Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25+12=0\\ \Leftrightarrow4x+38=0\\ \Leftrightarrow x=-\dfrac{19}{2}\)
\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=-12\\ \Leftrightarrow4x=-38\Leftrightarrow x=-\dfrac{19}{2}\)
\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25+12=0\\ \Leftrightarrow4x+38=0\\ \Leftrightarrow x=-\dfrac{19}{2}\)
\(\Rightarrow3x^2+2x+x^2+2x+1-4x^2+25=-12\)
\(\Rightarrow4x=-38\Rightarrow x=-\dfrac{19}{2}\)
Bài 1:
$2x(x+3)+(2x+3)(5-x)=2$
$\Leftrightarrow 2x^2+6x+(10x-2x^2+15-3x)=2$
$\Leftrightarrow 2x^2+6x+7x-2x^2+15=2$
$\Leftrightarrow 13x+15=2$
$\Leftrightarrow 13x=2-15=-13$
$\Leftrightarrow x=-13:13=-1$
Bài 2:
$x-y=4\Rightarrow x=y+4$. Thay vào $xy=5$ thì:
$(y+4)y=5$
$\Leftrightarrow y^2+4y-5=0$
$\Leftrightarrow (y-1)(y+5)=0$
$\Leftrightarrow y=1$ hoặc $y=-5$
Nếu $y=1$ thì $x=y+4=5$. Khi đó $x^3+y^3=5^3+1^3=126$
Nếu $y=-5$ thì $x=y+4=-1$. Khi đó: $x^3+y^3=(-1)^3+(-5)^3=-126$
a: \(\Leftrightarrow\left(x+2\right)\left(x+2-2x+10\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)
\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)
=>\(x\left(x^2-25\right)-\left(x^3+8\right)=3\)
=>\(x^3-25x-x^3-8=3\)
=>-25x-8=3
=>-25x=11
=>\(x=\dfrac{11}{-25}=\dfrac{-11}{25}\)
x(x+5)(x−5)−(x+2)(x2−2x+4)=3
=>x3−25x−x3+2x2−4x−2x2+4x−8=3
=>−25x= 11
=> x= −1125
a) \(\Leftrightarrow x^2-4x-x^2+6x-9=0\\ \Leftrightarrow2x=9\\ \Leftrightarrow x=4,5\)
b) \(\Leftrightarrow x^2-3x-10=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(5x+10\right)=0\\ \Leftrightarrow x\left(x+2\right)-5\left(x+2\right)=0\\ \left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
c) \(\Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\\ \Leftrightarrow\left(2x-10\right)\left(2x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
d) \(\Leftrightarrow\left(2x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)
\(x\left(x-5\right)\left(x+5\right)-\left(x-2\right)\left(x^2+2x+4\right)=-17\)
\(\Leftrightarrow x^3-25x-x^3+8=-17\)
\(\Leftrightarrow-25x=-25\)
hay x=1
a) 5(2x -1) - 4(8 - 3x) = 7
<=> 10x - 5 - 32 + 12x = 7
<=> 22x = 44
<=> x =2
Vậy x = 2 là nghiệm phương trình
b) 7(2x - 5) - 5(7x - 2) + 2(5x - 7) = (x - 2) - (x + 4)
<=> 14x - 35 - 35x + 10 + 10x - 14 = x - 2 - x - 4
<=> -11x - 39 = - 6
<=> -11x = 33
<=> x = -3
Vậy x = -3 là nghiệm phương trình
\(a,10x-5-32+12x=7\)
\(22x=44\)
\(x=2\)
\(b,14x-35-35x+10+10x-14=x-2-x-4\)
\(-11x-39=-6\)
\(-11x=-33\)
\(x=3\)
phân tích hằng đẳng thức ra em
(2x-5)2=4x2 -2.2.5x + 52=4x2-20x+25
(x-2)2=x2- 4x +4
---->> 4x2-20x+25=x2-4x+4--->3x2-16x +21=0 --->x=3 và x=7/3
hoặc cách khác em nhé :
2x-5=x-2--->x=3
2x-5=-x+2--->3x=7--->x=7/3