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a: =>4x-3x=1-2
=>x=-1
b: =>3x=12
=>x=4
c: =>2(x^2-6)=x(x+3)
=>2x^2-12-x^2-3x=0
=>x^2-3x-12=0
=>\(x=\dfrac{3\pm\sqrt{57}}{2}\)
a: =>4x-3x=1-2
=>x=-1
b: =>3x=12
=>x=4
c: =>2(x^2-6)=x(x+3)
=>2x^2-12=x^2+3x
=>x^2-3x-12=0
=>\(x=\dfrac{3\pm\sqrt{57}}{2}\)
Ta có: \(\dfrac{2x}{x^2-x+1}-\dfrac{x}{x^2+x+1}=\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{2x\left(x^2+x+1\right)-x\left(x^2-x+1\right)}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{2x^3+2x^2+2x-x^3+x^2-x}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{x^3+3x^2+x}{\left(x^2+1\right)^2-x^2}=\dfrac{5}{3}\)
\(\Leftrightarrow3x^3+9x^2+3x=5\left(x^4+2x^2+1-x^2\right)\)
\(\Leftrightarrow3x^3+9x^2+3x=5x^4+5x^2+5\)
\(\Leftrightarrow5x^4+5x^2+5-3x^3-9x^2-3x=0\)
\(\Leftrightarrow5x^4-3x^3-4x^2-3x+5=0\)
\(\Leftrightarrow5x^4-5x^3+2x^3-2x^2-2x^2+2x-5x+5=0\)
\(\Leftrightarrow5x^3\left(x-1\right)+2x^2\left(x-1\right)-2x\left(x-1\right)-5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x^3+2x^2-2x-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x^3-5x^2+7x^2-7x+5x-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[5x^2\left(x-1\right)+7x\left(x-1\right)+5\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(5x^2+7x+5\right)=0\)
mà \(5x^2+7x+5>0\forall x\)
nên x-1=0
hay x=1
Bài 1)1)\(x^2+5x+6=x^2+3x+2x+6\)=0
=x(x+3)+2(x+3)=(x+2)(x+3)=0
Dễ rồi
2)\(x^2-x-6=0=x^2-3x+2x-6=0\)
=x(x-3)+2(x-3)=0
=(x+2)(x-3)=0
Dễ rồi
3)Phương trình tương đương:\(\left(x^2+1\right)\left(x+2\right)^2=0\)
Vì \(x^2+1>0\)
=>\(\left(x+2\right)^2=0\)
Dễ rồi
4)Phương trình tương đương\(x^2\left(x+1\right)+\left(x+1\right)\)=0
=> \(\left(x^2+1\right)\left(x+1\right)=0Vì\) \(x^2+1>0\)
=>x+1=0
=>..................
5)\(x^2-7x+6=x^2-6x-x+6\) =0
=x(x-6)-(x-6)=0
=(x-1)(x-6)=0
=>.....
6)\(2x^2-3x-5=2x^2+2x-5x-5\)=0
=2x(x+1)-5(x+1)=0
=(2x-5)(x+1)=0
7)\(x^2-3x+4x-12\)=x(x-3)+4(x-3)=(x+4)(x-3)=0
Dễ rồi
Nghỉ đã hôm sau làm mệt
giải các Phương trình sau
a) (5x+3)(x2+1)(x-1)=0
b) (4x-1)(x-3)-(x-3)(5x+2)=0
c) (x+6)(3x-1)+x2-36 =0
a: =>(5x+3)(x-1)=0
=>x=1 hoặc x=-3/5
b: =>(x-3)(4x-1-5x-2)=0
=>(x-3)(-x-3)=0
=>x=-3 hoặc x=3
c: =>(x+6)(3x-1+x-6)=0
=>(x+6)(4x-7)=0
=>x=7/4 hoặc x=-6
\(a,x^2-10x=-25\)
\(< =>x^2-10x+25=0\)
\(< =>\left(x-5\right)^2=0< =>x=5\)
b, \(4x^2-4x=-1\)
\(< =>4x^2-4x+1=0\)
\(< =>\left(2x-1\right)^2=0< =>x=\frac{1}{2}\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(x+3\right)=192\)
\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x+1\right)=192\)
\(\Leftrightarrow\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3-192=0\)
\(\Leftrightarrow\left(x^2+2x\right)^2-2\left(x^2+2x\right)-195=0\)
\(\Leftrightarrow\left(x^2+2x-15\right)\left(x^2+2x+13\right)=0\)
=>(x+5)(x-3)=0
=>x=3 hoặc x=-5