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b) \(\left(x+3\right)^2-5x-15=0\\ \Leftrightarrow\left(x+3\right)^2-5\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x+3-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là : \(S=\left\{-3;2\right\}\)
c) \(2x^5-4x^3+2x=0\\ \Leftrightarrow2x\left(x^4-2x^2+1\right)=0\\ \Leftrightarrow2x\left(x^2-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\\left(x^2-1\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy tập nghiệm của pt là : \(S=\left\{0;1;-1\right\}\)
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Sửa đề bài 1 : k => x P/s : đề sai r :))
\(A=\left(3-2x\right)3x^2-8+\left(2x+5\right)\left(3x-2\right)-20x\)
\(=9x^2-6x^3-8+6x^2-4x+15x-10-20x=15x^2-6x^3-18-9x\)
Vậy biểu thức phụ thuộc biến x
\(B=\left(3-5x\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=6x+33-10x^2-55x-6x^2-14x-9x-21=-72x+12-16x^2\)
Vậy biểu thức phụ thuộc biến x
Bài 2 :
a, \(2x\left(x-1\right)-x^2+6=0\Leftrightarrow2x^2-2x-x^2+6=0\)
\(\Leftrightarrow x^2-2x+6=0\)( vô nghiệm )
b, \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)-x\left(x-2\right)\left(x+2\right)=15\)
\(\Leftrightarrow x^2-9-x\left(x^2-4\right)=15\Leftrightarrow x^2-9-x^3+12=15\)
\(\Leftrightarrow-x^3+x^2-12=0\Leftrightarrow x=2\)
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a) 4x(x+1)=8(x+1)
<=>4x(x+1)-8(x+1)=0
<=>(4x-8)(x+1)=0
<=>\(\left[\begin{array}{} 4x-8=0\\ x+1=0 \end{array} \right.\)
<=>\(\left[\begin{array}{} x=2\\ x=-1 \end{array} \right.\)
Vậy...
b)x(x-1)-2(1-x)=0
<=>(x+2)(x-1)=0
<=>\(\left[\begin{array}{} x+2=0\\ x-1=0 \end{array} \right.\)
<=>\(\left[\begin{array}{} x=-2\\ x=1 \end{array} \right.\)
Vậy...
c)5x(x-2)-(2-x)=0
<=>(5x+1)(x-2)=0
<=>\(\left[\begin{array}{} 5x+1=0\\ x-2 \end{array} \right.\)
<=>\(\left[\begin{array}{} x=-1/5\\ x=2 \end{array} \right.\)
d)5x(x-200)-x+200=0
<=>(5x-1)(x-200)=0
<=>\(\left[\begin{array}{} 5x-1=0\\ x-200=0 \end{array} \right.\)
<=>\(\left[\begin{array}{} x=1/5\\ x=200 \end{array} \right.\)
e)\(x^3+4x=0 \)
\(\Leftrightarrow x(x^2+4)=0 \)
\(\Leftrightarrow \left[\begin{array}{} x=0\\ x^2+4=0 (loại vì x^2+4>=0 với mọi x) \end{array} \right.\)
Vậy x=0
f)\((x+1)=(x+1)^2\)
\(\Leftrightarrow (x+1)-(x+1)^2=0\)
\(\Leftrightarrow (x+1)(1-x-1)=0\)
\(\Leftrightarrow (x+1)(-x)=0\)
\(\Leftrightarrow \left[\begin{array}{} x=-1\\ x=0 \end{array} \right.\)
Vậy....
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a,5x(4x-3)+6-8x=0
=>5x(4x-3)-8x+6=0
=>5x(4x-3)-2(4x-3)=0
=>(5x-2)(4x-3)=0
=>\(\orbr{\begin{cases}5x-2=0\\4x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{2}{5}\\x=\frac{3}{4}\end{cases}}}\)
b, 5x(x-9)-x+9=0
=>5x(x-9)-(x-9)=0
=>(5x-1)(x-9)=0
=>\(\orbr{\begin{cases}5x-1=0\\x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=9\end{cases}}}\)
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1.
a.\(\Leftrightarrow7x-5x=3+12\)
\(\Leftrightarrow2x=15\Leftrightarrow x=\dfrac{15}{2}\)
b.\(\Leftrightarrow6x-10-7x-7=2\)
\(\Leftrightarrow x=-19\)
c.\(\Leftrightarrow1-3x=4x-3\)
\(\Leftrightarrow7x=2\Leftrightarrow x=\dfrac{2}{7}\)
d.\(\Leftrightarrow8x^2-4x+12x-6-8x^2-8x-2=12\)
\(\Leftrightarrow-2=12\left(voli\right)\)
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Bài làm :
a) x( 2x - 7 ) - 4x + 14 = 0
<=> x( 2x - 7 ) - 2( 2x - 7 ) = 0
<=> ( 2x - 7 )( x - 2 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)
b) Sửa đề : 5x3 + x2 - 4x + 9 = 0
<=>( 5x3 + 5 ) + (x2 - 4x +4)=0
<=> 5(x3 + 1) + (x-2)2 = 0
<=> 5(x+1)(x2 - x +1) + (x+2)2 =0
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
c) 3x3 - 7x2 + 6x - 14 = 0
<=> 3x2( x - 7/3 ) + 6( x - 7/3 ) = 0
<=> ( x - 7/3 )( 3x2 + 6 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{7}{3}=0\\3x^2+6=0\end{cases}}\Leftrightarrow x=\frac{7}{3}\)
d) 5x2 - 5x = 3( x - 1 )
<=> 5x( x - 1 ) - 3( x - 1 ) = 0
<=> ( x - 1 )( 5x - 3 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{3}{5}\end{cases}}\)
e) 4x2 - 25 - ( 4x - 10 ) = 0
<=> ( 2x - 5 )( 2x + 5 ) - 2( 2x - 5 ) = 0
<=> ( 2x - 5 )( 2x + 5 - 2 ) = 0
<=> ( 2x - 5 )( 2x + 3 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\2x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)
f) x3 + 27 + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 ) + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 + x - 9 ) = 0
<=> ( x + 3 )( x2 - 2x ) = 0
<=> x( x + 3 )( x - 2 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}\\\end{cases}}\begin{cases}x=0\\x=-3\\x=2\end{cases}\)
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a) x( 2x - 7 ) - 4x + 14 = 0
<=> x( 2x - 7 ) - 2( 2x - 7 ) = 0
<=> ( 2x - 7 )( x - 2 ) = 0
<=> \(\orbr{\begin{cases}2x-7=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)
b) 5x3 + x2 - 4x - 9 = 0 ( đề sai )
c) 3x3 - 7x2 + 6x - 14 = 0
<=> 3x2( x - 7/3 ) + 6( x - 7/3 ) = 0
<=> ( x - 7/3 )( 3x2 + 6 ) = 0
<=> \(\orbr{\begin{cases}x-\frac{7}{3}=0\\3x^2+6=0\end{cases}}\Leftrightarrow x=\frac{7}{3}\)( do 3x2 + 6 ≥ 6 > 0 với mọi x )
d) 5x2 - 5x = 3( x - 1 )
<=> 5x( x - 1 ) - 3( x - 1 ) = 0
<=> ( x - 1 )( 5x - 3 ) = 0
<=> \(\orbr{\begin{cases}x-1=0\\5x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{3}{5}\end{cases}}\)
e) 4x2 - 25 - ( 4x - 10 ) = 0
<=> ( 2x - 5 )( 2x + 5 ) - 2( 2x - 5 ) = 0
<=> ( 2x - 5 )( 2x + 5 - 2 ) = 0
<=> ( 2x - 5 )( 2x + 3 ) = 0
<=> \(\orbr{\begin{cases}2x-5=0\\2x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)
f) x3 + 27 + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 ) + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 + x - 9 ) = 0
<=> ( x + 3 )( x2 - 2x ) = 0
<=> x( x + 3 )( x - 2 ) = 0
<=> x = 0 hoặc x + 3 = 0 hoặc x - 2 = 0
<=> x = 0 hoặc x = -3 hoặc x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
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a) \(\Rightarrow x\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)
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\(4x\left(x-5\right)-5x+25=0\Leftrightarrow4x\left(x-5\right)-5\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(4x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x-5=0\\4x-5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=\frac{5}{4}\end{cases}}\)
4x( x - 5 ) - 5x + 25 = 0
<=> 4x( x - 5 ) - 5( x - 5 ) = 0
<=> ( x - 5 )( 4x - 5 ) = 0
<=> \(\orbr{\begin{cases}x-5=0\\4x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{5}{4}\end{cases}}\)