VẾ TRÁI LÀ:

A=1/2.3+1/3.2+1/4.5+...+1/x[x+1]

A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/n-1/n+1

A=1-1/n+1

1-1/n+1=2015/2016

1/n+1=1-2015/2016

1/n+1=1/2016

n=2016-1

n=2015

Vế trái là 

S=1/2+1/6+1/12+1/20+..+1/x(x+1)

S=1/1.2+1/2.3+1/3.4+1/4.5+...+1/x.(x+1)

S=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/x.(x+1)

S=1-1/(x+1)=Vế phải=2015/2016

<=>1-1/(x+1)=2015/2016

1/(x+1)=1/2016

=>x+1=2016

x=2015

Ủng hộ mk mha Chí Tiến

1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)

\(\Leftrightarrow5x+20+12x-28=7x+2\)

\(\Leftrightarrow17x-7x=2+8=10\)

hay x=1

2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)

\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)

\(\Leftrightarrow6x+4-12x=-3x+3\)

\(\Leftrightarrow-6x+3x=3-4\)

hay \(x=\dfrac{1}{3}\)

3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)

\(\Leftrightarrow4x-12-x-2=6x-3\)

\(\Leftrightarrow3x-14-6x+3=0\)

\(\Leftrightarrow-3x=11\)

hay \(x=-\dfrac{11}{3}\)

4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)

\(\Leftrightarrow3x-6-8x-12=x+6\)

\(\Leftrightarrow-5x-x=6+18\)

hay x=-4

5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)

\(\Leftrightarrow6x-3+2x-6=-1\)

\(\Leftrightarrow8x=8\)

hay x=1

Ta có :1.2+2.4+3.6+4.8+5.10/3.4+6.8+9.12+12.16+15.20=[1.2(1+4+9+...+25)]/[3.4(1+4+9+16)]

=(1.2)/(3.4)=2/12=1/6

CHÚC BẠN HỌC TỐT!!

cho tớ nhé!!!!

=2+8+18+32+50/12+48+108+192+300

=110/660

=0.166666667\

x*(1/2+1/6+1/12+1/20)=4/5

x*21/20=4/5

x=4/5:21/20

x=16/105

Dấu . là dấu nhân 

\(x.\frac{1}{2}+x.\frac{1}{6}+x.\frac{1}{12}+x.\frac{1}{20}=\frac{4}{5}\)

\(\Rightarrow x.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\right)=\frac{4}{5}\)

\(\Rightarrow x.\left(\frac{30}{60}+\frac{10}{60}+\frac{5}{60}+\frac{3}{60}\right)=\frac{4}{5}\)

\(\Rightarrow x.\frac{48}{60}=\frac{4}{5}\)

\(\Rightarrow x=\frac{4}{5}:\frac{48}{60}\)

\(\Rightarrow x=\frac{4}{5}.\frac{60}{48}\)

\(\Rightarrow x=1\)

Vậy \(x=1\)

Chúc bạn học tốt !!! 

\(x.\frac{1}{2}+x.\frac{1}{6}+x.\frac{1}{12}+x.\frac{1}{20}=\frac{4}{5}\)

\(\Rightarrow x.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\right)=\frac{4}{5}\)

\(\Rightarrow x.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\right)=\frac{4}{5}\)

\(\Rightarrow x.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)=\frac{4}{5}\)

\(\Rightarrow x.\left(1-\frac{1}{5}\right)=\frac{4}{5}\Rightarrow x.\frac{4}{5}=\frac{4}{5}\)

\(\Rightarrow x=1\)

x²-4x+7 = 0 ⇔ x² -4x + 4 + 3 = 0 

⇔ (x-2)²+3=0 ⇔ (x-2)²=-3 (vô lí)

Vậy pt vô nghiệm

*Chứng minh phương trình \(x^2-4x+7=0\) vô nghiệm

Ta có: \(x^2-4x+7=0\)

\(\Leftrightarrow x^2-4x+4+3=0\)

\(\Leftrightarrow\left(x-2\right)^2+3=0\)

mà \(\left(x-2\right)^2+3\ge3>0\forall x\)

nên \(x\in\varnothing\)(đpcm)