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x(x - 5)(x + 5) - (x + 2)(x2 - 2x + 4) = 17
=> x(x2 - 25) - (x3 + 23) = 17
=> x3 - 25x - x3 - 8 = 17
=> 25x - 8 = 17
=> 25x = 17 + 8
=> 25x = 25
=> x = 1
\(\left(3x+2\right)\left(x-1\right)-3\left(x+1\right)\left(x-2\right)=4\)
\(\Rightarrow3x^2-3x+2x-2-\left(3x+3\right)\left(x-2\right)=4\)
\(\Rightarrow3x^2-3x+2x-2-\left(3x^2-6x+3x-6\right)=4\)
\(\Rightarrow3x^2-3x+2x-2-3x^2+6x-3x+6=4\)
\(\Rightarrow2x+4=4\)
\(\Rightarrow x=0\)
a) \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2x+10-x^2-5x=0\)
\(\Leftrightarrow-x^2-3x+10=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow x^2-2x+5x-10=0\)
\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)
b) \(x^3-6x^2+12x-8=0\)
\(\Leftrightarrow\left(x^3-8\right)-\left(6x^2-12x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-6x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4-6x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
c)\(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x\right)^2-\left[3\left(x+1\right)\right]^2=0\)
\(\Leftrightarrow\left(4x-3x-1\right)\left(4x+3x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\7x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}}\)
d) \(x^3+x=0\)
\(\Leftrightarrow x^2\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
e)\(x^2-2x-3=0\)
\(\Leftrightarrow x^2+x-3x-3=0\)
\(\Leftrightarrow x\left(x+1\right)-3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)
xét :
|2x - 1| = 2x - 1 nếu 2x - 1 >0 hay x > \(\frac{1}{2}\)
=> 4(2x - 1) - x = 2 <=> 8x - 4 -x = 2 <=> 7x = 6 => x = \(\frac{6}{7}\)
( thỏa mãn ĐK )
|2x -1| = 1 - 2x nếu 2x - 1 < 0 hay x < \(\frac{1}{2}\)
=> 4.( 1 - 2x) - x = 2 <=> 4 - 8x -x = 2<=> 2 =9x => x = \(\frac{2}{9}\) (thỏa mãn ĐK)
vậy phương trình có nghiệm s = { \(\frac{1}{2},\frac{2}{9}\) }
a) 4(x+2) - 7(2x - 1) + 9(3x - 4)=30
⇔4x+8 - 14x + 7 + 27x - 36 = 30
⇔ 17x = 51
⇔ x = 3
b) 2(5x - 8) - 3(4x - 5) = 4(3x - 4) + 11
⇔ 10x - 16 - 12x + 15 = 12x - 16 + 11
⇔ -14x = -4
⇔ x= \(\frac{2}{7}\)
c) 5x(1 - 2x) - 3x(x + 18) = 0
⇔ 5x - 10x\(^2\) - 3x\(^2\) -54x =0
⇔ -13x\(^2\) -49 x = 0
⇔ -x ( 13x + 49 ) =0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\13x+49=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-49}{13}\end{matrix}\right.\)
d) 5x - 3{4x - 2[4x - 3(5x - 2)]} = 182
⇔ 5x - 3[ 4x - 2( 4x - 15x + 6 ) ]= 182
⇔5x - 3 ( 4x - 8x + 30x - 12 ) = 182
⇔ 5x - 3 ( 26x - 12 ) = 182
⇔ 5x - 78x + 36 = 182
⇔ - 73x = 146
⇔ x = -2
x(x-5).(x+5)-(x+2).(x^2-2x+4)=17
\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3-2x^2+4x+2x^2-4x+8\right)=17\)
\(\Leftrightarrow x^3-25x-x^3+2x^2-4x-2x^2+4x-8=17\)
\(\Leftrightarrow-25x=17+8\)
\(\Leftrightarrow-25x=25\)
\(\Leftrightarrow x=-1\)
#)Giải :
\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)
\(\Rightarrow x\left(x^2-25\right)-\left(x^3-2x^2+4x+2x^2-4x+8\right)=17\)
\(\Rightarrow x^3-25-\left(x^3+8\right)=17\)
\(\Rightarrow x^3-25x-x^3-8=17\)
\(\Rightarrow-25x=25\Rightarrow x=-1\)
Vậy x = -1