1+1+1=?.Thưc khuy
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
12 tháng 4 2020
a/ Đkxđ: \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
Vậy phân thức được xác định khi \(\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
b/ \(A=\left[1+\frac{1}{x}+\frac{2}{x+1}\left(1+\frac{1}{x}\right)\right]:\frac{x^3+27}{2x}\)
\(=\left[1+\frac{1}{x}+\frac{2}{x+1}+\frac{2}{\left(x+1\right)x}\right]:\frac{\left(x+3\right)\left(x^2-3x+9\right)}{2x}\)
\(=\left[\frac{x\left(x+1\right)+\left(x+1\right)+2x+2}{\left(x+1\right)x}\right].\frac{2x}{\left(x+3\right)\left(x^2-3x+9\right)}\)
\(=\frac{x^2+4x+3}{\left(x+1\right)x}.\frac{2x}{\left(x+3\right)\left(x^2-3x+9\right)}=\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)x}.\frac{2x}{\left(x+3\right)\left(x^2-3x+9\right)}\)
\(=\frac{2}{x^2-3x+9}\)
25 tháng 6 2019
4,\(6x^2+10x-9x-15=6x^2+12x\)
\(6x^2+x-15-6x^2-12x\) =0
11x-15=0
11x=15
x=\(\frac{15}{11}\)
vậy.......
hc tốt
25 tháng 6 2019
\(a,\left(2x-3\right)\left(3x+5\right)+3=6x\left(x+2\right)\)
\(\Rightarrow6x^2+2x-15+3=6x^2+12x\)
\(\Rightarrow10x=-12\)
\(\Rightarrow x=-\frac{5}{7}\)
\(b,\)Sai đề không ?
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 4 2020
ĐKXĐ: \(x\ne\left\{-\frac{1}{2};\frac{1}{2};-1\right\}\)
\(B=\left(\frac{x\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\frac{4x+1}{\left(2x-1\right)\left(2x+1\right)}\right).\left(\frac{2x-1}{\left(x+1\right)\left(x^2-x+1\right)}\right)\)
\(=\frac{\left(2x^2+3x+1\right)}{\left(2x+1\right)\left(2x-1\right)}.\frac{\left(2x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{\left(x+1\right)\left(2x+1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)\left(x+1\right)\left(x^2-x+1\right)}=\frac{1}{x^2-x+1}\)
NM
3
H9
HT.Phong (9A5)
CTVHS
16 tháng 11 2023
1. a) \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=14x^5+21x^7\)
b) \(\left(x^3-x^2+x-1\right):\left(x-1\right)=\dfrac{x^3-x^2+x-1}{x-1}\)
\(=\dfrac{x^2\left(x-1\right)+\left(x-1\right)}{x-1}=\dfrac{\left(x-1\right)\left(x^2+1\right)}{x-1}=x^2+1\)
16 tháng 11 2023
2: \(x^2-8x+7=0\)
=>\(x^2-x-7x+7=0\)
=>\(x\left(x-1\right)-7\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x-7\right)=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)
1:
a: \(7x^2\left(2x^3+3x^5\right)=7x^2\cdot2x^3+7x^2\cdot3x^5=21x^7+14x^5\)
b: \(\dfrac{x^3-x^2+x-1}{x-1}=\dfrac{x^2\left(x-1\right)+\left(x-1\right)}{\left(x-1\right)}\)
\(=x^2+1\)
25 tháng 2 2020
a) A có nghĩa <=> \(\left\{{}\begin{matrix}2x-2\ne0\\2-2x^2\ne0\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}x-1\ne0\\\left(1-x\right)\left(x+1\right)\ne0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x\ne1\\x\ne\pm1\end{matrix}\right.\)
b) Ta có:
A = \(\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}\)
A = \(\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x^2-1\right)}\)
A = \(\frac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
A = \(\frac{x^2+x-x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
A = \(\frac{x+1}{2\left(x-1\right)\left(x+1\right)}=\frac{1}{2\left(x-1\right)}\)
c) A = -1/2
<=> \(\frac{1}{2\left(x-1\right)}=-\frac{1}{2}\)
<=> 2(x - 1) = -2
<=> x - 1 = -1
<=> x = 0 (tmđk)
Vậy x = 0
MH
25 tháng 9 2021
c) \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x-1\right)^3-\left(x-1\right)\left(x^2+x.1+1^2\right)\)
\(=\left(x-1\right)^3-\left(x-1\right)^3\)
\(=0\)
MH
25 tháng 9 2021
d) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2\)
\(=\left(x-3\right)^3-\left(x-3\right)\left(x^2+x.3+3^2\right)+6\left(x+1\right)^2\)
\(=\left(x-3\right)^3-\left(x-3\right)^3+6\left(x+1\right)^2\)
\(=0+6\left(x+1\right)^2\)
\(=6\left(x+1\right)^2\)
25 tháng 2 2021
x2( x + 1 )2 + 2x2 + 2x - 8
= [ x( x + 1 ) ]2 + 2( x2 + x ) - 8
= ( x2 + x )2 + 2( x2 + x ) - 8 (*)
Đặt a = x2 + x
(*) = a2 + 2a - 8
= a2 - 2a + 4a - 8
= a( a - 2 ) + 4( a - 2 )
= ( a - 2 )( a + 4 )
= ( x2 + x - 2 )( x2 + x + 4 )
= ( x2 - x + 2x - 2 )( x2 + x + 4 )
= [ x( x - 1 ) + 2( x - 1 ) ]( x2 + x + 4 )
= ( x - 1 )( x + 2 )( x2 + x + 4 )
25 tháng 2 2021
ta co
\(x^2\left(x+1\right)^2+2x^2+2x-8\)
=\(\left(x\left(x+1\right)\right)^2+2x\left(x+1\right)+1-9\)
=\(\left(x^2+x+1\right)^2-9\)
=\(\left(x^2+x-8\right)\left(x^2+x+10\right)\)
=\(\left(x^2+2x\frac{1}{2}+\frac{1}{4}-\frac{33}{4}\right)\left(x^2+x+10\right)\)
=\(\left(\left(x+\frac{1}{2}\right)^2-\frac{33}{4}\right)\left(x^2+x+10\right)\)
=\(\left(x+\frac{1}{2}-\sqrt{\frac{33}{4}}\right)\left(x+\frac{1}{2}+\sqrt{\frac{33}{4}}\right)\left(x^2+x+10\right)\)
1 + 1 + 1 = 2 + 1 = 5 - 2 = 90 - 87 = ... = ... = ... = ... = ... = ... = ...= ... = ... = ... = ... = ... = ... = ...=... = ... = ... = ... = ... = ... = ...=... = ... = ... = ... = ... = ... = ...=3
học tốt
&YOUTUBER&
= 3
hk tốt !