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PN
13 tháng 6 2019
a) \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)-3=-3\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3-3=-3\)
\(\Leftrightarrow14x=0\)
\(\Leftrightarrow x=0\)
Vậy pt có nghiệm duy nhất x = 0.
b) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=\left(x+2\right)-\left(x-5\right)\)
\(\Leftrightarrow6x^2+19x-7-6x^2-x+5=7\)
\(\Leftrightarrow18x-2=7\)
\(\Leftrightarrow18x=9\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy pt có nghiệm duy nhất \(x=\frac{1}{2}\)
c) \(\left(6x-2\right)^2+\left(5x-2\right)^2-4\left(3x-1\right)\left(5x-2\right)=0\)
\(\Leftrightarrow36x^2-24x+4+25x^2-20x+4-60x^2+33x-8=0\)
\(\Leftrightarrow x^2-11x=0\)
\(\Leftrightarrow x\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=11\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{0;11\right\}\)
d) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow x^2-6x+9-x^2-4x+32=1\)
\(\Leftrightarrow41-10x=1\)
\(\Leftrightarrow-10x=40\)
\(\Leftrightarrow x=-4\)
Vậy pt có nghiệm duy nhất x = -4.
e) \(3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\)
\(\Leftrightarrow3\left(x^2+4x+4\right)+4x^2-4x+1-7x^2+36=36\)
\(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2=0\)
\(\Leftrightarrow8x=-13\)
\(\Leftrightarrow x=-\frac{13}{8}\)
Vậy pt có nghiệm duy nhất \(x=-\frac{13}{8}\)
12 tháng 6 2019
a) \(2x\left(x-3\right)-x\left(2x+1\right)-3\left(x+5\right)=11\)
\(\Rightarrow2x^2-6x-2x^2-x-3x-15=11\)
\(\Rightarrow-10x=26\Rightarrow x=-2,6\)
Vậy ...........
b) \(x\left(x-1\right)-\left(x^2+3x-5\right)-2\left(x+3\right)=17\)
\(\Rightarrow x^2-x-x^2-3x+5-2x-6=17\)
\(\Rightarrow-6x=18\Rightarrow x=-3\)
c) \(5x\left(x-7\right)-\left(5x+1\right)x-\left(x+3\right)2=13\)
\(\Rightarrow5x^2-35x-5x^2-x-2x-6=13\)
\(\Rightarrow-38x=19\Rightarrow x=-\frac{1}{2}\)
d) \(\left(2x^2-3x+5\right)-2x\left(x-3\right)+\left(x-1\right)\left(-2\right)=10\)
\(\Rightarrow2x^2-3x+5-2x^2+6x-2x+2=10\)
\(\Rightarrow x=3\)
20 tháng 7 2020
Tìm x
a) Ta có: \(3\left(1-4x\right)\left(x-1\right)+4\left(3x+2\right)\left(x+3\right)=38\)
\(\Leftrightarrow3\left(x-1-4x^2+4x\right)+4\left(3x^2+9x+2x+6\right)=38\)
\(\Leftrightarrow3\left(-4x^2+5x-1\right)+4\left(3x^2+11x+6\right)-38=0\)
\(\Leftrightarrow-12x^2+15x-3+12x^2+44x+24-38=0\)
\(\Leftrightarrow59x-17=0\)
\(\Leftrightarrow59x=17\)
hay \(x=\frac{17}{59}\)
Vậy: \(x=\frac{17}{59}\)
b) Ta có: \(5\left(2x+3\right)\left(x+2\right)-2\left(5x-4\right)\left(x-1\right)=75\)
\(\Leftrightarrow5\left(2x^2+4x+3x+6\right)-2\left(5x^2-5x-4x+4\right)-75=0\)
\(\Leftrightarrow5\left(2x^2+7x+6\right)-2\left(5x^2-9x+4\right)-75=0\)
\(\Leftrightarrow10x^2+35x+30-10x^2+18x-8-75=0\)
\(\Leftrightarrow53x-53=0\)
\(\Leftrightarrow53x=53\)
hay x=1
Vậy: x=1
c) Ta có: \(2x^2+3\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\)
\(\Leftrightarrow2x^2+3x^2-3=5x^2+5x\)
\(\Leftrightarrow5x^2-3-5x^2-5x=0\)
\(\Leftrightarrow-3-5x=0\)
\(\Leftrightarrow-5x=-3\)
hay \(x=\frac{3}{5}\)
Vậy: \(x=\frac{3}{5}\)
d) Ta có: \(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow8x+16-5x^2-10x+4\left(x^2+x-2x-2\right)+2\left(x^2-4\right)=0\)
\(\Leftrightarrow-5x^2-2x+16+4x^2-4x-8+2x^2-8=0\)
\(\Leftrightarrow x^2-6x=0\)
\(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: \(x\in\left\{0;6\right\}\)
b) 5( 2x + 3 )( x + 2 ) - 2( 5x - 4 )( x - 1 ) = 75
<=> 5( 2x2 + 7x + 6 ) - 2( 5x2 - 9x + 4 ) = 75
<=> 10x2 + 35x + 30 - 10x2 + 18x - 8 = 75
<=> 53x + 22 = 75
<=> 53x = 53
<=> x = 1
c) 2x2 + 3( x - 1 )( x + 1 ) = 5x( x + 1 )
<=> 2x2 + 3( x2 - 1 ) = 5x2 + 5x
<=> 2x2 + 3x2 - 3 - 5x2 - 5x = 0
<=> -3 - 5x = 0
<=> -5x = 3
<=> x = -3/5
b, \(5\left(2x+3\right)\left(x+2\right)-2\left(5x-4\right)\left(x-1\right)=75\)
\(\Leftrightarrow10x^2+20x+15x+30-10x^2+10x+8x-8=75\)
\(\Leftrightarrow53x+22=75\Leftrightarrow x=1\)