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\(a,\Leftrightarrow\left(x-4\right)\left(x^2+5\right)>0\\ \Leftrightarrow x-4>0\left(x^2+5\ge5>0\right)\\ \Leftrightarrow x>4\\ b,\Leftrightarrow\left(x-y\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=y\left(vô.lí.do.x\ne y\right)\\x=\dfrac{5}{3}\left(tm\right)\end{matrix}\right.\\ \Leftrightarrow S=x^2-x=\dfrac{25}{9}-\dfrac{5}{3}=\dfrac{10}{9}\)
a: Ta có: \(40x^4+5x=0\)
\(\Leftrightarrow5x\left(8x^3+1\right)=0\)
\(\Leftrightarrow x\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
b: Ta có: \(8x^2-2x-1=0\)
\(\Leftrightarrow8x^2-4x+2x-1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)
a, 4x2 - 49 = 0
⇔⇔ (2x)2 - 72 = 0
⇔⇔ (2x - 7)(2x + 7) = 0
⇔{2x−7=02x+7=0⇔⎧⎪ ⎪⎨⎪ ⎪⎩x=72x=−72⇔{2x−7=02x+7=0⇔{x=72x=−72
b, x2 + 36 = 12x
⇔⇔ x2 + 36 - 12x = 0
⇔⇔ x2 - 2.x.6 + 62 = 0
⇔⇔ (x - 6)2 = 0
⇔⇔ x = 6
e, (x - 2)2 - 16 = 0
⇔⇔ (x - 2)2 - 42 = 0
⇔⇔ (x - 2 - 4)(x - 2 + 4) = 0
⇔⇔ (x - 6)(x + 2) = 0
⇔{x−6=0x+2=0⇔{x=6x=−2⇔{x−6=0x+2=0⇔{x=6x=−2
f, x2 - 5x -14 = 0
⇔⇔ x2 + 2x - 7x -14 = 0
⇔⇔ x(x + 2) - 7(x + 2) = 0
⇔⇔ (x + 2)(x - 7) = 0
⇔{x+2=0x−7=0⇔{x=−2x=7
Bài 1:
a: \(=6x^3-10x^2+6x\)
b: \(=-2x^3-10x^2-6x\)
Bài 4:
a: =>3x+10-2x=0
=>x=-10
c: =>3x2-3x2+6x=36
=>6x=36
hay x=6
Bài 1:
\(a,=6x^3-10x^2+6x\\ b,=-2x^3-10x^2-6x\)
Bài 4:
\(a,\Leftrightarrow3x+10-2x=0\Leftrightarrow x=-10\\ b,\Leftrightarrow x\left(2x^2+9x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\\ \Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\\ \Leftrightarrow-6x=8\Leftrightarrow x=-\dfrac{4}{3}\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\)
Bài 1:
\(a,=7xy\left(2x-3y+4xy\right)\\ b,=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\\ c,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ d,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\\ =2x\left(4x+2\right)=4x\left(2x+1\right)\\ e,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x^2+8x-x-8=\left(x+8\right)\left(x-1\right)\\ g,\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\\ h,=x^2+3x+x+3=\left(x+3\right)\left(x+1\right)\)
a: \(\left(x+5\right)^2-\left(x-5\right)^2-2x+1=0\)
=>\(x^2+10x+25-\left(x^2-10x+25\right)-2x+1=0\)
=>\(x^2+8x+26-x^2+10x-25=0\)
=>18x+1=0
=>\(x=-\dfrac{1}{18}\)
b: \(\left(2x-7\right)^2-\left(x+3\right)^2=3x^2+6\)
=>\(4x^2-28x+49-\left(x^2+6x+9\right)-3x^2-6=0\)
=>\(x^2-28x+43-x^2-6x-9=0\)
=>34-34x=0
=>34x=34
=>x=1
c: \(\left(3x+2\right)^2-9\left(x-5\right)\left(x+5\right)=225-5x\)
=>\(9x^2+12x+4-9\left(x^2-25\right)-225+5x=0\)
=>\(9x^2+17x+4-225-9x^2+225=0\)
=>17x+4=0
=>x=-4/17
Lời giải:
a.
PT $\Leftrightarrow 3x^2+\frac{x}{2}-3x^2+3x+2=0$
$\Leftrightarrow \frac{7}{2}x+2=0$
$\Leftrightarrow \frac{7}{2}x=-2$
$\Leftrightarrow x=-2: \frac{7}{2}=\frac{-4}{7}$
b.
PT $\Leftrightarrow 5x^2-3-5x^2-6x=0$
$\Leftrightarrow -3-6x=0$
$\Leftrightarrow 6x=-3$
$\Leftrightarrow x=\frac{-3}{6}=\frac{-1}{2}$
a) 2x.(3x2 – 5x + 3)
=2x3-10x2+6x
b(-2x-1).( x2 + 5x – 3 ) – (x-1)3
=-2x3 - 10x2 + 6x - x2 - 5x + 3 - x3 + 3x2 - 3x + 1
= -3x3 - 8x2 - 2x + 4
d) (6x5y2 – 9x4y3 + 15x3y4) : 3x3y2
=2x2-3xy+5y2
3x2 - 5x -2 =0
3x2 - 6x + x -2=0
3x(x - 2) + (x-2)=0
(x-2)(3x+1)=0
\(\orbr{\begin{cases}x-2=0\\3x+1=0\end{cases}}\)
\(\orbr{\begin{cases}x=2\\x=-\frac{1}{3}\end{cases}}\)
Vậy x = 2 hoặc x = -1/3
3x^2 - 5x + 2 = 0
3x^2 - 6x + x - 2 = 0
3x*(x-2) + 1*(x-2) = 0
(x-2)*(3x+1) = 0
x - 2 = 0 hoặc 3x + 1 = 0
x = 0 + 2 hoặc 3x = -1
x = 2 hoặc x = -1/3