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\(a,x+5x^2=0\\ \Rightarrow a,x\left(1+5x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\\ b,\left(x+3\right)^2+\left(4+x\right)\left(4-x\right)=0\\ \Rightarrow x^2+6x+9+16-x^2=0\\ \Rightarrow6x+25=0\\ \Rightarrow6x=-25\\ \Rightarrow x=-\dfrac{25}{6}\)
\(c,5x\left(x-1\right)=x-1\\ \Rightarrow c,5x\left(x-1\right)-\left(x-1\right)\\ \Rightarrow\left(x-1\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ d,x^2-2x-3=0\\ \Rightarrow\left(x^2-3x\right)+\left(x-3\right)=0\\ \Rightarrow x\left(x-3\right)+\left(x-3\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
a)
<=> 10x - 35 + 16x - 10 = 5
<=> 10x + 16x = 5 + 35 + 10
<=> 26x = 50
<=> x = 50/26 = 25/13
ta có
a. (5x-7)(x-9)-(-x+3)(-5x+2)= 2x(x-4)-(x-1)(2x+3)
\(\Leftrightarrow5x^2-52x+63-\left(5x^2-17x+6\right)=2x^2-8x-\left(2x^2+x-3\right)\)
\(\Leftrightarrow-35x+57=-9x+3\Leftrightarrow26x=54\Leftrightarrow x=\frac{27}{13}\)
b. (x-3)(-x+10)+(x-8)(x+3)= (5x^2-1)(x+3)-5x^3-15x^2
\(\Leftrightarrow-x^2+13x-30+x^2-5x-24=5x^3+15x^2-x-3-5x^3-15x^2\)
\(\Leftrightarrow8x-54=-x-3\Leftrightarrow9x=51\Leftrightarrow x=\frac{17}{3}\)
a) x4 + 2x3 + 5x2 + 4x - 12 = 0
=> x4 - x3+ 3x3- 3x2 + 8x2 -8x + 12x - 12 = 0
=> x3( x - 1) + 3x2( x - 1) + 8x( x - 1) + 12 ( x - 1 ) = 0
=> ( x - 1)( x3 + 3x2 + 8x + 12 ) = 0
=> ( x - 1)( x3 + 2x2 + x2 + 2x + 6x + 12 ) = 0
=> ( x - 1)[ x2( x + 2) + x( x + 2) + 6( x + 2) ] = 0
=> ( x - 1)( x + 2)( x2 + x + 6 ) = 0
Ta thấy : x2 + x + 6
= x2 + 2.\(\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{1}{4}+6=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\text{≥}\dfrac{23}{4}>0\text{∀}x\)
=> ( x - 1)( x + 2 ) = 0
=> x = 1 hoặc x = -2
Vậy,....
b) ( x + 1)3 + ( x - 2)3 = ( 2x - 1)3
=>x3+ 3x2 + 3x + 1 + x3 - 6x2 + 12x - 8 - ( 8x3 - 12x2 + 6x - 1)=0
=> 2x3 - 3x2 + 15x - 7 - 8x3 + 12x2 - 6x + 1 = 0
=> 9x2 - 6x3 + 9x - 6 = 0
=> 9x( x + 1) -6( x3 + 1 ) = 0
=> 9x( x + 1) - 6( x + 1)( x2 - x + 1) = 0
=> 3( x + 1)( 3x - 2x2 + 2x - 2) = 0
=> 3( x + 1)( - 2x2 + 5x - 2) = 0
=> 3( x + 1)( - 2x2 + x + 4x - 2) = 0
=> 3( x + 1)[ x( 1 - 2x ) - 2( 1 - 2x ) ] = 0
=> 3( x + 1)( 1 - 2x )( x - 2) = 0
Suy ra :
* x + 1 = 0 => x = -1
* 1 - 2x = 0 => x = \(\dfrac{1}{2}\)
* x - 2 = 0 => x = 2
Vậy,.....
1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)
\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)
\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)
\(\Leftrightarrow-41x=-115\)
hay \(x=\dfrac{115}{41}\)
2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)
\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)
\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)
\(\Leftrightarrow x^3=64\)
hay x=4
3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)
\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)
\(\Leftrightarrow-5x-15=10x-20\)
\(\Leftrightarrow-5x-10x=-20+15\)
\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)
a)4(x+2)-7(2x-1)+9(3x-4)=30 b)2(5x-8)-3(4x-5)=4(3x-4)+11
<=>4x+8-14x+7+27x-36=30 <=>10x-16-12x+15=12x-16+11
<=>17x-21=30 <=> -14x=-4 <=>x=2/7
<=>17x=51
<=>x=3
a) pt <=> (2x-1)(2y+3)=7
TH1: 2x-1=7 và 2y+3=1
<=> x = 4 và y = -1
TH2: 2x - 1 = -7 và 2y + 3 = -1
<=> x = -3 và y = -2
TH3: 2x-1=1 và 2y+3=7
<=> x = 1 và y=2
TH4: 2x-1=-1 và 2y+3=-7
<=> x=0 và y=-5
Bài làm
a, ( x + 3 )10 = 45
=> ( x + 3 )10 = ( 22)5
=> ( x + 3 )10 = 210
=> x + 3 = 2
=> x = -1
b, x15 = 2710
=> ( x3 )5 = ( 272 )5
=> x3 = 272
=> x3 = 729
=> x3 = 93
=> x = 9
Vậy x = 9
c, ( 4 - 5x )3 = 27
=> ( 4 - 5x )3 = 33
=> 4 - 5x = 3
=> 5x = 4 - 3
=> 5x = 1
=> x = \(\frac{1}{5}\)
Vậy x = \(\frac{1}{5}\)
d, ( 1 - x )3 = 82
=> ( 1 - x )3 = ( 23 )2
=> ( 1 - x )3 = 26
=> ( 1 - x )3 = ( 22 )3
=> ( 1 - x )3 = 43
=> 1 - x = 4
=> x = -3
Vậy x = -3
# Học tốt #
a. (x + 3)10 = 45
<=> (x + 3)10 = (22)5
<=> x + 3 = 2
<=> x = -1
b. x15 = 2710
<=> x15 = (33)10
<=> x15 = (32)15
<=> x = 9
c. (4 - 5x)3 = 27
<=> (4 - 5x)3 = 33
<=> 4 - 5x = 3
<=> x = \(\frac{1}{5}\)
d. (1 - x)3 = 82
<=> (1 - x)3 = (23)2
<=> (1 - x)3 = 43
<=> 1 - x = 4
<=> x = -3