Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(b,n^4-n^2=n^2\left(n^2-1\right)=n^2\left(n-1\right)\left(n+1\right)\)
\(=n.n\left(n-1\right)\left(n+1\right)\)
xét \(n=2k\)
\(n.n=4k⋮4\)
xét \(n=2k+1\)
\(\left(n-1\right)\left(n+1\right)=2k\left(2k+2\right)=4k\left(k+1\right)⋮4\)
\(< =>n.n\left(n-1\right)\left(n+1\right)⋮4\)
\(n^4-n^2⋮4< =>ĐPCM\)
Trả lời:
a, \(x+5x^2=0\)
\(\Leftrightarrow x\left(1+5x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\1+5x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{5}\end{cases}}}\)
Vậy x = 0; x = - 1/5 là nghiệm của pt.
b, \(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x-5=0\)
\(\Leftrightarrow x=5\)
Vậy x = 5 là nghiệm của pt.
a/ Tách 300 thành 100 chữ số 3 rồi chuyển vế dồn từng số 3 vào ( ) có \(\left(x^2-x-2\right)+\left(x^2-2x\right)+\left(x^2-3x+2\right)+...+\left(x^2-100x+196\right)\)
=0 \(\Leftrightarrow\left(x-2\right)\left(x+1\right)+x\left(x-3\right)+\left(x-1\right)\left(x-2\right)+...+\left(x-96\right)\left(x-4\right)+\left(x-97\right)\left(x-3\right)+\left(x-98\right)\left(x-2\right)\)=0\(\Leftrightarrow\left(x-2\right)\left(2x-97\right)+\left(x-3\right)\left(2x-97\right)+...=0\Rightarrow x=2\)
b tường đương \(x^2-4+\frac{4x^2}{x^2-4x+4}-1=0\Leftrightarrow\left(x-2\right)\left(x+2\right)+\frac{3x^2+4x-4}{\left(x-2\right)^2}=0\Leftrightarrow\left(x-2\right)\left(x+2\right)+\frac{\left(x+2\right)\left(3x-2\right)}{\left(x-2\right)^2}=0\Leftrightarrow\left(x-2\right)\left(x+2+\frac{3x-2}{\left(x-2\right)^2}\right)=0\Leftrightarrow x=2\)
Bài 1:
a/ \(x\ne1;2\)
\(\frac{x-2}{\left(x-1\right)\left(x-2\right)}-\frac{7\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow x-2-7x+7+1=0\)
\(\Leftrightarrow-6x+6=0\)
\(\Rightarrow x=1\) (loại)
Vậy pt vô nghiệm
b/ \(x\ne\frac{3}{2}\)
\(\frac{2x+3}{2x-3}-\frac{3}{2\left(2x-3\right)}-\frac{2}{5}=0\)
\(\Leftrightarrow\frac{10\left(2x+3\right)}{10\left(2x-3\right)}-\frac{15}{10\left(2x-3\right)}-\frac{4\left(2x-3\right)}{10\left(2x-3\right)}=0\)
\(\Leftrightarrow20x+30-15-8x+12=0\)
\(\Leftrightarrow12x+27=0\)
\(\Rightarrow x=-\frac{9}{4}\)
c/ \(x\ne\pm1\)
\(\frac{x+1}{x-1}-\frac{4}{x+1}+\frac{3-x^2}{x^2-1}=0\)
\(\Leftrightarrow\frac{\left(x+1\right)^2}{x^2-1}-\frac{4\left(x-1\right)}{x^2-1}+\frac{3-x^2}{x^2-1}=0\)
\(\Leftrightarrow x^2+2x+1-4x+4+3-x^2=0\)
\(\Leftrightarrow-2x+8=0\)
\(\Rightarrow x=4\)
Bài 1:
d/\(x\ne\pm3\)
\(\frac{x-1}{x+3}-\frac{x}{x-3}+\frac{7x-3}{x^2-9}=0\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{x^2-9}-\frac{x\left(x+3\right)}{x^2-9}+\frac{7x-3}{x^2-9}=0\)
\(\Leftrightarrow x^2-4x+3-x^2-3x+7x-3=0\)
\(\Rightarrow0=0\)
Vậy pt có vô số nghiệm \(x\ne\pm3\)
e/ \(x\ne\pm1\)
\(\frac{1}{x+1}+\frac{2}{x^2\left(x-1\right)-\left(x-1\right)}+\frac{3}{x^2-1}=0\)
\(\Leftrightarrow\frac{1}{x+1}+\frac{2}{\left(x^2-1\right)\left(x-1\right)}+\frac{3}{x^2-1}=0\)
\(\Leftrightarrow\frac{1}{x+1}+\frac{2}{\left(x+1\right)\left(x-1\right)^2}+\frac{3}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)^2}+\frac{2}{\left(x+1\right)\left(x-1\right)^2}+\frac{3\left(x-1\right)}{\left(x+1\right)\left(x-1\right)^2}=0\)
\(\Leftrightarrow x^2-2x+1+2+3x-3=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\left(l\right)\end{matrix}\right.\)
Trả lời:
a) x2 + 4y2 + 4xy = x2 + 2.x.2y + (2y)2 = ( x + 2y )2
b) \(\frac{1}{64}-27x^3=\left(\frac{1}{4}\right)^3-\left(3x\right)^3=\left(\frac{1}{4}-3x\right)\left(\frac{1}{16}+\frac{3}{4}x+9x^2\right)\)
c) x3 - 6x2 + 12x - 8 = x3 - 3.x2.2 + 3.x.22 - 23 = ( x - 2 )3
d) x2 - x - y2 - y = ( x2 - y2 ) - ( x + y ) = ( x - y )( x + y ) - ( x + y ) = ( x + y )( x - y - 1 )
e) 5x - 5y + ax - ay = ( 5x - 5y ) + ( ax - ay ) = 5 ( x - y ) + a ( x - y ) = ( x - y )( 5 + a )
c) Ta có: x4 - 2x2 + 1 =0
(x-1)2 =0
x=1
Vậy x=1
b) Ta có: (2 - x)3 + (3 + x)(9 - 3x + x2) + 6x(1 - x) = 17
8 - 12x + 6x2 - x3 + 27 + x3 + 6x - 6x2 = 17
35 - 6x = 17
x = 3
Vậy x = 3
a) Ta có: (x + 3)2 + (4 - x)(x + 4) = 1
(x + 3)2 - (x - 4)(x + 4) = 1
x2 + 6x + 9 - x2 + 16 = 1
6x + 25 = 1
x = -4
Vậy x = -4