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a) x3 + 3x2 + 3x + 1 = 64
=> (x + 1)3 = 64
=> (x + 1)3 = 43
=> x + 1 = 4 => x = 3
b) x3 + 6x2 + 9x = 4x
=> x3 + 6x2 + 9x - 4x = 0
=> x3 + 6x2 + 5x = 0
=> x3 + 5x2 + x2 + 5x = 0
=> x2(x + 5) + x(x + 5) = 0
=> (x + 5)(x2 + x) = 0
=> (x + 5)x(x + 1) = 0
=> \(\hept{\begin{cases}x=-5\\x=0\\x=-1\end{cases}}\)
c) 4(x - 2)2 = (x + 2)2
=> 4(x2 - 4x + 4) = x2 + 4x + 4
=> 4x2 - 16x + 16 = x2 + 4x + 4
=> 4x2 - 16x + 16 - x2 - 4x - 4 = 0
=> 3x2 - 20x + 12 = 0
=> 3x2 - 18x - 2x + 12 = 0
=> 3x(x - 6) - 2(x - 6) = 0
=> (x - 6)(3x - 2) = 0
=> \(\orbr{\begin{cases}x=6\\x=\frac{2}{3}\end{cases}}\)
d) x4 - 16x2 = 0
=> x2(x2 - 16) = 0
=> \(\orbr{\begin{cases}x^2=0\\x^2=16\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
e) x4 - 4x3 + x2 - 4x = 0
=> x4 + x2 - 4x3 - 4x = 0
=> x2(x2 + 1) - 4x(x2 + 1) = 0
=> (x2 - 4x)(x2 + 1) = 0
=> x(x - 4)(x2 + 1) = 0
=> \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)(vì x2 + 1 \(\ge\)1 > 0 \(\forall\)x)
f) x3 + x = 0 => x(x2 + 1) = 0 => x = 0 (vì x2 + 1 \(\ge1>0\forall\)x)
a) \(x^2-16=0\Rightarrow x^2=16\Rightarrow x^2=\pm4\)
b) \(4x^2-9=0\Rightarrow\left(2x-3\right)\left(2x+3\right)=0\Rightarrow x=\pm1,5\)
c) \(25x^2-1=0\Rightarrow\left(5x-1\right)\left(5x+1\right)=0\Rightarrow x=\pm0,2\)
d) \(4\left(x-1\right)^2-9=0\Rightarrow\left(2x-2-3\right)\left(2x-2+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-5=0\Rightarrow x=2,5\\2x+1=0\Rightarrow x=-0,5\end{matrix}\right.\)
e) \(25x^2-\left(5x+1\right)^2=0\Rightarrow\left(5x+5x+1\right)\left(5x-5x-1\right)=0\Rightarrow10x+1=0\Rightarrow x=-0,1\)
f) \(\dfrac{1}{4}-9\left(x-1\right)^2=0\Rightarrow\left(\dfrac{1}{2}+3x-3\right)\left(\dfrac{1}{2}-3x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{7}{6}\end{matrix}\right.\)
g) \(\dfrac{1}{16}-\left(2x+\dfrac{3}{4}\right)^2=0\Rightarrow\left(\dfrac{1}{4}+2x+\dfrac{3}{4}\right)\left(\dfrac{1}{4}-2x-\dfrac{3}{4}\right)=0\Rightarrow\left[{}\begin{matrix}x=-0,5\\x=-0,25\end{matrix}\right.\)
h) \(\dfrac{1}{9}x^2-\dfrac{2}{3}x+1=0\Rightarrow\left(\dfrac{1}{3}x-1\right)^2=0\Rightarrow\dfrac{1}{3}x=1\Rightarrow x=3\)
k) \(4\left(x-3\right)^2-\left(2-3x\right)^2=0\Rightarrow\left(2x-6+2-3x\right)\left(2x-6-2+3x\right)=0\Rightarrow\left[{}\begin{matrix}-x-4=0\Rightarrow x=-4\\5x-8=0\Rightarrow x=1,6\end{matrix}\right.\)
l) \(x^2-x-12=0\Rightarrow x^2-4x+3x-12=0\Rightarrow x\left(x-4\right)+3\left(x-4\right)=0\Rightarrow\left(x+3\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
Bài 1:Tìm x,y biết:
a)\(x^2-6x+y^2+10y+34\)
=>\(\left(x^2-2.x.3+3^2\right)+\left(y^2+2.y.5+5^2\right)=0\)
=>\(\left(x-3\right)^2+\left(y+5\right)^2=0\)
=>\(\left\{{}\begin{matrix}x-3=0\\y+5=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3\\y=-5\end{matrix}\right.\)
\(x^3+9x=0\)
<=> \(x\left(x^2+9\right)=0\)
<=> \(\orbr{\begin{cases}x=0\\x^2+9=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=0\\x\in\varnothing\end{cases}}\)
<=> \(x=0\)
\(9x^2-4-2\left(3x-2\right)^2=0\)
<=> \(\left(9x^2-4\right)-2\left(3x-2\right)^2=0\)
<=> \(\left[\left(3x\right)^2-2^2\right]-2\left(3x-2\right)^2=0\)
<=> \(\left(3x-2\right)\left(3x+2\right)-2\left(3x-2\right)^2=0\)
<=> \(\left(3x-2\right)\left[\left(3x+2\right)-2\left(3x-2\right)\right]=0\)
<=> \(\left(3x-2\right)\left(3x+2-6x+4\right)=0\)
<=> \(\left(3x-2\right)\left(-3x+6\right)=0\)
<=> \(\left(3x-2\right)3\left(-x+2\right)=0\)
<=> \(3\left(3x-2\right)\left(2-x\right)=0\)
<=> \(\orbr{\begin{cases}3x-2=0\\2-x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}3x=2\\x=2\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{2}{3}\\x=2\end{cases}}\)
\(\left(x^3-x^2\right)-4x+8x-4=0\)
<=> \(\left(x^3-x^2\right)+\left(4x-4\right)=0\)
<=> \(x^2\left(x-1\right)+4\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(x^2+4\right)=0\)
<=> \(\orbr{\begin{cases}x-1=0\\x^2+4=0\end{cases}}\)
<=> \(x=1\)
\(\left(25x^2-10x\right):\left(-5x\right)-3\left(x-2\right)=4\)
<=> \(5x\left(5x-2\right)\left(-\frac{1}{5x}\right)-3\left(x-2\right)=4\)
<=> \(-\left(5x-2\right)-3\left(x-2\right)=4\)
<=> \(\left(5x-2\right)+3\left(x-2\right)=-4\)
<=> \(5x-2+3x-6=-4\)
<=> \(8x-8=-4\)
<=> \(8\left(x-1\right)=-4\)
<=> \(x-1=-\frac{1}{2}\)
<=> \(x=-\frac{3}{2}\)
\(a,\)\(x^4-4x^3+4x^2=0\)
\(\Leftrightarrow x^2.\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x^2.\left(x^2-2.x.2+2^2\right)=0\)
\(\Leftrightarrow x^2.\left(x-2\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(b,\)\(x^2+5x+4=0\)
\(\Leftrightarrow x^2+x+4x+4=0\)
\(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)
\(c,\)\(9x-6x^2-3=0\)
\(\Leftrightarrow-3.\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow2x^2-2x-x+1=0\)
\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right).\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
\(d,\)\(2x^2+5x+2=0\)
\(\Leftrightarrow2x^2+4x+x+2=0\)
\(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\2x=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}}\)
\(a,x^3-13x=0\)
\(x.\left(x^2-13\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=13\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\sqrt{13}\end{cases}}}\)
\(b,2-25x^2=0\)
\(\Rightarrow25x^2=2\Rightarrow x^2=\frac{2}{25}\Rightarrow x=\sqrt{\frac{2}{25}}\)
\(c,x^2-x+\frac{1}{4}=0\)
\(\left(x-\frac{1}{2}\right)^2=0\Rightarrow x=\frac{1}{2}\)
a, x 3 - 13 x = 0
=> x ( x 2 - 13 ) = 0
=> \(\orbr{\begin{cases}x=0\\x^2=13\end{cases}\Rightarrow[\begin{cases}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{cases}}\)
b, 2 - 25 x 2 = 0
=> 25 x 2 = 2
=> x 2 = 0,08
=> \(\orbr{\begin{cases}x=\frac{\sqrt{2}}{5}\\x=\frac{-\sqrt{2}}{5}\end{cases}}\)
x, x 2 - x + \(\frac{1}{4}\)= 0
=> \(\left(x-\frac{1}{2}\right)^2=0\)
=> \(x-\frac{1}{2}=0\)
=> \(x=\frac{1}{2}\)
Bài làm :
\(a\text{)}3x^2+4x=0\Leftrightarrow x\left(3x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\3x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{4}{3}\end{cases}}\)
\(b\text{)}25x^2-0,64=0\Leftrightarrow\left(5x-0,8\right)\left(5x+0,8\right)=0\Leftrightarrow\orbr{\begin{cases}5x-0,8=0\\5x+0,8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0,16\\-0,16\end{cases}}\)
\(c\text{)}x^4-16x^2=0\Leftrightarrow\left(x^2-4x\right)\left(x^2+4x\right)=0\Leftrightarrow\orbr{\begin{cases}x^2-4x=0\\x^2+4x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x-4\right)=0\\x\left(x+4\right)=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
\(d\text{)}x^2+x=6\Leftrightarrow x^2+x-6=0\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
Bài làm :
\(a)3x^2+4x=0\)
\(\Rightarrow x\left(3x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\3x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-4}{3}\end{cases}}\)
Vậy x = 0 hoặc \(x=\frac{-4}{3}\) .
\(b)25x^2-0,64=0\)
\(\Rightarrow\left(5x\right)^2=\frac{16}{25}\)
\(\Rightarrow\left(5x\right)^2=\left(\frac{4}{5}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}5x=\frac{4}{5}\\5x=\frac{-4}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{4}{25}\\x=\frac{-4}{25}\end{cases}}\)
Vậy \(x=\frac{4}{25}\) hoặc \(x=\frac{-4}{25}\) .
\(c)x^4-16x^2=0\)
\(\Rightarrow x^2\left(x^2-16\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2=0\\x^2-16=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=4^2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
Vậy x = 0 hoặc \(x=\pm4\) .