K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

2 tháng 8 2021

=872,9909

2 tháng 8 2021

30-0,01=29,99 bn ơi

ko bằng 29,09 đc haha

22 tháng 12 2021

\(a,\dfrac{x}{3x+6}=\dfrac{x}{3\left(x+2\right)}=\dfrac{x\left(x+2\right)}{3\left(x+2\right)^2}\\ \dfrac{5}{x^2+4x+4}=\dfrac{5}{\left(x+2\right)^2}=\dfrac{15}{3\left(x+2\right)^2}\\ b,\dfrac{5}{x^2-y^2+2x+1}=\dfrac{5}{\left(x-y+1\right)\left(x+y+1\right)}=\dfrac{5x}{x\left(x-y+1\right)\left(x+y+1\right)}\\ \dfrac{6}{x\left(x+y+1\right)}=\dfrac{6\left(x-y+1\right)}{x\left(x-y+1\right)\left(x+y+1\right)}\)

\(c,\dfrac{7x}{x^4-1}=\dfrac{7x}{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)}=\dfrac{7x\left(x^2+1\right)}{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)}\\ \dfrac{5x}{x^4+2x^2+1}=\dfrac{5x}{\left(x^2+1\right)^2}=\dfrac{5x\left(x-1\right)\left(x+1\right)}{\left(x^2+1\right)^2\left(x-1\right)\left(x+1\right)}\)

8 tháng 8 2019

\(13,5x5,8-8,3x4,2-5,8x8,3+4,2x13,5\)

\(=13,5x\left(5,8+4,2\right)-8,3x\left(4,2+5,8\right)\)

\(=13,5x10-8,3x10\)

\(=135-83\)

\(=52\)

\(x^2+4xy-3x+4y^2-6y\)

\(=x^2+4xy+4y^2-3x-6y\)

\(=\left(x+2y\right)^2-3\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x+2y-3\right)\)

\(x^2+4xy-3x+4y^2-6y\)

\(=x^2+4xy+4y^2-3x-6y\)

\(=\left(x+2y\right)^2-3\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x+2y-3\right)\)

24 tháng 10 2021

a: Ta có: \(\left(x+3\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2-9-x^2-3x+10=6\)

\(\Leftrightarrow-3x=5\)

hay \(x=-\dfrac{5}{3}\)

c: \(4x^2-9=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

24 tháng 10 2021

\(a,\Leftrightarrow x^2-9-x^2-3x+10=6\\ \Leftrightarrow-3x=5\Leftrightarrow x=-\dfrac{5}{3}\\ b,\Leftrightarrow2x^2+3x^2-3=5x^2+5x\\ \Leftrightarrow5x=-3\Leftrightarrow x=-\dfrac{3}{5}\\ c,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\\ d,\Leftrightarrow\left(5-2x\right)^2-4=0\\ \Leftrightarrow\left(5-2x-2\right)\left(5-2x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{7}{2}\end{matrix}\right.\\ e,\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

\(f,\Leftrightarrow\left(2x+9\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{9}{2}\end{matrix}\right.\\ g,\Leftrightarrow\left(x^2-4\right)\left(3x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=\dfrac{4}{3}\end{matrix}\right.\\ h,\Leftrightarrow\left(x+1\right)\left(x^4+x^2+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^4+2x^2+1-x^2\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=-1\)

AH
Akai Haruma
Giáo viên
27 tháng 11 2023

Lời giải:

a. $99^3+1+3(99^2+99)=99^3+3.99^2.1+3.99.1^2+1^3=(99+1)^3=100^3=1000000$

b. $11^3-1-3(11^2-11)=11^3-3.11^2.1+3.11.1^2-1^3=(11-1)^3=10^3=1000$

\(x^2-5\)

\(=x^2-\left(\sqrt{5}\right)^2\)

\(=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)

31 tháng 12 2018

\(56^2+44^2+2.44.56\)

\(=\left(56+44\right)^2\)

\(=100^2=10000\)

31 tháng 12 2018

= (56+44)^2=100^2=10000

hok tốt na

5 tháng 11 2021

 tròn nhé

Chọn C