3x+4+3x+2-20=53 nhân 2
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LT
1
20 tháng 9 2019
\(53^2+53.94+47^2=53^2+2.53.47+47^2=\left(53+47\right)^2=100^2=10000\)
\(\left(3x-2\right)^2-2\left(3x-2\right).\left(3x-5\right)+\left(5-3x\right)^2=\left(3x-2-3x+5\right)^2=3^2=9\)
trong đó\(\left(5-3x\right)^2=\left(3x-5\right)^2\)
27 tháng 7 2017
1, x^2 +5x+4x+20
= x(x+4)+5(x+4)
= (x+5).(x+4)
2,x^2 -x -20
= x^2 -5x+4x-20
= x(x-5)+4(x-5)
=(x+4).(x-5)
3,3x^2 +3x -2x -2
= 3x(x+1)-2(x+1)
= (3x-2)(x+1)
4, 2x^2-4x-x-2
=2x(x-2)-x-2
=(2x-1)(x+2)
5,6x^2-2x+9x-3
=2x(3x-1)+3(3x-1)
=(2x+3)(3x-1)
6,3x^2 +2x+9x+6
=3x(x+3)+2(x+3)
=(3x+2)(x+3)
7,= 2(x^2-10x+3)
8,=x^4 +4^3
mk làm luôn ko chép đề nha bn
DD
26 tháng 7 2017
\(x^2+9x+20\)
\(\Leftrightarrow x^2+4x+5x+20\)
\(\Leftrightarrow\left(x^2+4x\right)+\left(5x+20\right)\)
\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)\)
\(\Leftrightarrow\left(x+4\right)\left(x+5\right)\)
\(x^2-x-20\)
\(\Leftrightarrow x^2-5x+4x-20\)
\(\Leftrightarrow\left(x^2-5x\right)+\left(4x-20\right)\)
\(\Leftrightarrow x\left(x-5\right)+4\left(x-5\right)\)
\(\Leftrightarrow\left(x-5\right)\left(x+4\right)\)
\(3x^2+x-2\)
\(\Leftrightarrow3x^2+3x-2x-2\)
\(\Leftrightarrow\left(3x^2+3x\right)-\left(2x+2\right)\)
\(\Leftrightarrow3x\left(x+1\right)-2x\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(3x-2x\right)\)
\(\)
M
0
KV
17 tháng 12 2023
Bài 1
a) 5x²y - 20xy²
= 5xy(x - 4y)
b) 1 - 8x + 16x² - y²
= (1 - 8x + 16x²) - y²
= (1 - 4x)² - y²
= (1 - 4x - y)(1 - 4x + y)
c) 4x - 4 - x²
= -(x² - 4x + 4)
= -(x - 2)²
d) x³ - 2x² + x - xy²
= x(x² - 2x + 1 - y²)
= x[(x² - 2x+ 1) - y²]
= x[(x - 1)² - y²]
= x(x - 1 - y)(x - 1 + y)
= x(x - y - 1)(x + y - 1)
e) 27 - 3x²
= 3(9 - x²)
= 3(3 - x)(3 + x)
f) 2x² + 4x + 2 - 2y²
= 2(x² + 2x + 1 - y²)
= 2[(x² + 2x + 1) - y²]
= 2[(x + 1)² - y²]
= 2(x + 1 - y)(x + 1 + y)
= 2(x - y + 1)(x + y + 1)
17 tháng 12 2023
Bài 2:
a: \(x^2\left(x-2023\right)+x-2023=0\)
=>\(\left(x-2023\right)\left(x^2+1\right)=0\)
mà \(x^2+1>=1>0\forall x\)
nên x-2023=0
=>x=2023
b:
ĐKXĐ: x<>0
\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)
=>\(-x\left(x-4\right)+2x^2-4x-9=0\)
=>\(-x^2+4x+2x^2-4x-9=0\)
=>\(x^2-9=0\)
=>(x-3)(x+3)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
c: \(x^2+2x-3x-6=0\)
=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)
=>\(x\left(x+2\right)-3\left(x+2\right)=0\)
=>(x+2)(x-3)=0
=>\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
d: 3x(x-10)-2x+20=0
=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)
=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)
=>\(\left(x-10\right)\left(3x-2\right)=0\)
=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)
Câu 1:
a: \(5x^2y-20xy^2\)
\(=5xy\cdot x-5xy\cdot4y\)
\(=5xy\left(x-4y\right)\)
b: \(1-8x+16x^2-y^2\)
\(=\left(16x^2-8x+1\right)-y^2\)
\(=\left(4x-1\right)^2-y^2\)
\(=\left(4x-1-y\right)\left(4x-1+y\right)\)
c: \(4x-4-x^2\)
\(=-\left(x^2-4x+4\right)\)
\(=-\left(x-2\right)^2\)
d: \(x^3-2x^2+x-xy^2\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x^2-2x+1\right)-y^2\right]\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-1-y\right)\left(x-1+y\right)\)
e: \(27-3x^2\)
\(=3\left(9-x^2\right)\)
\(=3\left(3-x\right)\left(3+x\right)\)
f: \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)-y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x+1+y\right)\left(x+1-y\right)\)
NA
3
15 tháng 6 2016
\(^{x^4+3x^3+x^2-12x-20}\)
= x^4 + 2X^3 + x^3 + 2x^2 - X^2 - 2X -10X - 20
= X^3( x + 2 ) + X^2( x+2) -x(x+2) - 10(x+2)
= ( x^3 + x^2 - x -10) (x+2 )
= (x-2)( x^2 + 3x+5)(x+2)
15 tháng 6 2016
x^4+3x^2+x^2-12x-20
=x^4+2x^3+x^3+2x^2-x^2-2x-10x-20
=x^3(x+2)+x^2(x+2)-x(x+2)-10(x+2)
=(x+2)(x^3+x^2-x-10)
=(x+2)(x^3-2x^2+3x^2-6x+5x-10)
=(x+2)[x^2(x-2)+3x(x-2)+5(x-2)]
=(x+2)(x-2)(x^2+3x+5)
9 tháng 8 2015
a) x2 + x - 20 = x2 - 4x + 5x - 20 = x(x - 4) + 5(x - 4) = (x - 4)(x + 5)
b) x2 - x - 20 = x2 + 4x - 5x - 20 = x(x + 4) - 5(x + 4) = (x + 4)(x - 5)
c) 2x2 - 3x - 2 = 2x2 - 4x + x - 2 = 2x(x - 2) + (x - 2) = (x - 2)(2x + 1)
d) 3x2 + x - 2 = 3x2 + 3x - 2x - 2 = 3x(x + 1) - 2(x + 1) = (x + 1)(3x - 2)
3x+4+3x+2-20=53.2
=> 3x.34+3x.32-20=125.2
=> 3x(34+32)=125.2+20
=> 3x.90=270
=> 3x=270:90
=> 3x=3
=> x=1