 0
⇔ t2 - 9 - 72 = 0
⇔ t2 - 81 = 0
⇔ ( t - 9 )( t + 9 ) = 0
⇔ ( x2 - 9x + 17 - 9 )( x2 - 9x + 17 + 9 ) = 0
⇔ ( x2 - 9x + 8 )( x2 - 9x + 26 ) = 0
⇔ ( x2 - 8x - x + 8 )( x2 - 9x + 26 ) = 0
⇔ [ x( x - 8 ) - ( x - 8 ) ]( x2 - 9x + 26 ) = 0
⇔ ( x - 8 )( x - 1 )( x2 - 9x + 26 ) = 0
⇔ x - 8 = 0 hoặc x - 1 = 0 hoặc x2 - 9x + 26 = 0
⇔ x = 8 hoặc x = 1 [ x2 - 9x + 26 = ( x2 - 9x + 81/4 ) + 23/4 = ( x - 9/2 )2 + 23/4 ≥ 23/4 > 0 ∀ x ]
\(x^3-2x^2-8x=x\left(x^2-2x-8\right)=x\left(x^2-4x+2x-8\right)=x\left[x\left(x-4\right)+2\left(x-4\right)\right]\)
\(=x\left(x+2\right)\left(x-4\right)\)
\(x\left(x-1\right)-x^2+2x=x^2-x-x^2+2x=x=5\)
\(4x^3-36x=4x\left(x^2-9\right)=4x\left(x-3\right)\left(x+3\right)\Leftrightarrow x=0\text{ hoặc }x=3\text{ hoặc }x=-3\)
\(2x^2-2x=x^2-2x+1\Leftrightarrow x^2=1\Leftrightarrow x=-1\text{ hoặc }1\)
\(\left(x-7\right)\left(x-4\right)\left(x-5\right)\left(x-2\right)=72\Leftrightarrow\left(x^2-9x+14\right)\left(x^2-9x+20\right)=72\)
đến đây đặt x^2-9x+14=a r giải như thường
Mình giải từ cuối lên , mình giải dần -)
n, <=> x(2x-1)-3(2x-1)=0
<=> (x-3)(2x-1)=0
<=> x= 3 hoặc x= 1/2
m, <=> (x+2)(x2-3x+5)-x2(x+2)=0
<=> (x+2)(x2-3x+5-x2)=0
<=> (x+2)(5-3x)=0
=> x= -2 hoặc5/3
1: Ta có: \(\left(3-x\right)^2+\left(2x+1\right)^2-\left(2-x\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-3+x-2\right)=0\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
2: Ta có: \(\left(1-2x\right)^2-3\left(x-1\right)^2+\left(x+1\right)^2-\left(x-1\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow4x^2-4x+1-3x^2+6x-3+\left(x+1\right)^2-2\left(x-1\right)^2=0\)
\(\Leftrightarrow x^2+2x-2+x^2+2x+1-2\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow2x^2+4x+1-2x^2+4x-2=0\)
\(\Leftrightarrow x=\dfrac{1}{8}\)