\(\dfrac{1}{38}>\dfrac{1}{40}>\dfrac{1}{42}>...>\dfrac{1}{50}\)

=>\(\dfrac{1}{38}+\dfrac{1}{40}+\dfrac{1}{42}+\dfrac{1}{44}+\dfrac{1}{46}+\dfrac{1}{48}+\dfrac{1}{50}< 7\cdot\dfrac{1}{38}=\dfrac{7}{38}< 1\)

Vậy tổng trên bé hơn 1

A=-1-3-5-...-2017

=-(1+3+5+...+2017)

Xét tổng B=1+3+5+...+2017

Tổng B có:(2017-1):2+1=1009(số hạng)

Tổng B=\(\dfrac{\left(2017+1\right)\cdot1009}{2}=1009\cdot1009=1018081\)

=>A=-B=-1018081

a) x⁴ chia hết cho xⁿ \(\Rightarrow n\le4\Rightarrow n\in\left\{0;1;2;3;4\right\}\)

b) xⁿ chia hết cho x³ \(\Rightarrow n\ge3\Rightarrow n\in\left\{3;4;5;6;...\right\}\)

c) 5xⁿy³ chia hết cho 4x²y² \(\Rightarrow n\ge2\Rightarrow n\in\left\{2;3;4;5;...\right\}\)

1)a)(x+5)(2x+1)-x²+25=0

<=>(x+5)(2x+1)-(x-5)(x+5)=0

<=>(x+5)(2x+1-x+5)=0

<=>(x+5)(x+6)=0

<=>x+5=0 hoặc x+6=0

<=>x=-5 hoặc x=-6

Vậy...

b)\(\dfrac{3x}{x-2}-\dfrac{x}{x-5}+\dfrac{3x}{\left(x-2\right)\left(x-5\right)}=0\)

ĐKXĐ:\(x\ne2;x\ne5\)

=>3x²-15x-x²+2x+3x=0

<=>2x²-10x=0

<=>2x(x-5)=0

<=>2x=0 hoặc x-5=0

<=>x=0(TM) hoặc x=5(L)

Vậy...

2)\(\dfrac{x+1}{x+2+m}=\dfrac{x+1}{x+2-m}\)

ĐKXĐ:\(x\ne-m-2;x\ne m-2\)

a)Với m=-3

ĐKXĐ: \(x\ne1;x\ne-5\)

\(\dfrac{x+1}{x-1}=\dfrac{x+1}{x+5}\)

=>(x+1)(x+5)=(x+1)(x-1)

<=>(x+1)(x+5)-(x+1)(x-1)=0

<=>(x+1)(x+5-x+1)=0

<=>6(x+1)=0

<=>x+1=0

<=>x=-1(TM)

Vậy...

b)Với x=3

ĐKXĐ:-m-2\(\ne\)3;m-2\(\ne\)3 hay m\(\ne\)-5;m\(\ne\)5

Thay x=3 vào phương trình ta có:

\(\dfrac{4}{5+m}=\dfrac{4}{5-m}\)

=>4(5-m)=4(5+m)

<=>5-m=5+m

<=>2m=0

<=>m=0(TM)

Vậy m=0 thì phương trình có nghiệm x=3

cứ vc

ko ........ qan ........ tâm ! ~ 

Ta có:a⁴+16-2a³-8a

=(a⁴-8a²+16)-(2a³-8a²+8a)

=(a²-4)²-2a(a-2)²

=(a-2)²[(a+2)²-2a]

=(a-2)²(a²+4a+4-2a)

=(a-2)²(a²+2a+4)

=(a-2)²[(a+1)²+3]\(\)\(\ge\)0 với mọi a

=>a⁴+16-2a³-8a \(\ge\)0

<=>a⁴+16\(\ge\)2a³+8a

Công thức tổng quát: \(\dfrac{1}{n\left(n+1\right)}< \dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)

=>\(\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}< K< \dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{98.99}\)

=>\(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}< K< \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}\)

=>\(\dfrac{1}{4}-\dfrac{1}{100}< K< \dfrac{1}{3}-\dfrac{1}{100}\)

=>\(\dfrac{1}{4}< K< \dfrac{1}{3}\)

=>\(\dfrac{1}{5}< K< \dfrac{1}{3}\left(do\dfrac{1}{4}>\dfrac{1}{5}\right)\)

Ta có:f'(x)=4x-1

=>f'(x)\(\sqrt{x^2+1}=2x^2+2x+1\)

<=>(4x-1)\(\sqrt{x^2+1}=2x^2+2x+1\)

Nhận xét: vế phải > 0 nên đk để phương trình có nghiệm:x>\(\dfrac{1}{4}\)

Từ điều kiện trên phương trình

<=>(16x²-8x+1)(x²+1)=4x⁴+8x³+8x²+4x+1

<=>16x⁴+16x²-8x³-8x+x²+1=4x⁴+8x³+8x²+4x+1

<=>12x⁴-16x³+9x²-12x=0

<=>x(12x³-16x²+9x-12)=0

<=>x(3x-4)(4x²+3)=0

<=>x=0 hoặc x=\(\dfrac{4}{3}\)(do 4x²+3>0)

Vậy...

a)Q(x)=2x²+3x

Cho Q(x)=0

<=>2x²+3x=0

<=>x(2x+3)=0

<=>x=0 hoặc 2x+3=0

<=>x=0 hoặc x=-\(\dfrac{3}{2}\)

Vậy...

b)Cho x²+4x+5=0

<=>(x²+4x+4)+1=0

<=>(x+2)²+1=0(1)

Do (x+2)²\(\ge\)0 với mọi x

=>(x+2)²+1\(\ge\)1>0 với mọi x

=>(1) vô nghiệm

Vậy...

D=\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)

=>3D=1+\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

=>3D-D=(1+\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\))-(\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\))

=>2D=1-\(\dfrac{1}{3^{100}}< 1\)

=>D<\(\dfrac{1}{2}\)

Vậy...

ĐKXĐ:x khác 0 y khác 0

\(\left\{{}\begin{matrix}x-y=-1\\\dfrac{2}{x}+\dfrac{3}{y}=2\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}y=x+1\left(1\right)\\2y+3x=2xy\left(2\right)\end{matrix}\right.\)

Thay 1 vào 2 ta có:

2(x+1)+3x=2x(x+1)

<=>5x+2=2x²+2x

<=>2x²-3x+2=0

<=>2x²-3x+\(\dfrac{9}{8}\)+\(\dfrac{7}{8}\)=0

<=>2(x-\(\dfrac{3}{4}\))²+\(\dfrac{7}{8}\)=0(vô lí do \(2\left(x-\dfrac{3}{4}\right)^2\ge0\forall x\))

Vậy hệ vô nghiệm