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a) (x-2)3+6(x+1)2-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6(x2+2x+1)-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6x2+12x+6-x3+12=0
\(\Rightarrow\)24x+10=0
\(\Rightarrow\)24x=-10
\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)
b)(x-5)(x+5)-(x+3)2+3(x-2)2=(x+1)2-(x-4)(x+4)+3x2
\(\Rightarrow\)x2-25-(x2+6x+9)+3(x2-4x+4)=x2+2x+1-(x2-16)+3x2
\(\Rightarrow\)x2-25-x2-6x-9+3x2-12x+12=x2+2x+1-x2+16+3x2
\(\Rightarrow\)3x2-18x-22=3x2+2x+17
\(\Rightarrow\)3x2-18x-22-3x2-2x-17=0
\(\Rightarrow\)-20x-39=0
\(\Rightarrow\)-20x=39
\(\Rightarrow\)x=\(-\dfrac{39}{20}\)
1) \(\Rightarrow16x^2+24x+9+9x^2-24x+16+4-25x^2=x\)
\(\Rightarrow x=29\)
2)
a) \(=x^2-9-x^2+6x-9=6x-18\)
b) \(=\left(3x-1+2x+1\right)^2=\left(5x\right)^2=25x^2\)
1.
a) \(=x^2-6x+9+3x^2-15x=4x^2-21x+9\)
b) \(=9x^2+12x+4-x^2+9=8x^2+12x+13\)
2.
a) \(\Leftrightarrow x^2+8x+16-x^2+4-5=0\\ \Leftrightarrow8x=-15\\ \Leftrightarrow x=-\dfrac{15}{8}\)
b) \(\Leftrightarrow9x^2-6x+1-8x^2+12x-2x+3-5-x^2=0\\ \Leftrightarrow4x=1\\ \Leftrightarrow x=\dfrac{1}{4}\)
`a)|2x+1|=5`
`<=>` \(\left[ \begin{array}{l}2x+1=5\\2x+1=-5\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-6\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\)
`b)|2x+1|=0`
`<=>2x+1=0`
`<=>2x=-1`
`<=>x=-1/2`
`c)|2x+1|=7`
`<=>` \(\left[ \begin{array}{l}2x+1=7\\2x+1=-7\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=6\\2x=-8\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=4\\x=-4\end{array} \right.\)
`d)|2x+5|=|3x-7|`
`<=>` \(\left[ \begin{array}{l}2x+5=3x-7\\2x+5=7-3x\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=12\\5x=2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=12\\x=\dfrac25\end{array} \right.\)
`e)|2x+7|=1`
`<=>` \(\left[ \begin{array}{l}2x+7=1\\2x+7=-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=-6\\2x=-8\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=3\\x=-4\end{array} \right.\)
`g)|x-2|+|2x-3|=2`
Nếu `x>=2=>|x-2|=x-2,|2x-3|=2x-3`
`pt<=>x-2+2x-3=2`
`<=>3x-5=2`
`<=>3x=7`
`<=>x=7/3(tm)`
Nếu `x<=3/2=>|x-2|=2-x,|2x-3|=3-2x`
`pt<=>2-x+3-2x=2`
`<=>5-3x=2`
`<=>3x=3`
`<=>x=1(tm)`
Nếu `3/2<=x<=2=>|x-2|=2-x,|2x-3|=2x-3`
`pt<=>2-x+2x-3=2`
`<=>x-1=2`
`<=>x=3(l)`
`h)|x+2|+|1-x|=3x+2`
Vì `VT>=0=>3x+2>=0=>x>=-2/3`
`=>|x+2|=x+2`
`pt<=>x+2+|1-x|=3x+2`
`<=>|1-x|=2x(x>=0)`
`<=>` \(\left[ \begin{array}{l}2x=1-x\\2x=x-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}3x=1\\x=-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac13(TM)\\x=-1(KTM)\end{array} \right.\)
a.
$|2x+1|=5$
\(\Leftrightarrow \left[\begin{matrix}
2x+1=5\\
2x+1=-5\end{matrix}\right.\Leftrightarrow \left[\begin{matrix}
x=2\\
x=-3\end{matrix}\right.\)
b.
$|2x+1|=0$
$\Leftrightarrow 2x+1=0$
$\Leftrightarrow x=-\frac{1}{2}$
c.
$|2x+1|=7$
\(\Leftrightarrow \left[\begin{matrix} 2x+1=7\\ 2x+1=-7\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=3\\ x=-4\end{matrix}\right.\)
\(a,\Rightarrow4x^2-20x-4x^2+3x+4x-3=5\\ \Rightarrow-13x=8\Rightarrow x=-\dfrac{8}{13}\\ b,\Rightarrow3x^2-10x+8-3x^2+27x=-3\\ \Rightarrow17x=-11\Rightarrow x=-\dfrac{11}{17}\\ c,\Rightarrow\left(x+3\right)\left(2-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ d,\Rightarrow2x\left(4x^2-25\right)=0\\ \Rightarrow2x\left(2x-5\right)\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ e,Sửa:\left(4x-3\right)^2-3x\left(3-4x\right)=0\\ \Rightarrow\left(4x-3\right)^2+3x\left(4x-3\right)=0\\ \Rightarrow\left(4x-3\right)\left(7x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
a.
4x(x-5) - (x-1)(4x-3)-5=0
4x^2-20x-4x^2+3x+4x+3=0
(4x^2-4x^2)+(-20x+3x+4x)+3=0
13x+3 = 0
13x=-3
x=-3/13
b,
(3x-4)(x-2)-3x(x-9)+3=0
3x^2-6x-4x+8 - 3x^2+27x+3=0
(3x^2-3x^2)+(-6x-4x+27x)+(8+3)=0
17x+11=0
17x=-11
x=-11/17
c, 2(x+3)-x^2-3x=0
2(x+3) - x(x+3)=0
(x+3)(2-x)=0
TH1: x+3 = 0; x=-3
TH2: 2-x=0;x=2
a) \(\left(x-1\right)^3\)
\(=x^3-3x^2+3x-1\)
b) \(\left(2x-3y\right)^3\)
\(=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^3+\left(3y\right)^3\)
\(=8x^3-36x^2y+54xy^2-27y^3\)
Bài 3:
a: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=5\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=5\)
\(\Leftrightarrow12x=13\)
hay \(x=\dfrac{13}{12}\)
b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=4\)
\(\Leftrightarrow x^3-1-x^3+4x=4\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)
\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)
\(\Leftrightarrow24x=-13\)
hay \(x=-\dfrac{13}{24}\)
1: =>x^2+4x-21=0
=>(x+7)(x-3)=0
=>x=3 hoặc x=-7
2: =>(2x-5-4)(2x-5+4)=0
=>(2x-9)(2x-1)=0
=>x=9/2 hoặc x=1/2
3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15
=>-9x^2+27x+9x^2+18x+9=15
=>18x=15-9-27=-21
=>x=-7/6
6: =>4x^2+4x+1-4x^2-16x-16=9
=>-12x-15=9
=>-12x=24
=>x=-2
7: =>x^2+6x+9-x^2-4x+32=1
=>2x+41=1
=>2x=-40
=>x=-20
\(\frac{x^2-x}{x+3}-\frac{x^2}{x-3}=\frac{7x^2-3x^2}{9-x^2}\) ĐKXĐ : \(x\ne\pm3\)
\(\Leftrightarrow\frac{\left(x^2-x\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\frac{x^2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{3x^2-7x^2}{\left(x+3\right)\left(x-3\right)}\)
\(\Leftrightarrow x^3-3x^2-x^2+3x-x^3-3x^2=3x^2-7x^2\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(-3x^2-x^2-3x^2-3x^2+7x^2\right)-3x=0\)
\(\Leftrightarrow-3x^2-3x=0\)
\(\Leftrightarrow-3x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-3x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
KL : nghiệm của PT là : \(S=\left\{0;-1\right\}\)
\(\frac{x-4}{x-1}+\frac{x+4}{x+1}=2\) DKXĐ : \(x\ne\pm1\)
\(\Leftrightarrow\frac{\left(x-4\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{\left(x+4\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=2\)
\(\Leftrightarrow x^2+x-4x-4+x^2-x+4x-4=2\)
\(\Leftrightarrow\left(x^2+x^2\right)\left(x-4x-x+4x\right)+\left(-4-4\right)=2\)
\(\Leftrightarrow2x^2-8=2\)
\(\Leftrightarrow2x^2=10\)
.....
a.
x^3+x^2 +x -x^2 -x -1 - x(x^2 -4)=5
=> x^3 +x^2 -x^2 -x -1 -x^3 +4x =5
=> 3x -1 =5
=> 3x= 5+1 =6
=> x= 6/3 =2