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\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)
\(\Rightarrow8x+16-5x^2-10x+4x^2+4x-8x-8=x^2-4\)
\(\Rightarrow-6x-x^2-8-x^2+4=0\)
\(\Rightarrow-6x-2x^2-4=0\)
\(\Rightarrow-2\left(3x+x^2+2\right)=0\)
\(\Rightarrow\left(x+1,5\right)^2-0,25=0\)
\(\Rightarrow\left(x+2\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=-1\end{cases}}}\)
a) Ta có: \(8x\left(2x-3\right)-4x\left(4x+3\right)=72\)
\(\Leftrightarrow16x^2-24x-16x^2-12x=72\)
\(\Leftrightarrow-36x=72\)
hay x=-2
b) Ta có: \(\left(x+2\right)\left(x+4\right)-x\left(x+2\right)=104\)
\(\Leftrightarrow x^2+6x+8-x^2-2x=104\)
\(\Leftrightarrow4x=96\)
hay x=24
c) Ta có: \(\left(x-1\right)\left(x+4\right)-x\left(x-1\right)=308\)
\(\Leftrightarrow x^2+3x-4-x^2+x=308\)
\(\Leftrightarrow4x=312\)
hay x=78
d) Ta có: \(15x\left(2x-3\right)-\left(5x+2\right)\left(6x-5\right)=-22\)
\(\Leftrightarrow30x^2-45x-30x^2+25x-12x+10=-22\)
\(\Leftrightarrow-32x=-32\)
hay x=1
\(\dfrac{x-5}{2012}+\dfrac{x-4}{2013}=\dfrac{x-3}{2014}+\dfrac{x-2}{2015}\)
\(\Rightarrow\left(\dfrac{x-5}{2012}-1\right)+\left(\dfrac{x-4}{2013}-1\right)=\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-2}{2015}-1\right)\)
\(\Leftrightarrow\dfrac{x-2017}{2012}+\dfrac{x-2017}{2013}=\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}\)
\(\Leftrightarrow\dfrac{x-2017}{2012}+\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}-\dfrac{x-2017}{2015}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\right)=0\)
\(\Rightarrow x-2017=0\Leftrightarrow x=2017\)
Vậy x = 2017
Ta có: \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+5x-10=3x^2-12x-5x+20\)
\(\Leftrightarrow-2x-22+17x-20=0\)
\(\Leftrightarrow15x=42\)
hay \(x=\dfrac{14}{5}\)
( x+2)(x+5)(x+4)(x+3) = 24
<=> (x2 + 5x + 2x + 10)( x2 + 3x+4x+12 ) = 24
<=> ( x2 +7x+10)(x2+7x+12) = 24
Đặt x2 + 7x = t
Thay t vào phương trình , ta có
( t + 10)(t+12) = 24
<=> t2 + 12t + 10t + 120 - 24 = 0
<=> t2 + 22t + 96 = 0
<=> t2 + 6t + 16t + 96 = 0
<=> t( t+6)+16(t+6) = 0
<=> (t+16)(t+6) = 0
=> t+ 16 = 0 => t= -16
hoặc t+6=0 => t= - 6
rồi từ đó giải phương trình x2+ 7x = -16 và phương trình x2+7x = -6
x là tất cả các giá trị tìm được
a: \(\Leftrightarrow\left(2-x\right)\left(x-3\right)+\left(x-1\right)\left(x+3\right)=-4x\)
\(\Leftrightarrow2x-6-x^2+3x+x^2+3x-x-3=-4x\)
=>7x-9=-4x
=>11x=9
hay x=9/11
b: \(\Leftrightarrow\left(5-x\right)\left(x-4\right)+\left(x+2\right)\left(x+4\right)=-3x\)
\(\Leftrightarrow5x-20-x^2+4x+x^2+6x+8=-3x\)
=>15x-12=-3x
=>18x=12
hay x=2/3
\(x^4+x^3+x+1=x^5-1=\left(x-1\right)\left(x^4+x^3+x^2+x+1\right)\\ \)
có vẻ VP bạn ghi thiếu hệ số x^2 xem lại xem đúng chưa