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a, \(2x\left(x-3\right)-15+5x=0\\ \Rightarrow2x\left(x-3\right)-\left(15-5x\right)=0\\ \Rightarrow2x\left(x-3\right)-5\left(3-x\right)=0\\ \Rightarrow\left(2x+5\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=3\end{matrix}\right.\)
b, \(x^3-7x=0\\ \Rightarrow x\left(x^2-7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm7\end{matrix}\right.\)
c, \(\left(2x-3\right)^2-\left(x+5\right)^2=0\\ \Rightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\\ \Rightarrow\left(x-8\right)\left(3x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Xem lại đề câu d
Bài 2 :
\(A=4x^2-2.2x.2+4+1\)
\(=\left(2x-2\right)^2+1\)
Thấy : \(\left(2x-2\right)^2\ge0\)
\(A=\left(2x-2\right)^2+1\ge1\)
Vậy \(MinA=1\Leftrightarrow x=1\)
\(B=\left(5x\right)^2-2.5x.1+1-4\)
\(=\left(5x-1\right)^2-4\)
Thấy : \(\left(5x-1\right)^2\ge0\)
\(\Rightarrow B=\left(5x-1\right)^2-4\ge-4\)
Vậy \(MinB=-4\Leftrightarrow x=\dfrac{1}{5}\)
\(C=\left(7x\right)^2-2.7x.2+4-5\)
\(=\left(7x-2\right)^2-5\)
Thấy : \(\left(7x-2\right)^2\ge0\)
\(\Rightarrow C=\left(7x-2\right)^2-5\ge-5\)
Vậy \(MinC=-5\Leftrightarrow x=\dfrac{2}{7}\)
\(1.\)
\(A=-x^2-10x+1=-\left(x^2+10x-1\right)\)
\(=-\left(x^2+2.5x+5^2-5^2-1\right)=-\left[\left(x+5\right)^2-26\right]\)
\(=-\left(x+5\right)^2+26\le26\) dấu "=" xảy ra<=>x=-5
\(B=-4x^2-6x-5=-4\left(x^2+\dfrac{6}{4}x+\dfrac{5}{4}\right)\)
\(=-4\left(x^2+2.\dfrac{3}{4}x+\dfrac{9}{16}+\dfrac{11}{16}\right)\)\(=-4\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{6}\right]\le-\dfrac{11}{4}\)
\(C=-16x^2+8x-1=-16\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)\)
\(=-16\left(x^2-2.\dfrac{1}{4}x+\dfrac{1}{16}\right)=-16\left(x-\dfrac{1}{4}\right)^2\le0\)
dấu"=" xảy ra<=>x=1/4
a,4x^2-4x+1=0
4x^2-2x-2x+1=0
2x (2x-1)-(2x-1)=0
(2x-1)(2x-1)=0
(2x-1)^2=0
=>2x-1=0 <=> x=1/2
a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)
\(\Leftrightarrow\left(x-3\right)^2-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)
\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
hay \(x=-\dfrac{1}{4}\)
c) Ta có: \(8x^3-50x=0\)
\(\Leftrightarrow2x\left(4x^2-25\right)=0\)
\(\Leftrightarrow x\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
Lời giải:
PT $\Leftrightarrow 8x^3-16x^2+6x+2=0$
$\Leftrightarrow (8x^3-8x^2)-(8x^2-8x)-(2x-2)=0$
$\Leftrightarrow 8x^2(x-1)-8x(x-1)-2(x-1)=0$
$\Leftrightarrow (x-1)(8x^2-8x-2)=0$
$\Leftrightarrow 2(x-1)(4x^2-4x-1)=0$
$\Leftrightarrow x-1=0$ hoặc $4x^2-4x-1=0$
Nếu $x-1=0\Leftrightarrow x=1$
Nếu $4x^2-4x-1=0$
$\Leftrightarrow (2x-1)^2-2=0$
$\Leftrightarrow (2x-1-\sqrt{2})(2x-1+\sqrt{2})=0$
$\Leftrightarrow x=\frac{1\pm \sqrt{2}}{2}$
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
a: ĐKXD: x<>0
\(\dfrac{14x^3+12x^2-14x}{2x}=\left(x+2\right)\left(3x-4\right)\)
=>\(\dfrac{2x\left(7x^2+6x-7\right)}{2x}=\left(x+2\right)\left(3x-4\right)\)
=>\(7x^2+6x-7=3x^2-4x+6x-8\)
=>\(7x^2+6x-7=3x^2+2x-8\)
=>\(4x^2+4x+1=0\)
=>\(\left(2x+1\right)^2=0\)
=>2x+1=0
=>x=-1/2(nhận)
b: \(\left(4x-5\right)\left(6x+1\right)-\left(8x+3\right)\left(3x-4\right)=15\)
=>\(24x^2+4x-30x-5-\left(24x^2-32x+9x-12\right)=15\)
=>\(24x^2-26x-5-24x^2+23x+12=15\)
=>-3x+7=15
=>-3x=8
=>\(x=-\dfrac{8}{3}\)
*) \(4x^2-4x=-1\Leftrightarrow4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\)
\(\Leftrightarrow2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\) vậy \(x=\dfrac{1}{2}\)
*)\(8x^3+12x^2+6x+1=0\Leftrightarrow8x^3+8x^2+2x+4x^2+4x+1\)
\(\Leftrightarrow2x\left(4x^2+4x+1\right)+\left(x^2+4x+1\right)=0\Leftrightarrow\left(2x+1\right)\left(4x^2+4x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1\right)^2=0\Leftrightarrow\left(2x+1\right)^3=0\Leftrightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\) vậy \(x=\dfrac{-1}{2}\)
a,42 - 4x = -1
=> 4x2 - 4x + 1 = 0
=> (2x - 1)2 = 0
=> 2x -1 = 0
=> x = \(\dfrac{1}{2}\)
b, 8x3 +12x2 + 6x +1=0
=> ( 2x +1 )3 = 0
=> 2x + 1 = 0
=> x = \(\dfrac{-1}{2}\)
a, 4x^2 - 4x = -1
\(\Leftrightarrow\)4x^2 - 4x + 1 = 0
\(\Leftrightarrow\)(2x-1)2 =0
\(\Leftrightarrow\)2x - 1 = 0
\(\Leftrightarrow\)x = 1/2
b, \(\Leftrightarrow\)( 2x + 1)^3 = 0
\(\Leftrightarrow\)2x + 1 = 0
\(\Leftrightarrow\)x = -1/2
đúng thì
a) \(4x^2-4x=-1\)
\(\Leftrightarrow4x^2-4x+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
b) \(8x^3+12x^2+6x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^3=0\)
\(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow x=-\frac{1}{2}\)