A = \(\frac{4024x\left(2010+4\right)-2}{2011+2012x2010}\)= \(\frac{2024x2010+4024-2}{2011+2012x2010}\)=\(\frac{4024x2010+4022}{2011+2012x2010}\)= 2

câu B hình như sai đề bài . mk moi hoc lop 6 thoi nen cũng ko chắc .

A=2.002983591

B= -1.004960108

A=2.002983591

B=  -1

mình không biết kq =mấy

nhứng mình c/m kq =2 là sai

\(A-2=\dfrac{4024.2014-2}{Khongquantam}-2=\dfrac{4024.2014-2-2.2011-2.2012.2010}{Khongquantam}\)

\(A-2=\dfrac{2\left(2012.2014-2011-2012.2010-1\right)}{Khongquantam}=\dfrac{2\left[2012.\left(2014-2010\right)-2011-1\right]}{Khongquantam}\)

\(A-2=\dfrac{2\left[4.2012-2011-1\right]}{Khongquantam}=\dfrac{2\left[3.2011+3\right]}{Khongquantam}\)

\(A-2=\dfrac{2\left[3.\left(2011+1\right)\right]}{Khongquantam}=\dfrac{2.3.2012}{Khongquantam}\ne0\)\(A-2\ne0\)

\(\Rightarrow A\ne2\Rightarrow kq=2=sai\)

từng bước bao gồm cả lập luân luôn

a)\(\left(1+\frac{1}{2}+...+\frac{1}{2012}\right).503x=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\) (1)

\(A=\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\) (có 2011 số hạng)

nếu ta trừ một vào từng số hạng được tử số giống nhau

\(A-2011=\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4023}{2011}-1\right)+\left(\frac{4024}{2012}-1\right)\)

\(A-2011=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}=2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)

\(A-2011+2012=2012\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)công 2012 hai vế

\(A+1=VP=2012\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)

\(\left(1\right)\Leftrightarrow\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right).503x=2012\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\left(2\right)\)

Chia cả hai vế (2) cho: \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\Rightarrow503x=2012\)

\(x=\frac{2012}{503}\)

Ta có: \(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}=\frac{x+4}{2011}+\frac{x+5}{2010}+\frac{x+6}{2009}\)

\(\Rightarrow\frac{x+1}{2014}+1+\frac{x+2}{2013}+1+\frac{x+3}{2012}+1=\frac{x+4}{2011}+1+\frac{x+5}{2010}+1+\frac{x+6}{2009}+1\)

\(\Rightarrow\frac{2015+x}{2014}+\frac{2015+x}{2013}+\frac{2015+x}{2012}=\frac{2015+x}{2011}+\frac{2015+x}{2010}+\frac{2015+x}{2009}\)

\(\Rightarrow\left(2015+x\right)\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)

=> 2015 + x = 0

=> x = -2015

Các bạn check lại ở dáp án của Ngọc Vĩ nhé!

\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)\cdot503x=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4024}{2012}\)

\(\Leftrightarrow503x=\frac{1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4024}{2012}}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}\)

\(\Leftrightarrow503x=\frac{\frac{2014}{2}-1+\frac{2015}{3}-1+...+\frac{4024}{2012}-1+2012}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}\)

\(\Leftrightarrow503x=\frac{\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2012}+2012}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}\)

\(\Leftrightarrow503x=\frac{2012\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}\)

\(\Leftrightarrow503x=2012\)

\(\Leftrightarrow x=\frac{2012}{503}\)

Phân số \(\frac{2013}{2011}\) phải là \(\frac{4023}{2011}\)

Tử số = \(2012+\left(2015-2\right)x2014=2012+2014x2015-4028=2014x2015-\left(4028-2012\right)=2014x2015-2016\)

Tử số = mấu số nên kết quả =1