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a. 5-(x-6)=4(3-2x)
<=>5-x+6 = 12-8x
<=>-x+8x =-5-6+12
<=>7x=1
<=>x=\(\frac{1}{7}\)
Vậy phương trình có nghiệm là S= ( \(\frac{1}{7}\))
c.7 -(2x+4) =-(x+4)
<=> 7-2x-4=-x-4
<=>-2x+x= -7+4-4
<=> -x = -7
<=> x=7
Vậy phương trình có nghiệm là S=(7)
\(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)
\( < =>2\left[x\left(x^2+4x+4\right)-\left(2x\right)^2\right]=2\left(x^3-8\right)\)
\(< =>x^3+4x^2+4x-4x^2=x^3-8\)
\(< =>4x=-8< =>x=-2\)
Bài làm:
Ta có: \(B=2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)
\(\Leftrightarrow2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
\(\Leftrightarrow2x^3+8x^2+8x-8x^2=2x^3-16\)
\(\Leftrightarrow8x+16=0\)
\(\Leftrightarrow8x=-16\)
\(\Rightarrow x=-2\)
a.
\(=\left(x+1\right)\left(x+2\right)\left(x-2\right)\left(x-3\right)\)
b.
\(=\left(x+1\right)\left(x+1\right)\left(x^2+x+1\right)\)
c.
1) \(x^4-6x^3-x^2+54x-72=0\)
\(\Leftrightarrow x^3\left(x-2\right)-4x^2\left(x-2\right)-9x\left(x-2\right)+36\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-4x^2-9x+36\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-4\right)-9\left(x-4\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x-3\right)\left(x+3\right)=0\)
Tự làm nốt...
2) \(x^4-5x^2+4=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)-4\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
Tự làm nốt...
\(x^4-2x^3-6x^2+8x+8=0\)
\(\Leftrightarrow x^3\left(x-2\right)-6x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-6x-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+2\right)-2x\left(x+2\right)-2\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x^2-2x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left[\left(x-1\right)^2-\left(\sqrt{3}\right)^2\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1-\sqrt{3}\right)\left(x-1+\sqrt{3}\right)=0\)
...
\(2x^4-13x^3+20x^2-3x-2=0\)
\(\Leftrightarrow2x^3\left(x-2\right)-9x^2\left(x-2\right)+2x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^3-9x^2+2x+1\right)=0\)
Bí
\(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)
\(\Leftrightarrow2x\left(x^2+4x+4\right)-8x^2=2\left(x^2-8\right)\)
\(\Leftrightarrow2x^3+8x^2+8x=2x^3-16\)
\(\Leftrightarrow8x^2+8x+16=0\)
\(\Leftrightarrow7x^2+\left(x+4\right)^2=0\left(ktm\right)\)
Vậy tập nghiệm của phương trình là \(S=\varnothing\)
\(a,-x^3+x^2+4=0\)
\(-\left(x^3-x^2-4\right)=0\)
\(x^3-2x^2+x^2+2x-2x-4=0\)
\(x^2\left(x-2\right)+x\left(x+2\right)-2\left(x+2\right)=0\)
\(x^2\left(x-2\right)+\left(x+2\right)\left(x-2\right)=0\)
\(\left(x-2\right)\left(x^2+x+2\right)=0\)
Vì \(x^2+x+2>0\left(\forall x\right)\)
\(\Rightarrow x-2=0\)
\(\Rightarrow x=2\)
\(2x^2+2xy+y^2=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+x^2=0\)
\(\Leftrightarrow\left(x+y\right)^2+x^2=0\)
\(\Leftrightarrow x=y=0\)
a, ( 8x + 5 )( 4x + 3 )( 2x + 1 ) = 9
<=> ( 8x + 5 )[ 2( 4x+3)] [ 4 ( 2x+1 )] = 9* 2 * 4
<=> (8x+5)(8x+6)(8x+4) = 72
Đặt 8x+5 = y ta có phương trình tương đương :
y ( y -1 ) ( y+1) = 72
......................
b, Tương tự phần a nhé
c, x^3 + 5x^2 + 5x + 2=0
<=> x^3 + 1 + 5x^2 + 5x + 1 = 0
<=> (x+1)(x^2 - x +1) + 5x ( x+1 ) + 1 =0
<=> (x+1 ) ( x^2+4x + 1) + 1 = 0