A = | 2 + x | + | 2x - 1 | - x
B = x + 5 - | 3 + x |
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
TN
28 tháng 6 2017
\(1,\left(x-3\right)\left(x-1\right)-3\left(x-3\right)\)
\(=\left(x-3\right)\left(x-1-3\right)\)
\(=\left(x-3\right)\left(x-4\right)\)
\(2,6x+3-\left(2x-5\right)\left(2x+1\right)\)
\(=3\left(2x+1\right)-\left(2x-5\right)\left(2x+1\right)\)
\(=\left(2x+1\right)\left(3-2x+5\right)\)
\(=\left(2x+1\right)\left(-2-2x\right)\)
\(3,\left(x-1\right)\left(2x+1\right)+3\left(x-1\right)\left(x+2\right)\left(2x+1\right)\)\(=\left(x-1\right)\left(2x+1\right)\left(1+3x+6\right)\)
\(=\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\)
\(4,\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\left(3x-2\right)\left(3x-6\right)\)
\(5,\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)-\left(5-x\right)\left(2x+1\right)\)\(=\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)+\left(x-5\right)\left(2x+1\right)\)\(=\left(x-5\right)\left(x-5+x+5+2x+1\right)\)
\(=\left(x-5\right)\left(4x+1\right)\)
6, Tương tự
5 tháng 11 2017
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
30 tháng 9 2018
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
MP
2 tháng 5 2021
Bạn ơi xem lại hộ mk đề bài phần b với ạ, mk thấy có j đó sai sai😿😿
CW
18 tháng 6 2016
f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)
\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)
\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)
\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)
\(-x^3=27\)
\(x=-3\)
12 tháng 8 2021
\(a,\dfrac{5}{-x^2+5x-6}+\dfrac{x+3}{2-x}=0\left(x\ne2;x\ne3\right)\\ \Leftrightarrow\dfrac{5}{\left(x-3\right)\left(x-2\right)}-\dfrac{x+3}{x-2}=0\\\Leftrightarrow\dfrac{5-\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}=0 \\ \Leftrightarrow5-x^2+9=0\\ \Leftrightarrow14-x^2=0\\ \Leftrightarrow x^2=14\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{14}\\x=-\sqrt{14}\end{matrix}\right.\)
\(b,\dfrac{x}{2x+2}-\dfrac{2x}{x^2-2x-3}=\dfrac{x}{6-2x}\left(x\ne-1;x\ne3\right)\\ \Leftrightarrow\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x-3\right)\left(x+1\right)}=\dfrac{x}{2\left(3-x\right)}\\ \Leftrightarrow\dfrac{x\left(x-3\right)-2x\cdot2}{2\left(x-3\right)\left(x+1\right)}=\dfrac{-x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}\\ \Leftrightarrow x^2-3x-4x=-x^2-x\\ \Leftrightarrow2x^2-6x=0\\ \Leftrightarrow2x\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
\(c,\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\left(x\ne1\right)\\ \Leftrightarrow\dfrac{x^2+x+1-3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\\ \Leftrightarrow-2x^2+x+1=2x^2-2x\\ \Leftrightarrow4x^2-3x-1=0\\ \Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{4}\end{matrix}\right.\)
\(d,\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}=\dfrac{5-x}{2x^2+10x}\left(x\ne5;x\ne-5\right)\\ \Leftrightarrow\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}=\dfrac{5-x}{2x\left(x+5\right)}\\ \Leftrightarrow\dfrac{x^2+25x-2\left(x+5\right)^2}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{\left(5-x\right)\left(x-5\right)}{2x\left(x+5\right)\left(x-5\right)}\\ \Leftrightarrow x^2+25x-2\left(x^2+10x+25\right)=-\left(x^2-10x+25\right)\\ \Leftrightarrow x^2+25x-2x^2-20x-50=-x^2+10x-25\\ \Leftrightarrow-5x=25\\ \Leftrightarrow x=-5\)
Tick nha
TG
10 tháng 8 2020
a) \(A=\left(5-x\right)\left(x+5\right)-2\left(x-1\right)\left(x-3\right)-3\left(x-2\right)^2\)
\(=\left(5-x\right)\left(5+x\right)-\left(2x-2\right)\left(x-3\right)-3\left(x^2-2.2x+2^2\right)\)
\(=\left(5^2-x^2\right)-\left[2x\left(x-3\right)-2\left(x-3\right)\right]-3\left(x^2-4x+4\right)\)
\(=25-x^2-\left[\left(2x^2-6x\right)-\left(2x-6\right)\right]-3x^2+12x-12\)
\(=25-x^2-\left(2x^2-6x-2x+6\right)-3x^2+12x-12\)
\(=25-x^2-2x^2+6x+2x-6-3x^2+12x-12\)
\(=7+20x-6x^2\)
b/ \(B=\left(3-2x\right)\left(x-2\right)+\left(2x-5\right)^2-\left(x-4\right)\)
\(=3\left(x-2\right)-2x\left(x-2\right)+\left[\left(2x\right)^2-2.2x.5+5^2\right]-x+4\)
\(=3x-6-2x^2+4x+4x^2-20x+25-x+4\)
\(=23-14x+2x^2\)
c/ \(C=\left(x-4\right)\left(x-2\right)-3\left(x-2\right)\left(3-2x\right)-\left(2x+1\right)^2\)
\(=x\left(x-2\right)-4\left(x-2\right)-\left(3x-6\right)\left(3-2x\right)-\left[\left(2x\right)^2+2.2x.1+1^2\right]\)
\(=x^2-2x-4x+8-\left[3x\left(3-2x\right)-6\left(3-2x\right)\right]-4x^2-4x-1\)
\(=x^2-2x-4x+8-\left(9x-6x^2-18+12x\right)-4x^2-4x-1\)
\(=x^2-2x-4x+8-9x+6x^2+18-12x-4x^2-4x-1\)
\(=25-31x+3x^2\)
d/ \(D=2\left(x-1\right)^2-3\left(x-1\right)\left(x+2\right)-\left(2x+1\right)^2\)
\(=2.\left(x^2-2x+1\right)-\left(3x-3\right)\left(x+2\right)-\left[\left(2x\right)^2+2.2x+1\right]\)
\(=2x^2-4x+2-\left[3x\left(x+2\right)-3\left(x+2\right)\right]-\left(4x^2+4x+1\right)\)
\(=2x^2-4x+2-\left(3x^2+6x-3x-6\right)-\left(4x^2+4x+1\right)\)
\(=2x^2-4x+2-3x^2-6x+3x+6-4x^2-4x-1\)
\(=7-11x-5x^2\)
P/s: Ko chắc ạ!
21 tháng 8 2021
a) \(\left(x+3\right)^2-\left(x-2\right)^3=\left(x+5\right)\left(x^2-5x+25\right)-108\)
\(\Leftrightarrow x^2+6x+9-x^2+4x-4=x^3-5x^2+25x+5x^2-25x+125-108\)
\(\Leftrightarrow x^3-10x+12=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+6\right)=0\)
\(\Leftrightarrow x=2\)( do \(x^2+2x+6=\left(x+1\right)^2+4\ge4>0\))
11 tháng 5 2020
Trình bày đề bài cho dễ nhìn bạn eyy :v
Khó nhìn như này thì God cũng chịu -.-
10 tháng 9 2020
a) \(\left(2x-5\right)^2-\left(2x+3\right)\left(2x-3\right)=10\Leftrightarrow\left(4x^2-20x+25\right)-\left(4x^2-9\right)-10=0\)
\(\Leftrightarrow-20x+24=0\Leftrightarrow x=\frac{6}{5}\)
b) \(\left(4x-1\right)\left(x+2\right)-\left(2x+3\right)^2-5\left(x-1\right)=9\Leftrightarrow-10x-15=0\)
\(\Leftrightarrow x=\frac{-3}{2}\)
c) \(\left(x+1\right)^3-\left(x-1\right)^3-2=6\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-8=0\)
\(\Leftrightarrow6x^2-6=0\Leftrightarrow x=\pm1\)
d) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x+1\right)\left(x^2-x+1\right)-3\left(-x-2\right)=5\)
\(\Leftrightarrow\left(x^3+8\right)-\left(x^3+1\right)+3x+6=5\Leftrightarrow3x+8=0\Leftrightarrow x=\frac{-8}{3}\)