Bài 1: x2-x -1. Tính A=\(\frac{x^6-3x^5+3x^4-x^3+2015}{x^6-x^3+3x^2-3x+2015}\)
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MA
19 tháng 9 2016
\(\frac{x^6-3x^5+3x^4-x^3+2015}{x^6-x^3-3x^2-3x+2015}=\frac{x^6-3x^5+3x^4+3x^3+2015-4x^3}{x^6+3x^3-3x^2-3x+2015-4x^3}=\frac{x^6-3x^3\left(x^2-x-1\right)+2015-4x^3}{6+3x\left(x^2-x-1\right)+2015-4x^3}\)
Theo bài ra: \(x^2-x-1=0\)
\(\frac{x^6-3x^5+3x^4-x^3+2015}{x^6-x^3-3x^2-3x+2015}=\frac{x^6-3x^3\left(x^2-x-1\right)+2015-4x^3}{x^6+3x\left(x^2-x-1\right)+2015-4x^3}=\frac{x^6+2015-4x^3}{x^6+2015-4x^3}=1\)
Vậy:...
MA
19 tháng 9 2016
Mk nhầm đoạn số 6 bạn sửa lại thành x^6 nhé:
\(\frac{x^6-3x^5+3x^4-x^3+2015}{x^6-x^3-3x^2-3x+2015}=\frac{x^6-3x^5+3x^4+3x^3+2015-4x^3}{x^6+3x^3-3x^2-3x+2015-4x^3}=\frac{x^6-3x^3\left(x^2-x-1\right)+2015-4x^3}{x^6+3x\left(x^2-x-1\right)+2015-4x^3}\)
Theo bài ra: \(x^2-x-1=0\)
\(\Rightarrow\frac{x^6-3x^5+3x^4-x^3+2015}{x^6-x^3-3x^2-3x+2015}=\frac{x^6-3x^3\left(x^2-x-1\right)+2015-4x^3}{x^6+3x\left(x^2-x-1\right)+2015-4x^3}=\frac{x^6+2015-4x^3}{x^6+2015-4x^3}=1\)
Vậy:......
30 tháng 10 2020
Bài 1.
[ 4( x - y )5 + 2( x - y )3 - 3( x - y )2 ] : ( y - x )2 < sửa một lũy thừa rồi nhé >
= [ 4( x - y )5 + 2( x - y )3 - 3( x - y )3 ] : ( x - y )2
Đặt t = x - y
bthuc ⇔ ( 4t5 + 2t3 - 3t2 ) : t2
= 4t5 : t2 + 2t3 : t2 - 3t2 : t2
= 4t3 + 2t - 3
= 4( x - y )3 + 2( x - y ) - 3
Bài 2.
5x( x - 2 ) + 3x - 6 = 0
⇔ 5x( x - 2 ) + 3( x - 2 ) = 0
⇔ ( x - 2 )( 5x + 3 ) = 0
⇔ x - 2 = 0 hoặc 5x + 3 = 0
⇔ x = 2 hoăc x = -3/5
Bài 3.
A = x2 - 6x + 2023
= ( x2 - 6x + 9 ) + 2014
= ( x - 3 )2 + 2014 ≥ 2014 ∀ x
Dấu "=" xảy ra khi x = 3
=> MinA = 2014 <=> x = 3
Bài 4.
B = ( 3x + 5 )2 + ( 3x - 5 )2 - 2( 3x + 5 )( 3x - 5 )
= [ ( 3x + 5 ) - ( 3x - 5 ) ]2
= ( 3x + 5 - 3x + 5 )2
= 102 = 100
Vậy B không phụ thuộc vào x ( đpcm )
Bài 6.
C = 12 - 22 + 32 - 42 + 52 - 62 + ... + 20132 - 20142 + 20152
= ( 20152 - 20142 ) + ... + ( 52 - 42 ) + ( 32 - 22 ) + 1
= ( 2015 - 2014 )( 2015 + 2014 ) + ... + ( 5 - 4 )( 5 + 4 ) + ( 3 - 2 )( 3 + 2 ) + 1
= 4029 + ... + 9 + 5 + 1
= \(\frac{\left(4029+1\right)\left[\left(4029-1\right)\div4+1\right]}{2}\)
= 2 031 120
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
11 tháng 3 2019
a.\(\frac{1}{2}-\left(x-\frac{1}{3}\right)=\frac{1}{6}\)
\(x-\frac{1}{3}=\frac{1}{2}-\frac{1}{6}\)
\(x-\frac{1}{3}=\frac{1}{3}\)
\(x=\frac{2}{3}\)
11 tháng 3 2019
\(a.\frac{1}{2}-\left(x-\frac{1}{3}\right)=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{2}-x+\frac{1}{3}=\frac{1}{6}\)
\(\Leftrightarrow\frac{5}{6}-x=\frac{1}{6}\)
\(\Leftrightarrow\frac{5}{6}-\frac{1}{6}=x\)
\(\Leftrightarrow x=\frac{2}{3}\)
\(b.||3x+2|-2x-5|=3x-\left(-1\right)^{2015}\)
\(\Leftrightarrow||3x+2|-2x-5|=3x+1\)
\(\Leftrightarrow\orbr{\begin{cases}|3x+2|-2x-5=3x+1\\|3x+2|-2x-5=-3x-1\end{cases}\Leftrightarrow\orbr{\begin{cases}|3x+2|=5x+6\left(n\right)\\|3x+2|=-\left(x-4\right)\left(l\right)\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x+2=5x+6\\3x+2=-5x-6\end{cases}\Leftrightarrow\orbr{\begin{cases}-2x=4\\8x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2\\x=-1\end{cases}}}\)
V...\(x=-1;x=-2\)