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c, x+x4=0
=>x(x+3)=0
=>x=0 hoặc x+3=0
=>x=0 hoặc x = -3
\(a,2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+5=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-5\\x=3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{2}\\x=3\end{cases}}\)
Vậy .........
\(b,\left(x^2-4\right)+\left(x-2\right)\left(3-2x=0\right)\)
\(\Leftrightarrow x^2-4-2x^2+7x-6=0\)
\(\Leftrightarrow-x^2+7x-10=0\)
\(\Leftrightarrow-\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=2\end{cases}}\)
Vậy ..................
\(c,x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x=1\)
\(d,x\left(2x-7\right)-4x+14=0\)
\(\Leftrightarrow2x^2-7x-4x+14=0\)
\(\Leftrightarrow2x^2-11x+14=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)
Vậy ............
\(e,\left(2x-5\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow4x^2-20x+25-x^2-4x-4=0\)
\(\Leftrightarrow3x^2-24x+21=0\)
\(\Leftrightarrow3\left(x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=1\end{cases}}\)
Vậy .....................
\(f,x^2-x-\left(3x-3\right)=0\)
\(\Leftrightarrow x^2-x-3x+3=0\)
\(\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
Vậy ..............
a, (3x-1)2 - (x+3)2 = 0
<=> [(3x-1)-(x+3)][(3x-1)+(x+3)] = 0
<=> (3x-1-x-3)(3x-1+x+3) = 0
<=> (2x-4)(4x+2) = 0
=> 2x-4=0 hoặc 4x+2=0
=> 2x =4 hoặc 4x = -2
=> x = 2 hoặc x = \(\frac{-1}{2}\)
\(\begin{array}{l} a){\left( {3x - 1} \right)^2} - {\left( {x + 3} \right)^2} = 0\\ \Leftrightarrow \left( {3x - 1 + x + 3} \right)\left[ {3x - 1 - x - 3} \right] = 0\\ \Leftrightarrow \left( {4x + 2} \right)\left( {2x - 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} 4x + 2 = 0\\ 2x - 4 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - \dfrac{1}{2}\\ x = 2 \end{array} \right.\\ b){x^3} - \dfrac{x}{{49}} = 0\\ \Leftrightarrow 49{x^3} - x = 0\\ \Leftrightarrow x\left( {49{x^2} - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = 0\\ 49{x^2} - 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \pm \dfrac{1}{7} \end{array} \right.\\ c){x^2} - 7x + 12 = 0\\ \Leftrightarrow {x^2} - 3x - 4x + 12 = 0\\ \Leftrightarrow x\left( {x - 3} \right) - 4\left( {x - 3} \right) = 0\\ \Leftrightarrow \left( {x - 3} \right)\left( {x - 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - 3 = 0\\ x - 4 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 3\\ x = 4 \end{array} \right.\\ d)4{x^2} - 3x - 1 = 0\\ \Leftrightarrow 4{x^2} + x - 4x - 1 = 0\\ \Leftrightarrow x\left( {4x + 1} \right) - \left( {4x + 1} \right) = 0\\ \Leftrightarrow \left( {4x + 1} \right)\left( {x - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} 4x + 1 = 0\\ x - 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - \dfrac{1}{4}\\ x = 1 \end{array} \right.\\ e){x^3} - 2x - 4 = 0\\ \Leftrightarrow {x^3} - 4x + 2x - 4 = 0\\ \Leftrightarrow x\left( {{x^2} - 4} \right) + 2\left( {x - 2} \right) = 0\\ \Leftrightarrow x\left( {x - 2} \right)\left( {x + 2} \right) + 2\left( {x - 2} \right) = 0\\ \Leftrightarrow \left( {x - 2} \right)\left[ {x\left( {x + 2} \right) + 2} \right] = 0\\ \Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 2x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - 2 = 0\\ {x^2} + 2x + 2 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 2\\ {x^2} + 2x + 2x = 0\left( {VN} \right) \end{array} \right.\\ f){x^3} + 8{x^2} + 17x + 10 = 0\\ \Leftrightarrow \left( {x + 1} \right)\left( {{x^2} + 7x + 10} \right) = 0\\ \Leftrightarrow \left( {x + 1} \right)\left( {{x^2} + 5x + 2x + 10} \right) = 0\\ \Leftrightarrow \left( {x + 1} \right)\left[ {x\left( {x + 5} \right) + 2\left( {x + 5} \right)} \right] = 0\\ \Leftrightarrow \left( {x + 1} \right)\left( {x + 5} \right)\left( {x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x + 1 = 0\\ x + 5 = 0\\ x + 2 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - 1\\ x = - 5\\ x = - 2 \end{array} \right. \end{array}\)
a)<=>\(\left(x^3+x^2-2x\right)+\left(3x^2+3x-6\right)=0\)
<=>\(x\left(x^2+x-2\right)+3\left(x^2+x-2\right)=0\)
<=>\(\left(x^2+x-2\right)\left(x+3\right)=0\)
Phương trình trên bạn tự bấm máy tính nha
<=>\(\left(x-1\right)\left(x+2\right)\left(x+3\right)=0\)
Đến đây tự làm đc rồi
Vậy x=1 hoặc -2 hoặc -3
b)<=>\(\left(x^3-4x^2+4x\right)+\left(x^2-4x+4\right)=0\)
<=>\(x\left(x^2-4x+4\right)+\left(x^2-4x+4\right)=0\)
<=>\(\left(x+1\right)\left(x^2-4x+4\right)=0\)
<=>\(\left(x+1\right)\left(x-2\right)^2=0\)
<=>\(\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
c)Câu c mik chưa làm đc
Đáp án câu C:
\(x^3-4x^2+5x=0\)
\(\Leftrightarrow x\left(x^2-4x^2+5x\right)=0\)
\(Tacó:x^2-4x+5=x^2-4x+2^2+1\)
\(=\left(x-2\right)^2+1\)
\(Mà\left(x-2\right)^2\ge0\)
\(Nên\left(x-2\right)^2+1\ge1\)
\(Khiđó:x\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x=0\)
n) \(\left|3-x\right|+x^2-x\left(x+4\right)=0\)
\(\Rightarrow\left|3-x\right|+x^2-x^2-4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x-4x=0\left(đk:3-x\ge0\right)\\-\left(3-x\right)-4x=0\left(đk:3-x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(đk:x\le3\right)\\x=-1\left(đk:x>3\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x=\dfrac{3}{5}\)
m) \(\left(x-1\right)^2+\left|x+21\right|-x^2-13=0\)
\(\Rightarrow x^2-2x+1+\left|x+21\right|-x^2-13=0\)
\(\Leftrightarrow-2x-12+\left|x+21\right|=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x-12+x+21=0\left(đk:x+21\ge0\right)\\-2x-12-\left(x+21\right)=0\left(đk:x+21< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\left(đk:x\ge-21\right)\\x=-11\left(đk:x< -21\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=9\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x=9\)
e) \(\left|5x\right|=3x-2\)
\(\Rightarrow5\cdot\left|x\right|=3x-2\)
\(\Leftrightarrow5\cdot\left|x\right|-3x=-2\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-3x=-2\left(đk:x\ge0\right)\\5\cdot\left(-x\right)-3x=-2\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(đk:x\ge0\right)\\x=\dfrac{1}{4}\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x\in\varnothing\)
g) \(\left|-2,5x\right|=x-12\)
\(\Rightarrow2,5\cdot\left|x\right|=x-12\)
\(\Leftrightarrow2x5\cdot\left|x\right|-x=-12\)
\(\Leftrightarrow\left[{}\begin{matrix}2,5x-x=-12\left(đk:x\ge0\right)\\2,5\cdot\left(-x\right)-x=-12\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-8\left(đk:x\ge0\right)\\x=\dfrac{24}{7}\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x\in\varnothing\)
Ta có : (x + 1)(x + 2)(x + 3)(x + 4) = 3x2
=> [(x + 1)(x + 4)][(x + 2)(x + 3)] = 3x2
=> (x2 + 5x + 4) (x2 + 5x + 6) = 3x2
Đặt x2 + 5x + 5 = a
Thay vào biểu thức ta có : (a - 1)(a + 1) = 3x2
<=> a2 - 1 = 3a2
<=> (x2 + 5x + 5)2 = 3x2
<=> x4 + 10x2 + 15 = 3x2
=> x4 + 10x2 + 15 - 3x2 = 0
<=> x4 + 7x2 + 15 = 0
<=> (x2 + 3,5)2 + 2,75 = 0
=> sai đề
a) pt <=> ( x - 1 )3 + x2( x - 1 ) = 0
<=> ( x - 1 )[ ( x - 1 )2 + x2 ] = 0
<=> x = 1
Vậy pt có nghiệm x = 1
b) x2 + x - 12 = 0
<=> x2 - 3x + 4x - 12 = 0
<=> x( x - 3 ) + 4( x - 3 ) = 0
<=> ( x - 3 )( x + 4 ) = 0
<=> x = 3 hoặc x = -4
Vậy S = { 3 ; -4 }
c) x + x4 = 0
<=> x( x3 + 1 ) = 0
<=> x( x + 1 )( x2 - x + 1 ) = 0
<=> x = 0 hoặc x = -1
Vậy S = { 0 ; -1 }
a,\(x^3-3x^2+3x-1+x\left(x^2-x\right)=0\)
\(\Leftrightarrow\left(x^3-3x^2+3x-1\right)+x\left(x^2-x\right)=0\)
\(\Leftrightarrow\left(x-1\right)^3+x^2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(x-1\right)^2+x^2\right]=0\)
\(\Leftrightarrow x=1\)