Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,x^2+4x-21-x^2-4x+5=-16\\ b,=\left(x+8-x+2\right)^2=10^2=100\\ c,=x^2\left(x^2-16\right)-\left(x^4-1\right)\\ =x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)
\(a,=x^2+4x-21-x^2-4x+5=-16\\ b,=\left(x+8-x+2\right)^2=10^2=100\\ c,=x^2\left(x^2-16\right)-\left(x^4-1\right)\\ =x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)
a: \(\dfrac{x+5}{x-1}+\dfrac{8}{x^2-4x+3}=\dfrac{x+1}{x-3}\)
=>(x+5)(x-3)+8=x^2-1
=>x^2+2x-15+8=x^2-1
=>2x-7=-1
=>x=3(loại)
b: \(\dfrac{x-4}{x-1}-\dfrac{x^2+3}{1-x^2}+\dfrac{5}{x+1}=0\)
=>(x-4)(x+1)+x^2+3+5(x-1)=0
=>x^2-3x-4+x^2+3+5x-5=0
=>2x^2+2x-6=0
=>x^2+x-3=0
=>\(x=\dfrac{-1\pm\sqrt{13}}{2}\)
e: =>x^2-2x+1+2x+2=5x+5
=>x^2+3=5x+5
=>x^2-5x-2=0
=>\(x=\dfrac{5\pm\sqrt{33}}{2}\)
g: (x-3)(x+4)*x=0
=>x=0 hoặc x-3=0 hoặc x+4=0
=>x=0;x=3;x=-4
Bài 1:
a) \(x\left(x+1\right)+x\left(x-1\right)-2x^2\)
\(=x^2+x+x^2-x-2x^2\)
\(=2x^2-2x^2\)
\(=0\)
b) \(\left(x+2\right)\left(x^2-x+1\right)-\left(x-2\right)\left(x^2+x+1\right)\)
\(=x^3-x^2+x+2x^2-2x+2-x^3-x^2-x+2x^2+2x+2\)
\(=\left(x^3-x^3\right)+\left(-x^2+2x^2-x^2+2x^2\right)+\left(x-2x-x+2x\right)+\left(2+2\right)\)
\(=2x^2+4\)
c) \(\left(3-x\right)^2+2\left(x-3\right)\left(x+7\right)+\left(x+7\right)^2\)
\(=\left(x-3\right)^2+2\left(x-3\right)\left(x+7\right)+\left(x+7\right)^2\)
\(=\left[\left(x-3\right)+\left(x+7\right)\right]^2\)
\(=\left(x-3+x+7\right)^2\)
\(=\left(2x+4\right)^2\)
Lời giải:
a. $2x^2+3(x-1)(x+1)=5x(x+1)$
$\Leftrightarrow 2x^2+3x^2-3=5x^2+5x$
$\Leftrightarrow 5x^2-3=5x^2+5x$
$\Leftrightarrow 5x=-3$
$\Leftrightarrow x=\frac{-3}{5}$
b.
PT $\Leftrightarrow (-5x^2-2x+16)+4(x^2-x-2)=4-x^2$
$\Leftrightarrow -x^2-6x+8=4-x^2$
$\Leftrightarrow -6x+8=4$
$\Leftrightarrow -6x=-4$
$\Leftrightarrow x=\frac{2}{3}$
c.
PT $\Leftrightarrow 4(x^2+4x-5)-(x^2+7x+10)=3(x^2+x-2)$
$\Leftrightarrow 4x^2+16x-20-x^2-7x-10=3x^2+3x-6$
$\Leftrightarrow 3x^2+9x-30=3x^2+3x-6$
$\Leftrightarrow 6x=24$
$\Leftrightarrow x=4$
a) ( x + 2 )( x + 3 ) - ( x - 2 )( x + 5 )
= x2 + 5x + 6 - ( x2 + 3x - 10 )
= x2 + 5x + 6 - x2 - 3x + 10
= 2x + 16
b) ( 8 - 5x )( x + 2 ) + 4( x - 2 )( x + 1 ) + 2( x - 2 )( x + 2 ) + 10
= -5x2 - 2x + 16 + 4( x2 - x - 2 ) + 2( x2 - 4 ) + 10
= -5x2 - 2x + 16 + 4x2 - 4x - 8 + 2x2 - 8 + 10
= x2 - 6x + 10
c) 4( x - 1 )( x + 5 ) - ( x + 2 )( x + 5 ) - 3( x - 1 )( x + 2 )
= 4( x2 + 4x - 5 ) - ( x2 + 7x + 10 ) - 3( x2 + x - 2 )
= 4x2 + 16x - 20 - x2 - 7x - 10 - 3x2 - 3x + 6
= 6x - 24
d) ( x - 1 )( x5 + x4 + x3 + x2 + x + 1 )
= x6 + x5 + x4 + x3 + x2 + x - x5 - x4 - x3 - x2 - x - 1
= x6 - 1
a)
\(\left(3x^2-x+1\right)\left(x-1\right)+x^2\left(4-3x\right)=\frac{5}{2}\)
\(\Leftrightarrow3x^3-x^2+x-3x^2+x-1+4x^2-3x^3=\frac{5}{2}\)
\(\Leftrightarrow2x-1=\frac{5}{2}\Leftrightarrow2x=1+\frac{5}{2}=\frac{7}{2}\Leftrightarrow x=\frac{7}{4}\)
b)
\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)
\(\Leftrightarrow4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)=11\)
\(\Leftrightarrow4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)
\(\Leftrightarrow8x+4-4x+1+8=11\Leftrightarrow4x+13=11\Leftrightarrow4x=-2\Leftrightarrow x=-\frac{1}{2}\)
c)
\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-7^2\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow-4x+1+6x+9+245=0\Leftrightarrow2x+255=0\Leftrightarrow x=-\frac{255}{2}\).
a ) ( 3x2 - x + 1 ) ( x + 1 ) + x2 ( 4 - 3x ) = 5/2
=> 3x3 + 3x2 - x2 - x + x + 1 + 4x2 - 3x3 = 5/2
=> 6x2 + 1 = 5/2
=> 6x2 = 1,5
=> x2 = 0,25
=> x = 0,5
Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn. Viết đề như thế này gây khó đọc.
c,\((x^3+1^3)-(x^3-1^3)\)
a) \(\left(x+8\right)^2-2\left(x+8\right)\left(x-2\right)+\left(x-2\right)^2\)
\(=\left[\left(x+8\right)-\left(x-2\right)\right]^2\)
\(=\left(x+8-x+2\right)^2\)
\(=10^2\)
\(=100\)