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Bài 1: Giải các phương trình sau:
a) 3(2,2-0,3x)=2,6 + (0,1x-4)
<=> 6.6 - 0.9x = 2,6 + 0,1x - 4
<=> - 0.9x - 0,1x = -6.6 -1,4
<=> -x = -8
<=> x = 8
Vậy x = 8
b) 3,6 -0,5 (2x+1) = x - 0,25(22-4x)
<=> 3,6 - x - 0,5 = x - 5,5 + x
<=> - x - 3,1 = -5,5
<=> - x = -2.4
<=> x = 2.4
Vậy x = 2.4
\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1\right)=\left(3x-2\right)\left(3x+2\right)\left(x+1\right)\)
\(\Leftrightarrow x-1=3x-2\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
c: =>x-3=0
hay x=3
d: \(\Leftrightarrow\left(3x-1\right)\cdot\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
\(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right).\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0.\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0.\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(-2x+1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0.\\x+1=0.\\-2x+1=0.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}.\\x=-1.\\x=\dfrac{1}{2}.\end{matrix}\right.\)
c: =>(x-3)(x2+3x+5)=0
=>x-3=0
hay x=3
d: =>(3x-1)(x2+2-7x+10)=0
=>(3x-1)(x-3)(x-4)=0
hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)
b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)
c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)
d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)
a) \(x^2-4x+4=25\\ \Rightarrow\left(x-2\right)^2=25\\ \Rightarrow\left[{}\begin{matrix}x-2=-5\\x-2=5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
b) \(\left(5-2x\right)^2-16=0\\ \Rightarrow\left(5-2x\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}5-2x=-4\\5-2x=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4,5\\0,5\end{matrix}\right.\)
c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\\ \Rightarrow\left(x-3\right)^3-\left(x-3\right)^3+9\left(x+1\right)^2=15\\ \Rightarrow9\left(x+1\right)^2=15\\ \Rightarrow\left(x+1\right)^2=\dfrac{5}{3}\\ \Rightarrow\left[{}\begin{matrix}x+1=-\sqrt{\dfrac{5}{3}}\\x+1=\sqrt{\dfrac{5}{3}}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3+\sqrt{15}}{3}\\x=\dfrac{-3+\sqrt{15}}{3}\end{matrix}\right.\)
a)\(\Leftrightarrow\)\(x^2-4x-21=0\)
\(\Leftrightarrow\)\(x^2-7x+3x-21=0\)
\(\Leftrightarrow\)\(x(x-7)+3(x-7)=0\)
\(\Leftrightarrow\)\((x-7)(x+3)=0\)
\(\Leftrightarrow\)\(\left[\begin{array}{} x=7\\ x=-3 \end{array} \right.\)
b)\(\Leftrightarrow\)\((5-2x)^2-4^2=0\)
\(\Leftrightarrow\)\((5-2x-4)(5-2x+4)=0\)
\(\Leftrightarrow\)\((-2x+1)(-2x+9)=0\)
\(\Leftrightarrow\)\(\left[\begin{array}{} x=\dfrac{1}{2}\\ x=\dfrac{9}{2} \end{array} \right.\)
a) Ta có: \(2\left(3x+1\right)-4\left(5-2x\right)>2\left(4x-3\right)-6\)
\(\Leftrightarrow6x+2-20+8x>8x-6-6\)
\(\Leftrightarrow14x-18-8x+12>0\)
\(\Leftrightarrow6x-6>0\)
\(\Leftrightarrow6x>6\)
hay x>1
Vậy: S={x|x>1}
b) Ta có: \(9x^2-3\left(10x-1\right)< \left(3x-5\right)^2-21\)
\(\Leftrightarrow9x^2-30x+3< 9x^2-30x+25-21\)
\(\Leftrightarrow9x^2-30x+3-9x^2+30x-4< 0\)
\(\Leftrightarrow-1< 0\)(luôn đúng)
Vậy: S={x|\(x\in R\)}
a) Ta có: \(\left|x^2-x+2\right|-3x-7=0\)
\(\Leftrightarrow\left|x^2-x+2\right|=3x+7\)
\(\Leftrightarrow x^2-x+2=3x+7\)(Vì \(x^2-x+2>0\forall x\))
\(\Leftrightarrow x^2-x+2-3x-7=0\)
\(\Leftrightarrow x^2-4x-5=0\)
\(\Leftrightarrow x^2-5x+x-5=0\)
\(\Leftrightarrow x\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
Vậy: S={5;-1}
bạn giải giúp mình câu b nữa với
mai mình phải nộp bài rồi!!!
b: 4(x+1)^2-9(x-1)^2=0
=>(2x+2)^2-(3x-3)^2=0
=>(2x+2-3x+3)(2x+2+3x-3)=0
=>(-x+5)(5x-1)=0
=>x=1/5 hoặc x=5
c: (x-1)^3+x^3+(x+1)^3=(x+2)^3
=>x^3-3x^2+3x-1+x^3+x^3+3x^2+3x+1=x^3+6x^2+12x+8
=>3x^3+6x-x^3-6x^2-12x-8=0
=>2x^3-6x^2-6x-8=0
=>x^3-3x^2-3x-4=0
=>x^3-4x^2+x^2-4x+x-4=0
=>(x-4)(x^2+x+1)=0
=>x-4=0
=>x=4