Giảipt
\(\left(x+1\right)^4-\left(x^2+2\right)^2\)
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5(x² + xy + y²) = 7(x + 2y)
<=> 5[ x^2/4 + xy + y^2 + 3x^2/4] = 7.(x+2y)
<=> 5[ (x/2 +y)^2 + 3x^2/4 ] = 7.(x+2y)
<=> 5.[(x+2y)^2 +3x^2 ] = 28(x+2y)
đặt a = x+2y ta có:
5[ a^2 +3x^2 ] = 28a
<=> 15x^2 = 28a - 5a^2
nhận thấy Vế trái >= 0 => Vế Phải >=0 => 28a - 5a^2 >=0 => a(5a-28) <=0 => 0 <= a<= 28/5
=> 0<= a<=5
5[ a^2 +3x^2 ] = 28a : dễ thấy 28a chia hết cho 5 mà do(28;5) = 1 => a chia hết cho 5
=> a = 5 hoặc a = 0
nếu a = 0 ; x+ 2y = 0 thì 5[ a^2 +3x^2 ] = 28a <=> 3x^2 = 0 <=> x = 0 => y = 0
nếu a = 5 ; x+2y = 5 rhì 5(25 + 3x^2) = 5 <=> 3x^2 +24 = 0 vô lý vì 3x^2 + 24 >0
vậy pt có nghiệm nguên duy nhất x = y = 0
\(1,2\left(x-3\right)+1=2\left(x+1\right)-9\\ \Rightarrow2x-6+1=2x+2-9\\ \Rightarrow2x-5=2x-7\\ \Rightarrow-2=0\left(vô.lí\right)\)
\(2,\dfrac{5-x}{2}=\dfrac{3x-4}{6}\\ \Rightarrow30-6x=6x-8\\ \Rightarrow12x=38\\ \Rightarrow x=\dfrac{19}{6}\)
\(3,\left(x-1\right)^2+\left(x+2\right)\left(x-2\right)=\left(2x+1\right)\left(x-3\right)\\ \Rightarrow x^2-2x+1+x^2-4=2x^2-6x+x-3\\ \Rightarrow2x^2-2x-3=2x^2-5x-3\\ \Rightarrow3x=0\\ \Rightarrow x=0\)
\(4,\left(x+5\right)\left(x-1\right)-\left(x+1\right)\left(x+2\right)=1\\ \Rightarrow x^2+5x-x-5-x^2-2x-x-2=1\\ \\ \Rightarrow x-7=1\\ \Rightarrow x=8\)
\(5,\dfrac{6x-1}{15}-\dfrac{x}{5}=\dfrac{2x}{3}\\ \Rightarrow\dfrac{6x-1}{15}-\dfrac{3x}{15}=\dfrac{10x}{15}\\ \Rightarrow6x-1-3x=10x\\ \Rightarrow3x-1=10x\\ \Rightarrow7x=-1\\ \Rightarrow x=\dfrac{-1}{7}\)
\(6,\dfrac{5\left(x-2\right)}{2}-\dfrac{x+5}{3}=1-\dfrac{4\left(x-3\right)}{5}\\ \Rightarrow\dfrac{75\left(x-2\right)}{30}-\dfrac{10\left(x+5\right)}{30}=\dfrac{30}{30}-\dfrac{24\left(x-3\right)}{30}\\ \Rightarrow75\left(x-2\right)-10\left(x+5\right)=30-24\left(x-3\right)\\ \Rightarrow75x-150-10x-50=30-24x+72\\ \Rightarrow65x-200=102-24x\\ \Rightarrow89x=302\\ \Rightarrow x=\dfrac{320}{89}\)
a: Ta có: \(\left(3x-1\right)^2-2\left(5x-2\right)^2-2\left(x^2+x-1\right)\left(x-1\right)\)
\(=9x^2-6x+1-2\left(25x^2-20x+4\right)-2\left(x^3-x^2+x^2-x-x+1\right)\)
\(=9x^2-6x+1-50x^2+40x-8-2\left(x^3-2x+1\right)\)
\(=-41x^2+34x-7-2x^3+4x-2\)
\(=-2x^3-41x^2+38x-9\)
b: Ta có: \(\left(3a+1\right)^2+2\left(9a^2-1\right)+\left(3a-1\right)^2\)
\(=\left(3a+1+3a-1\right)^2\)
\(=36a^2\)
a: Ta có: \(\left(7x+4\right)^2-\left(7x-4\right)\left(7x+4\right)\)
\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)
\(=8\left(7x+4\right)\)
=56x+32
b: Ta có: \(8\left(x-2\right)^2-3\left(x^2-4x-5\right)-5x^2\)
\(=8x^2-32x+32-3x^2+12x+15-5x^2\)
\(=-20x+47\)
c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(x+1\right)\)
\(=x^3+3x^2+3x+1-x^3+1-3x^2-3x\)
=2
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
\(\left(x+1\right)^4-\left(x^2+2\right)^2=0\)
\(\Leftrightarrow4x^3+2x^2+4x-3=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x^2+2x+3\right)=0\)
Mà \(2x^2+2x+3\ne0\) nên:
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy: nghiệm phương trình là \(x=\frac{1}{2}\)