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a)\(x^2+x-x^2+2=0\)\(\Rightarrow x+2=0\)\(\Rightarrow x=-2\)
b)\(2\left(3x+2\right)-2\left(x+6\right)=0\)
\(\Rightarrow2\left(3x+2-x-6\right)=0\)
\(\Rightarrow2\left(2x-4\right)=0\)
\(\Rightarrow2x-4=0\Rightarrow x=2\)
c)\(4x^4-6x^3-4x^4+6x^3-2x^2=0\)
\(\Rightarrow-2x^2=0\Rightarrow x=0\)
d)\(\left(3x^2-x-2\right)-3\left(x^2-x-2\right)=4\)
\(\Rightarrow3x^2-x-2-3x^2+3x+6=4\)
\(\Rightarrow2x+4=4\Rightarrow2x=0\Rightarrow x=0\)
a) (x + 6)(3x + 1) + x2 - 36 = 0
<=> 3x2 + x + 18x + 6 + x2 - 36 = 0
<=> 4x2 + 19x - 30 = 0
<=> 4x2 + 24x - 5x - 30 = 0
<=> 4x(x + 6) - 5(x + 6) = 0
<=> (x + 6)(4x - 5) = 0
<=> x + 6 = 0 hoặc 4x - 5 = 0
<=> x = -6 hoặc x = 5/4
Bài 1 mình đã làm xong rồi, anh em nào giúp mình bài 2 với!
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rồi tìm x
nha bn
a) \(\left(2x-3\right)^2-\left(2x+5\right)^2=10\)
\(\Leftrightarrow4x^2-12x+9-4x^2-20x-25-10=0\)
\(\Leftrightarrow-32x-26=0\)
\(\Leftrightarrow-32x=26\)
\(\Rightarrow x=-\frac{13}{16}\)
b) \(4\left(x+1\right)^2+\left(2x-1\right)^2+8\left(x-1\right)\left(x+1\right)=11\)
\(\Leftrightarrow4x^2+8x+4+4x^2-4x+1+8x^2-8=0\)
\(\Leftrightarrow16x^2+4x-3=0\)
\(\Leftrightarrow4\left(4x^2+x+\frac{1}{16}\right)-\frac{13}{4}=0\)
\(\Leftrightarrow\left[2\left(2x+\frac{1}{4}\right)\right]^2-\left(\frac{\sqrt{13}}{2}\right)^2=0\)
\(\Leftrightarrow\left(4x+\frac{1}{2}-\frac{\sqrt{13}}{2}\right)\left(4x+\frac{1}{2}+\frac{\sqrt{13}}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x+\frac{1-\sqrt{13}}{2}=0\\4x+\frac{1+\sqrt{13}}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}-1}{8}\\x=\frac{-1-\sqrt{13}}{8}\end{cases}}\)
c) \(\left(x+5\right)^2=45+x^2\)
\(\Leftrightarrow x^2+10x+25-x^2-45=0\)
\(\Leftrightarrow10x-20=0\)
\(\Leftrightarrow10x=20\)
\(\Rightarrow x=2\)
d) \(\left(2x-3\right)^2-\left(2x-1\right)^2=-3\)
\(\Leftrightarrow4x^2-12x+9-4x^2+4x-1+3=0\)
\(\Leftrightarrow-8x+11=0\)
\(\Leftrightarrow-8x=-11\)
\(\Rightarrow x=\frac{11}{8}\)
e) \(\left(x-1\right)^2-\left(5x-3\right)^2=0\)
\(\Leftrightarrow\left(x-1-5x+3\right)\left(x-1+5x-3\right)=0\)
\(\Leftrightarrow\left(-4x+2\right)\left(6x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-4x+2=0\\6x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{2}{3}\end{cases}}\)
\(A\))\(\left(x-1\right)^2+\left(x-3\right)^2-2x^2+1=0\)
\(x^2-2x+1+x^2-6x+9-2x^2+1=0\)
\(11-8x=0\)
\(\Rightarrow x=\frac{11}{8}\)
\(B\))\(\left(x-1\right)\left(x^2+x+1\right)-\left(x+1\right)\left(x^2-x+1\right)+2x=0\)
\(x^3-1-x^3-1+2x=0\)
\(2x-2=0\)
\(\Rightarrow x=1\)
\(A=\left(x-1\right)^2+\left(x-3\right)^2-2x^2+1=0\)
\(\Rightarrow x^2-2x+1+x^2-6x+9-2x^2+1=0\)
\(\Rightarrow\left(x^2+x^2-2x^2\right)+\left(-2x-6x\right)+\left(1+9+1\right)\)
\(\Rightarrow-8x+12=0\Leftrightarrow x=\frac{-11}{-8}=\frac{11}{8}\)
\(B=\left(x-1\right).\left(x^2+x-1\right)-\left(x+1\right).\left(x^2-x+1\right)+2x=0\)
\(\Rightarrow x.\left(x^2+x-1\right)-x^2-x+1-x.\left(x^2-x+1\right)-x^2+x-1+2x=0\)
\(\Rightarrow x^3+x^2-1-x^2-x+1-x^3+x^2-x-x^2+x-1+2x=0\)
\(\Rightarrow\left(x^3-x^3\right)+\left(x^2-x^2+x^2-x^2\right)+\left(-1+1-1\right)+\left(-x-x+x\right)+2x=0\)
\(\Rightarrow-1+x=0\Leftrightarrow x=1\)
\(C=\left(x-5\right).\left(x-5\right)+\left(2x+1\right)^2-3x^2=0\)
\(\Rightarrow x.\left(x-5\right)-5.\left(x-5\right)+\left(2x\right)^2+2.2x.1+1^2-3x^2=0\)
\(\Rightarrow x^2-5x-5x+25+4x^2+4x+1-3x^2=0\)
\(\Rightarrow\left(x^2-3x^2+4x^2\right)+\left(-5x-5x+4x\right)+26=0\)
\(\Rightarrow2x^2-6x+26=0\Leftrightarrow x=\)
\(D=\left(x-1\right)-9=0\Leftrightarrow x-1=9\Leftrightarrow x=10\)
a) x(x-1) + (x - 2)2 = 1
<=> x2 -x + x2 - 4x + 4 =1
<=>2x2 -5x + 3 =0
<=> 2x2 - 2x - 3x +3 =0
<=> 2x(x-1) - 3(x-1)=0
<=> (x - 1)(2x - 3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy
b) (x + 5)(x + 3) - (x - 2)2 = -1
<=> x2 + 3x + 5x + 15 - x2 +4x - 4 = -1
<=>12x + 12= 0
<=> 12(x - 1)=0
<=> x -1 = 0
<=.x=1
Vậy
c) x(2x-2) - (x-2)(2x+3)=0
<=>2x2 - 2x - 2x2 -3x + 4x +6 =0
<=> -x + 6 = 0
<=> x=6
vậy
d) 3x3 - 27x =0
<=> 3x(x2 - 9)=0
<=> 3x(x-3)(x+3)=0
<=>\(\left[{}\begin{matrix}3x=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
vậy