1: \(\Leftrightarrow5x^2+4x-1-2x^2+12x-18=3x^2+5x-2-x^2-8x-16+x^2-x\)
\(\Leftrightarrow3x^2+16x-19=3x^2-4x-18\)
=>20x=1
hay x=1/20
2: \(\Leftrightarrow5x^2-20x-41=x^2-10x+25+4x^2+4x+1-\left(x^2-2x\right)+\left(x-1\right)^2\)
\(\Leftrightarrow5x^2-20x-41=4x^2-4x+26+x^2-2x+1\)
\(\Leftrightarrow-20x-41=-6x+27\)
=>-14x=68
hay x=-34/7
a) ( 3x - 1 ) ( 2x + 7 ) - ( x + 1 ) ( 6x + 5 ) = 16
<=> 6x2 + 21x - 2x - 7 - ( 6x2 - 5x + 6x - 5) = 16
<=> 6x2 + 21x - 2x - 7 - ( 6x2 + x - 5 ) = 16
<=> 6x2+ 21x - 2x - 7 - 6x2 -x + 5 = 16
<=> 18x - 2 = 16
<=> 18x = 18
=> x = 1
Vậy....
a)
$(3x-1)(2x+7)-(x+1)(6x-5)=16$
$6x^2+19x-7-(6x^2+x-5)=16$
$18x-2=16$
$18x=18$
x = 1
b)
$(10x+9)x-(5x-1)(2x+3)=8$
$10x^2+9x-(10x^2+13x-3)=8$
$-4x+3=8$
$-4x=5$
x= -5/4
c) Hiểu đề là $(3x-5)(7-5x)+(5x+2)(2x+3)-2=0$.
$(3x-5)(7-5x)+(5x+2)(2x+3)-2=0$
$-15x^2+46x-35+10x^2+19x+6-2=0$
$-5x^2+65x-31=0$
$5x^2-65x+31=0$
$x=\frac{65\pm\sqrt{3605}}{10}$
d)
$x(x+1)(x+6)-x^3=5x$
$x(x^2+7x+6)-x^3=5x$
$x^3+7x^2+6x-x^3=5x$
$7x^2+x=0$
$x(7x+1)=0$
=> x = 0 hoặc x = -1/7
Giải:
1) \(\left(x-6\right)\left(x^2+6x+36\right)-\left(x+4\right)^3=\left(x-2\right)^3+\left(x+5\right)\left(x^2-10x+25\right)-\left(2x^3+6x^2\right)\)
\(\Leftrightarrow x^3-216-\left(x^3+12x^2+48x+64\right)=x^3-6x^2+12x-8+x^3+125-2x^3-6x^2\)
\(\Leftrightarrow x^3-216-x^3-12x^2-48x-64=x^3-6x^2+12x-8+x^3+125-2x^3-6x^2\)
\(\Leftrightarrow-280-12x^2-48x=-12x^2+12x+117\)
\(\Leftrightarrow-280-48x-12x-117=0\)
\(\Leftrightarrow-397-60x=0\)
\(\Leftrightarrow-60x=397\)
\(\Leftrightarrow x=-\dfrac{397}{60}\)
Vậy ...
2) \(\left(2x+3\right)^3-\left(2x+5\right)\left(4x^2-10x+25\right)=\left(6x-1\right)^2-\left(x-2\right)\left(x^2+2x+4\right)+x^3\)
\(\Leftrightarrow8x^3+36x^2+54x+27-\left(8x^3+125\right)=36x^2-12x+1-\left(x^3-8\right)+x^3\)
\(\Leftrightarrow8x^3+36x^2+54x+27-8x^3-125=36x^2-12x+1-x^3+8+x^3\)
\(\Leftrightarrow54x-98=-12x+9\)
\(\Leftrightarrow54x+12x=9+98\)
\(\Leftrightarrow66x=107\)
\(\Leftrightarrow x=\dfrac{107}{66}\)
Vậy ...
$\textbf{1)}$
$(4x-1)^2-2(4x-1)(3x-7)+(7-3x)^2$
$=(4x-1)^2-2(4x-1)(3x-7)+(3x-7)^2$
$=\left[(4x-1)-(3x-7)\right]^2$
$=(x+6)^2.$
Tại $x=44$: $(44+6)^2=50^2=2500.$
$\textbf{2)}$
$(2x-5)^2-2(2x-5)(3x-4)+(4-3x)^2$
$=(2x-5)^2-2(2x-5)(3x-4)+(3x-4)^2$
$=\left[(2x-5)-(3x-4)\right]^2$
$=(x+1)^2.$
Tại $x=24$: $(24+1)^2=625.$
\(\left(x^2+3\right)\left(3-x^2\right)\)
\(\left(x^2+3\right)\left(-x^2+3\right)\)
\(\left(-x^2+3\right).x^2+3\left(-x^2+3\right)\)
\(-x^2.x^2+3x^2+3\left(-x^2+3\right)\)
\(-x^2.x^2+3x^2-3x^2+9\)
\(-x^2.x^2+9\)
a) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(\Leftrightarrow\left(6x^2+21x-2x-7\right)-\left(6x^2-5x+6x-5\right)-16=0\)
\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5-16=0\)
\(\Leftrightarrow18x-18=0\)
\(\Leftrightarrow18x=18\)
\(\Leftrightarrow x=18:18\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
b) \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x-5\right)^2=x^2+6x+64\)
\(\Leftrightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2-\left(x^2+6x+64\right)=0\)
\(\Leftrightarrow\left(2x+3-2x+5\right)^2-x^2-6x-64=0\)
\(\Leftrightarrow8^2-x^2-6x-64=0\)
\(\Leftrightarrow64-x^2-6x-64=0\)
\(\Leftrightarrow-x^2-6x=0\)
\(\Leftrightarrow x\left(-x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=-6\)
a) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(\Leftrightarrow\left(6x^2+21x-2x-7\right)-\left(6x^2-5x+6x-5\right)-16=0\)
\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5-16=0\)
\(\Leftrightarrow18x-18=0\)
\(\Leftrightarrow18x=18\)
\(\Leftrightarrow x=18:18\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
b, \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x- 5\right)^2=x^2+6x+64\)
\(\Leftrightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2- \left(x^2+6x+64\right)=0\)
\(\Leftrightarrow\left(2x+3-2x+5\right)^2-x^2-6x-64=0\)
\(\Leftrightarrow8^2-x^2-6x-64=0\)
\(\Leftrightarrow64-x^2-6x-64=0\)
\(\Leftrightarrow-x^2-6x=0\)
\(\Leftrightarrow x\left(-x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=6\)
a) (3x−1)(2x+7)−(x+1)(6x−5)=16(3x−1)(2x+7)−(x+1)(6x−5)=16
⇔(6x2+21x−2x−7)−(6x2−5x+6x−5)−16=0⇔(6x2+21x−2x−7)−(6x2−5x+6x−5)−16=0
⇔6x2+21x−2x−7−6x2+5x−6x+5−16=0⇔6x2+21x−2x−7−6x2+5x−6x+5−16=0
⇔18x−18=0⇔18x−18=0
⇔18x=18⇔18x=18
⇔x=18:18⇔x=18:18
⇔x=1⇔x=1
Vậy x=1x=1
b) (2x+3)2−2(2x+3)(2x−5)+(2x−5)2=x2+6x+64(2x+3)2−2(2x+3)(2x−5)+(2x−5)2=x2+6x+64
⇔[(2x+3)−(2x−5)]2−(x2+6x+64)=0⇔[(2x+3)−(2x−5)]2−(x2+6x+64)=0
⇔(2x+3−2x+5)2−x2−6x−64=0⇔(2x+3−2x+5)2−x2−6x−64=0
⇔82−x2−6x−64=0⇔82−x2−6x−64=0
⇔64−x2−6x−64=0
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