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\(5x\left(x-3\right)\left(x+3\right)-\left(2x-3\right)^2-5\left(x+2\right)^2\)
\(+34x\left(x+2\right)=1\)
\(\Leftrightarrow5x\left(x^2-9\right)-\left(4x^2-12x+9\right)-5\left(x^2+4x+4\right)\)
\(+34x^2+68x=0\)
\(\Leftrightarrow5x^3-45x-4x^2+12x-9-5x^2-20x-20\)
\(+34x^2+68x=0\)
\(\Leftrightarrow5x^3+25x^2+15x-29=0\)
Giải nghiệm ta được ba nghiệm sau:
\(x_1\approx0,776\)
\(x_2\approx-1,96\)
\(x_3\approx-3,82\)
a)
<=> 10x - 35 + 16x - 10 = 5
<=> 10x + 16x = 5 + 35 + 10
<=> 26x = 50
<=> x = 50/26 = 25/13
\(2x\left(x-3\right)-x+3=0\)
<=> \(2x\left(x-3\right)-\left(x-3\right)=0\)
<=> \(\left(x-3\right)\left(2x-1\right)=0\)
<=> \(\orbr{\begin{cases}x=3\\x=\frac{1}{2}\end{cases}}\)
Vậy...
a) ( 3x - 1 ) ( 2x + 7 ) - ( x + 1 ) ( 6x + 5 ) = 16
<=> 6x2 + 21x - 2x - 7 - ( 6x2 - 5x + 6x - 5) = 16
<=> 6x2 + 21x - 2x - 7 - ( 6x2 + x - 5 ) = 16
<=> 6x2+ 21x - 2x - 7 - 6x2 -x + 5 = 16
<=> 18x - 2 = 16
<=> 18x = 18
=> x = 1
Vậy....
a) 5x( x - 1 ) = x - 1
<=> 5x2 - 5x = x - 1
<=> 5x2 - 5x - x + 1 = 0
<=> 5x2 - 6x + 1 = 0
<=> 5x2 - 5x - x + 1 = 0
<=> 5x( x - 1 ) - 1( x - 1 ) = 0
<=> ( x - 1 )( 5x - 1 ) = 0
<=> \(\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
b) 2( x + 5 ) - x2 - 5x = 0
<=> 2x + 10 - x2 - 5x = 0
<=> -x2 - 3x + 10 = 0
<=> -x2 - 5x + 2x + 10 = 0
<=> -x( x + 5 ) + 2( x + 5 ) = 0
<=> ( x + 5 )( 2 - x ) = 0
<=> \(\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
c) x2 - 2x - 3 = 0
<=> x2 + x - 3x - 3 = 0
<=> x( x + 1 ) - 3( x + 1 ) = 0
<=> ( x + 1 )( x - 3 ) = 0
<=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
d) 2x2 + 5x - 3 = 0
<=> 2x2 - x + 6x - 3 = 0
,<=> x( 2x - 1 ) + 3( 2x - 1 ) = 0
<=> ( 2x - 1 )( x + 3 ) = 0
<=> \(\orbr{\begin{cases}2x-1=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)
a) 5x ( x - 1 ) = x - 1 <=> 5x2 - 5x - x + 1 = 0
<=> 5x2 - 6x + 1 = 0 <=> 5x2 - x - ( 5x - 1 ) = 0
<=> x ( 5x - 1 ) - ( 5x - 1 ) = 0 <=> ( x - 1 )( 5x - 1 ) = 0
<=> x = 1 hoặc x = 1/5
b) 2 ( x + 5 ) - x2 - 5x = 0 <=> 2 ( x + 5 ) - x ( x + 5 ) = 0
<=> ( 2 - x ) ( x + 5 ) = 0 <=> x = 2 hoặc x = -5
c) x2 - 2x - 3 = 0 <=> x2 + x - 3x - 3 = 0
<=> x ( x + 1 ) - 3 ( x + 1 ) = 0 <=> ( x - 3 ) ( x + 1 ) = 0
<=> x = 3 hoặc x = -1
d) 2x2 + 5x - 3 = 0
Ta có : delta = 52 - 4.2.3 = 25 - 24 = 1
Khi đó : x = -1 hoặc x = 3/2
5x(x - 3)(x + 3) - (2x - 3)2 - 5(x + 2) + 34x(x + 2) = 1
\(\Leftrightarrow-\left(2x-3\right)^2+5x\left(x-3\right)\left(x+3\right)+34x\left(x+2\right)-5\left(x+2\right)-1=0\)
\(\Leftrightarrow\left[-\left(2x-3\right)^2+5x\left(x^2-9\right)-1\right]+\left(x+2\right)\left(34x-5\right)=0\)
\(\Leftrightarrow\left(-4x^2+12x-9+5x^3-45x-1\right)+\left(x+2\right)\left(34x-5\right)=0\)
\(\Leftrightarrow\left(5x^3-4x^2-33x-10\right)+\left(x+2\right)\left(34x-5\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(5x^2-14x-5\right)+\left(x+2\right)\left(34x-5\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(5x^2-14x-5+34x-5\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(5x^2+20x-10\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+4x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x^2+4x-2=0\left(1\right)\end{array}\right.\)
\(\left(1\right)\Leftrightarrow x^2+4x+4-6=0\)
\(\Leftrightarrow\left(x+2\right)^2=6\)
\(\Leftrightarrow x=\pm\sqrt{6}-2\)
Vậy pt có nghiệm là \(\left[\begin{array}{nghiempt}x=-2\\x=\pm\sqrt{6}-2\end{array}\right.\)
Để đẩu nữa mk về giải cho