giup mik voi!!
a) 3^x+x^2=1
b) (3x-14)^n=1
c) (5^2x.5^x+1):5=125
3^x+3^x+1+3^x+2=243.39
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`a,x(x-1)-(x+2)^2=1`
`<=>x^2-x-x^2-4x-4=1`
`<=>-5x=5`
`<=>x=-1`
`b,(x+5)(x-3)-(x-2)^2=-1`
`<=>x^2+2x-15-x^2+4x-4+1=0`
`<=>6x-18=0`
`<=>x-3=0`
`<=>x=3`
`c,x(2x-4)-(x-2)(2x+3)=0`
`<=>2x(x-2)-(x-2)(2x+3)=0`
`<=>(x-2)(2x-2x-3)=0`
`<=>-3(x-2)=0`
`<=>x-2=0`
`<=>x=2`
`d,x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12`
`<=>3x^2+2x+x^2+2x+1-4x^2+25=-12`
`<=>4x+26=-12`
`<=>4x=-38`
`<=>x=-19/2`
\(a)\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1.\left(x\ne2;4\right).\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1.\\ \Rightarrow x^2-4x-3x+12+x^2-4x+4+x^2-4x-2x+8=0.\\ \Leftrightarrow3x^2-17x+24=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}.\\x=3.\end{matrix}\right.\) (TM).
\(b)3x+12=0.\\ \Leftrightarrow3x=-12.\\ \Leftrightarrow x=-4.\)
\(c)5+2x=x-5.\\ \Leftrightarrow2x-x=-5-5.\\ \Leftrightarrow x=-10.\)
\(d)2x\left(x-2\right)+5\left(x-2\right)=0.\\ \Leftrightarrow\left(2x+5\right)\left(x-2\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{2}.\\x=2.\end{matrix}\right.\)
\(e)\dfrac{3x-4}{2}=\dfrac{4x+1}{3}.\\ \Rightarrow3\left(3x-4\right)-2\left(4x+1\right)=0.\\ \Leftrightarrow9x-12-8x-2=0.\\ \Leftrightarrow x=14.\)
\(f)\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1.\left(x\ne\pm1\right).\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x}{x^2-1}=1.\\ \Leftrightarrow x^2+3x-x^2+1=0.\\ \Leftrightarrow3x+1=0.\\ \Leftrightarrow x=\dfrac{-1}{3}.\)
\(g)\dfrac{2x}{x-1}+\dfrac{3-2x}{x+2}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}.\left(x\ne1;-2\right).\\ \Leftrightarrow\dfrac{2x^2+4x+\left(3-2x\right)\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}.\\ \Rightarrow2x^2+4x+3x-3-2x^2+2x-6=0.\\ \Leftrightarrow9x=9.\)
\(\Leftrightarrow x=1\left(koTM\right).\)
\(\left(x+1\right)^2-3\left(x+1\right)=\left(x+1\right)\left(x+1-3\right)=\left(x+1\right)\left(x-2\right)\)
\(2x\left(x-2\right)-\left(x-2\right)^2=\left(x-2\right)\left[2x-\left(x-2\right)\right]=\left(x-2\right)\left(2x-x+2\right)=\left(x-2\right)\left(x+2\right)\)
\(4x^2-20xy+25y^2=\left(2x\right)^2-2.2x.5y+\left(5y\right)^2=\left(2x-5y\right)^2\)
\(x^2+3x-x-3=x\left(x+3\right)-\left(x+3\right)=\left(x-1\right)\left(x+3\right)\)
\(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)
\(2y\left(x+2\right)-3x-6=2y\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(2y-3\right)\)
a: \(\left(x+\dfrac{1}{4}\right)+\left(3x-4\right)+2\left(x-3\right)=1\)
=>\(x+\dfrac{1}{4}+3x-4+2x-6=1\)
=>\(6x-\dfrac{39}{4}=1\)
=>\(6x=1+\dfrac{39}{4}=\dfrac{43}{4}\)
=>\(x=\dfrac{43}{4}:6=\dfrac{43}{24}\)
b: \(2\left(x-3\right)=3\left(x+2\right)-x+1\)
=>\(2x-6=3x+6-x+1\)
=>2x-6=2x+7
=>-6=7(vô lý)
c: \(x\left(x+3\right)+x\left(x-2\right)=2x\left(x-1\right)\)
=>\(x^2+3x+x^2-2x=2x^2-2x\)
=>3x-2x=-2x
=>3x=0
=>x=0
d: \(\left(x-1\right)\cdot3x-2\left(x+2\right)-2x=x\left(x-1\right)\)
=>\(3x^2-3x-2x-4-2x=x^2-x\)
=>\(3x^2-7x-4-x^2+x=0\)
=>\(2x^2-6x-4=0\)
=>\(x^2-3x-2=0\)
=>\(x=\dfrac{3\pm\sqrt{17}}{2}\)
b)x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12
<=> 3x^2 +2x +x^2+2x+1 - 4x^2 +25 +12=0
<=> 4x+38=0
=>4x= -38
=>x= -38/4= -19/2
a. 3^x=1-x^2
x=0 la nghiem
x>=1; VT>=3 VP<=0 vo nghiem
b. (de bai thieu n khac 0 vi neu n=0 dung voi moi x)
3x-14=1=> x=5
c.(5^2x5^x+1)=5^4
5^x+1=5^2=> x=1