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6 tháng 4 2019
964 - 1 = (932 + 1)(932 - 1) = ... = (932 + 1)(916 + 1)(98 + 1)(94 + 1)(92 + 1)(9 + 1)(9 - 1) > (932 + 1)(916 + 1)(98 + 1)(94 + 1)(92 + 1)(9 + 1)
6 tháng 4 2019
964=(932+1).(932-1)
=(932+1)(916+1)(916-1)
=(932+1)(916+1)(98+1)(98-1)
=(932+1)(916+1)(98+1)(94+1)(94-1)
=(932+1)(916+1)(98+1)(94+1)(92+1)(92-1)
=(932+1)(916+1)(98+1)(94+1)(92+1)(9+1)(9-1)
Vì (932+1)(916+1)(98+1)(94+1)(92+1)(9+1)(9-1)>(932+1)(916+1)(98+1)(94+1)(92+1)(9+1)
=>964-1>(932+1)(916+1)(98+1)(94+1)(92+1)(9+1)
26 tháng 2 2020
\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\left(x\ne1\right)\)
\(\Leftrightarrow\frac{1}{x-1}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x}{x^2+x+1}=0\)
\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}\left(x^2+x+1-3x^2-2x^2+2x\right)=0\)
\(\Leftrightarrow-4x^2+3x+1=0\left(\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}\ne0\right)\)
\(\Leftrightarrow-4x^2+4x-x+1=0\)
\(\Leftrightarrow-4x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-4x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\-4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\-4x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\left(loại\right)\\x=\frac{-1}{4}\end{cases}}}\)
Vậy \(x=\frac{-1}{4}\)
XO
20 tháng 6 2021
Ta có A = 2018.2020 + 2019.2021
= (2020 - 2).2020 + 2019.(2019 + 2)
= 20202 - 2.2020 + 20192 + 2.2019
= 20202 + 20192 - 2(2020 - 2019) = 20202 + 20192 - 2 = B
=> A = B
b) Ta có B = 964 - 1= (932)2 - 12
= (932 + 1)(932 - 1) = (932 + 1)(916 + 1)(916 - 1) = (932 + 1)(916 + 1)(98 + 1)(98 - 1)
= (932 + 1)(916 + 1)(98 + 1)(94 + 1)(94 - 1)
= (932 + 1)(916 + 1)(98 + 1)(94 + 1)(92 + 1)(92 - 1)
(932 + 1)(916 + 1)(98 + 1)(94 + 1)(92 + 1).80
mà A = (932 + 1)(916 + 1)(98 + 1)(94 + 1)(92 + 1).10
=> A < B
XO
20 tháng 6 2021
c) Ta có A = \(\frac{x-y}{x+y}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}=\frac{x^2-y^2}{x^2+2xy+y^2}< \frac{x^2-y^2}{x^2+xy+y^2}=B\)
=> A < B
d) \(A=\frac{\left(x+y\right)^3}{x^2-y^2}=\frac{\left(x+y\right)^3}{\left(x+y\right)\left(x-y\right)}=\frac{\left(x+y\right)^2}{x-y}=\frac{x^2+2xy+y^2}{x-y}< \frac{x^2-xy+y^2}{x-y}=B\)
=> A < B
NM
1
HM
20 tháng 11 2016
a) MSC= \(x^2-1\)
\(\Rightarrow\frac{x^2+1}{1}=\frac{\left(x^2+1\right)\left(x^2-1\right)}{1.\left(x^2-1\right)}=\frac{x^4-1}{x^2-1}\)
\(\frac{x^4}{x^2-1}=\frac{x^4}{x^2-1}\)
ND
9 tháng 2 2018
3,
\(9\left(x-3\right)^2=4\left(x+2\right)^2\)
\(\Leftrightarrow3^2\left(x-3\right)^2=2^2\left(x+2\right)^2\)
\(\Leftrightarrow\left(3x-9\right)^2=\left(2x+4\right)^2\)
\(\Leftrightarrow\left(3x-9\right)^2-\left(2x+4\right)^2=0\)
\(\Leftrightarrow\left(3x-9-2x-4\right)\left(3x-9+2x+4\right)=0\)
\(\Leftrightarrow\left(x-13\right)\left(5x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\5x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-13\\x=1\end{matrix}\right.\)
9 tháng 2 2018
1, \(\left(x+1\right)^2=4\left(x^2-2x+1\right)^2\)
\(\Leftrightarrow\left(x+1\right)^2-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2-\left(2x-2\right)^2=0\)
\(\Leftrightarrow\left(x+1-2x+2\right)\left(x+1+2x-2\right)=0\)
\(\Leftrightarrow\left(3-x\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
vậy tập nghiệm của phương trinh \(S=\left\{3;\dfrac{1}{3}\right\}\)
2, \(\left(x^2-9\right)^2-9\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x^2-9\right)^2-\left(3x-9\right)^2=0\)
\(\Leftrightarrow\left(x^2-9-3x+9\right)\left(x^2-9+3x-9\right)=0\)
\(\Leftrightarrow\left(x^2-3x\right)\left(x^2+3x-18\right)=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x^2+6x-3x-18\right)=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x+6\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-6\end{matrix}\right.\)
vậy tập nghiệm của phương trinh \(S=\left\{0;3;-6\right\}\)
3, \(9\left(x-3\right)^2=4\left(x+2\right)^2\)
\(\Leftrightarrow\left(3x-9\right)^2-\left(2x+2\right)^2=0\)
\(\Leftrightarrow\left(3x-9-2x-2\right)\left(3x-9+2x+2\right)=0\)
\(\Leftrightarrow\left(x-11\right)\left(5x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-11=0\\5x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=\dfrac{7}{5}\end{matrix}\right.\)
vậy tập nghiệm của phương trinh \(S=\left\{11;\dfrac{7}{5}\right\}\)
-8
1-9=-8