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T
11 tháng 6 2019
Phân tích đa thức thành nhân tử:(em làm luôn đấy,ko ghi lại đề)
\(\left(x^3+y^3\right)-\left(x+y\right)+3xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)+3xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)\(=\left(x+y\right)\left[\left(x+y\right)^2-1^2\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
DT
11 tháng 6 2019
\(8x^3+12x^2+6x+1=0.\)
\(\Leftrightarrow\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3=0\)
\(\Leftrightarrow\left(2x+1\right)^3=0\)
\(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow x=-\frac{1}{2}\)
\(2x^2+5x-3=0\Leftrightarrow\left(2x^2+6x\right)+\left(-x-3\right)=0\)
\(\Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)
\(x^2-2x-3=0\Leftrightarrow\left(x^2-3x\right)+\left(x-3\right)=0\)
\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}.}\)
\(\left(5x-1\right)+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)
\(=5x-1+2\left(4+5x-20x-25x^2\right)+25x^2+40x+16\)
\(=25x^2+45x+15+8+10x-40x-50x^2\)
\(=-25x^2+15x+23\)
\(\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)
\(=\left(x-y\right)^3-\left(x-y\right)^3+\left(x+y\right)^3-3x^2y-3xy^2\)
\(=\left(x+y\right)^3-3x^2y-3xy^2\)
\(=x^3+3x^2y+3xy^2+y^3-3xy^2-3x^2y\)
\(=x^3+y^3\)
NT
2
28 tháng 6 2018
\(a)\) \(\left(5x-1\right)^2-5x\left(5x-1\right)=0\)
\(\Leftrightarrow\)\(\left(5x-1\right)\left(5x-1-5x\right)=0\)
\(\Leftrightarrow\)\(\left(5x-1\right).\left(-1\right)=0\)
\(\Leftrightarrow\)\(5x-1=0\)
\(\Leftrightarrow\)\(5x=1\)
\(\Leftrightarrow\)\(x=\frac{1}{5}\)
Vậy \(x=\frac{1}{5}\)
\(b)\) \(x\left(x+1\right)\left(x+2\right)=0\)
Suy ra \(x=0\) hoặc \(x+1=0\) hoặc \(x+2=0\)
\(\Leftrightarrow\)\(x=0\) hoặc \(x=-1\) hoặc \(x=-2\)
Vậy \(x=0\) hoặc \(x=-1\) hoặc \(x=-2\)
\(c)\) \(\left(3x+2\right)x-3\left(3x+2\right)=0\)
\(\Leftrightarrow\)\(\left(3x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+2=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=0-2\\x=0+3\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=-2\\x=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=3\end{cases}}}\)
Vậy \(x=\frac{-2}{3}\) hoặc \(x=3\)
Chúc bạn học tốt ~
HH
28 tháng 6 2018
a/ \(\left(5x-1\right)^2-5x\left(5x-1\right)=0\)
<=> \(\left(5x-1\right)\left(5x-1-5x\right)=0\)
<=> \(-1\left(5x-1\right)=0\)
<=> \(5x-1=0\)
<=> \(5x=1\)
<=> \(x=\frac{1}{5}\)
b/ \(x\left(x+1\right)\left(x+2\right)=0\)
<=> \(x=0\) hoặc \(\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\)
<=> \(x=0\)hoặc \(\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
c/ \(\left(3x+2\right)x-3\left(3x+2\right)=0\)
<=> \(\left(3x+2\right)\left(x-3\right)=0\)
<=> \(\orbr{\begin{cases}3x+2=0\\x-3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}3x=-2\\x=3\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-\frac{2}{3}\\x=3\end{cases}}\)
24 tháng 9 2020
a) \(x^3=x^5\)
=> \(x^3-x^5=0\)
=> \(x^3\left(1-x^2\right)=0\)
=> \(\orbr{\begin{cases}x^3=0\\1-x^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
b) \(4x\left(x+1\right)=x+1\)
=> \(4x^2+4x-x-1=0\)
=> \(4x\left(x+1\right)-1\left(x+1\right)=0\)
=> \(\left(x+1\right)\left(4x-1\right)=0\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{4}\end{cases}}\)
c) \(x\left(x-1\right)-2\left(1-x\right)=0\)
=> \(x\left(x-1\right)-\left[-2\left(x+1\right)\right]=0\)
=> \(x\left(x-1\right)+2\left(x-1\right)=0\)
=> \(\left(x-1\right)\left(x+2\right)=0\Rightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)
d) Kết quả ?
e) \(\left(x-3\right)^2+3-x=0\)
=> \(x^2-6x+9+3-x=0\)
=> \(x^2-7x+12=0\)
=> \(x^2-3x-4x+12=0\)
=> \(x\left(x-3\right)-4\left(x-3\right)=0\)
=> (x - 4)(x - 3) = 0
=> \(\orbr{\begin{cases}x=4\\x=3\end{cases}}\)
f) Tương tự
DT
24 tháng 7 2016
\(5x.\left(x-1\right)=x-1\)
\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=1\end{cases}}}\)
\(S=\left\{\frac{1}{5};1\right\}\)
\(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2-x=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}}\)
\(S=\left\{2;-5\right\}\)
CH
1
IM
8 tháng 8 2016
a)
\(\Rightarrow x\left(x-5\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x-5=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)
b)
\(\Rightarrow3x\left(x-2\right)-2\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(3x-2\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-2=0\\3x-2=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{2}{3}\end{array}\right.\)
c)
\(\Rightarrow\left(3x-1\right)\left(5x+x-2\right)=0\)
\(\Rightarrow\left(3x-2\right)^2.2=0\)
\(\Rightarrow3x-2=0\)
\(\Rightarrow x=\frac{2}{3}\)
TK
6
26 tháng 8 2020
a) 5x( x - 1 ) = x - 1
<=> 5x2 - 5x = x - 1
<=> 5x2 - 5x - x + 1 = 0
<=> 5x2 - 6x + 1 = 0
<=> 5x2 - 5x - x + 1 = 0
<=> 5x( x - 1 ) - 1( x - 1 ) = 0
<=> ( x - 1 )( 5x - 1 ) = 0
<=> \(\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
b) 2( x + 5 ) - x2 - 5x = 0
<=> 2x + 10 - x2 - 5x = 0
<=> -x2 - 3x + 10 = 0
<=> -x2 - 5x + 2x + 10 = 0
<=> -x( x + 5 ) + 2( x + 5 ) = 0
<=> ( x + 5 )( 2 - x ) = 0
<=> \(\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
c) x2 - 2x - 3 = 0
<=> x2 + x - 3x - 3 = 0
<=> x( x + 1 ) - 3( x + 1 ) = 0
<=> ( x + 1 )( x - 3 ) = 0
<=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
d) 2x2 + 5x - 3 = 0
<=> 2x2 - x + 6x - 3 = 0
,<=> x( 2x - 1 ) + 3( 2x - 1 ) = 0
<=> ( 2x - 1 )( x + 3 ) = 0
<=> \(\orbr{\begin{cases}2x-1=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)
PN
26 tháng 8 2020
a) 5x ( x - 1 ) = x - 1 <=> 5x2 - 5x - x + 1 = 0
<=> 5x2 - 6x + 1 = 0 <=> 5x2 - x - ( 5x - 1 ) = 0
<=> x ( 5x - 1 ) - ( 5x - 1 ) = 0 <=> ( x - 1 )( 5x - 1 ) = 0
<=> x = 1 hoặc x = 1/5
b) 2 ( x + 5 ) - x2 - 5x = 0 <=> 2 ( x + 5 ) - x ( x + 5 ) = 0
<=> ( 2 - x ) ( x + 5 ) = 0 <=> x = 2 hoặc x = -5
c) x2 - 2x - 3 = 0 <=> x2 + x - 3x - 3 = 0
<=> x ( x + 1 ) - 3 ( x + 1 ) = 0 <=> ( x - 3 ) ( x + 1 ) = 0
<=> x = 3 hoặc x = -1
d) 2x2 + 5x - 3 = 0
Ta có : delta = 52 - 4.2.3 = 25 - 24 = 1
Khi đó : x = -1 hoặc x = 3/2
A
1
+++++\(x+5x^2=0\) \(\Leftrightarrow x.\left(1+5x\right)=0\) \(\Leftrightarrow\hept{\begin{cases}x=0\\1+5x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{-1}{5}\end{cases}}}\)
++++\(\left(x+1\right)^2=x+1\) \(\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0\) \(\Leftrightarrow\left(x+1\right).\left(x+1-1\right)=0\)
\(\Leftrightarrow x.\left(x+1\right)=0\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}}\)
++++\(5x.\left(x-1\right)=x-1\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\5x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}}\)
a) x + 5x^2 = 0
=> x(1+5x) = 0
=> \(\orbr{\begin{cases}x=0\\1+5x=0\end{cases}}\) => \(\orbr{\begin{cases}x=0\\5x=-1\end{cases}}\) => \(\orbr{\begin{cases}x=0\\x=\frac{-1}{5}\end{cases}}\)
Vậy: x=0 hoặc x=-1/5
b) (x+1)^2 = x+1
=> (x+1)(x+1) = x+1
=> x+1 = (x+1) : (x+1)
=> x+1 = 1
=> x = 0
Vậy: x = 0
c) 5x(x-1) = x-1
=> 5x = (x-1) : (x-1)
=> 5x = 1
=> x = 1/5
Vậy: x = 1/5