5x+5x+1+5x+2+.....+5x+2015=52019-125
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\(\Leftrightarrow5^x=\dfrac{5^{2019}}{5^{2010}\cdot5^2}=5^7\)
hay x=7
Phân tích đa thức thành nhân tử(tách hạng tử)
1)x^2+2x-3=x^2-x+3x-3=x(x-1)+3(x-1)=(x-1)(x+3)
2)x^2-5x+6=x^2-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)
3)x^2+7x+12=(x+3)(x+4)
4)x^2-x-12=(x-4)(x+3)
5)3x^2+3x-36=3[(x-3)(x+4)]
6)5x^2-5x-10=5[(x-2)(x+1) ]
7)3x^2-7x-6=(x-3)(3x+2)
8)4x^2+4x-3=4x^2+6x-2x-3=(2x-1)(2x+3)
9)8x^2-2x-3=8x^2+4x-6x-3=(4x-3)(2x+1)
1: \(x^2+2x-3=\left(x+3\right)\left(x-1\right)\)
2: \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)
3: \(x^2+7x^2+12x=4x\left(2x+3\right)\)
4: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)
5: \(3x^2+3x-36=3\left(x^2+x-12\right)=3\left(x+4\right)\left(x-3\right)\)
6: \(5x^2-5x-10=5\left(x^2-x-2\right)=5\left(x-2\right)\left(x+1\right)\)
a) x = 4
b) x = 5
c) x = 2
d) x = 2
e, x = 1
f, x = 0 hoặc x = 1
a) x = 4
b) x = 5
c) x = 2
d) x = 2
e) x = 1
f) x = 0 hoặc x = 1.
\(\left(5x-1\right)^2+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)
\(=\left(5x-1\right)^2-2\left(5x-1\right)\left(5x+4\right)+\left(5x+4\right)^2\)
\(=\left[\left(5x-1\right)-\left(5x+4\right)\right]^2\)
\(=\left(5x-1-5x-4\right)^2\)
\(=\left(-5\right)^2\)
\(=25\)
\(x=\dfrac{1}{2}\cdot\sqrt{\left(\sqrt{2}-1\right)^2}=\dfrac{\sqrt{2}-1}{2}\)
\(A=\left[4\cdot\left(\dfrac{\sqrt{2}-1}{2}\right)^4+4\cdot\left(\dfrac{\sqrt{2}-1}{2}\right)^3-5\cdot\left(\dfrac{\sqrt{2}-1}{2}\right)^2+5\cdot\dfrac{\sqrt{2}-1}{2}-2\right]^{2015}+2016\)
=-1,13+2016=2014,87
(5x-3^2015=(5x-3)^2013
=>(5x-3)^2015-(5x-3)^2013=0
=>(5x-3)^2013.(5x-3)^2-(5x-3)^2013=0
=>(5x-3)^2013.[(5x-3)^2-1]=0
=>(5x-3)^2013=0=>5x-3=0=>5x=3=>x=3/5
hoặc (5x-3)^2-1=0=>(5x-3)^2=1=>....(tự làm tiếp)
A = (5\(x\) + 1)2 + (5\(x\) - 1)2 - 2.( 5\(x\) +1).(5\(x\) - 1) tại \(x\) = 1
Thay \(x\) = 1 vào A ta có:
A = (5.1 + 1)2 + (5.1 - 1)2 - 2.(5.1 + 1).(5.1 - 1)
A = 62 + 42 - 2.6.4
A = 36 + 16 - 48
A = 52 - 48
A = 4