flash liên quân

k cho mk nha

x^4-2x^3+3x^2-2x+1

=(x^4-2x^3+x^2)+(x^2-2x+1)

=x^2(x^2-2x+1)+(x^2-2x+1)

=(x^2+1)(x^2-2x+1)

=(x^2+1)(x-1)^2

chết quên

mk mới phân tích thôi còn lại bạn lm nhé

ta có ; x-3/2015 -1 +x-2/2016 -1 = x-2016/2 -1 +x-2015/3-1

x-2018/2015 + x-2018/2016 = x-2018/2 +x-2018/3

(x-2018)*(1/2015+1/2016-1/2-1/3)=0

vi (1/2015+1/2016-1/2-1/3) luon khac 0

suy ra : x-2018 = 0 suy ra x=2018

\(\frac{x-3}{2015}+\frac{x-2}{2016}=\frac{x-2016}{2}+\frac{x-2015}{3}\)

trừ 2 vế với 2, ta có:

\(\frac{x-3}{2015}+\frac{x-2}{2016}-2=\frac{x-2016}{2}+\frac{x-2015}{3}-2\)

\(\left(\frac{x-3}{2015}-1\right)+\left(\frac{x-2}{2016}-1\right)=\left(\frac{x-2016}{2}-1\right)+\left(\frac{x-2015}{3}-1\right)\)

\(\frac{x-2018}{2015}+\frac{x-2018}{2016}=\frac{x-2018}{2}+\frac{x-2018}{3}\)

\(\left(x-2018\right)\frac{1}{2015}+\left(x-2018\right)\frac{1}{2016}=\left(x-2018\right)\frac{1}{2}+\left(x-2018\right)\frac{1}{3}\)

\(\left(x-2018\right)\left(\frac{1}{2015}+\frac{1}{2016}\right)=\left(x-2018\right)\left(\frac{1}{2}+\frac{1}{3}\right)\)

\(\left(x-2018\right)\left(\frac{1}{2015}+\frac{1}{2016}\right)-\left(x-2018\right)\left(\frac{1}{2}+\frac{1}{3}\right)=0\)

\(\left(x-2018\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2}-\frac{1}{3}\right)=0\)

Mà \(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2}-\frac{1}{3}\ne0\)

\(\Rightarrow x-2018=0\Leftrightarrow x=2018\)

Vậy tập nghiệm của PT là\(S=\left\{2018\right\}\)

t is English subject that I best in school. I English because it helps me open the big world. First of all, I can talk with people who are in the English-speaking world. Secondly, I can find many documents written by English for studying. Thirdly, I can watch movie and news without subtitle. Last but not least, it makes me interesting when studying. To study it well, I sometimes take a note and past it in somewhere in my room.

I always do all exercises before going to school. And I am very enthusias in the class. English is known to be very difficult and many students hate studying it. It my feeling that the problem is the methods teaching of Vietnam is not suitable. Student must study grammar hardly. I hope it will change in the future. In short, English is very necessary for modern life, I hope all students find in interesting.

Dự đoán dấu "=" khi x = 2 ; y= 1

Áp dụng bđt Cô-si cho 3 số và bđt \(\frac{a^2}{m}+\frac{b^2}{n}\ge\frac{\left(a+b\right)^2}{m+n}\) ta được

\(P=2x^2+y^2+\frac{28}{x}+\frac{1}{y}\)

    \(=\left(\frac{7x^2}{4}+\frac{14}{x}+\frac{14}{x}\right)+\left(\frac{y^2}{2}+\frac{1}{2y}+\frac{1}{2y}\right)+\left(\frac{x^2}{4}+\frac{y^2}{2}\right)\)

    \(\ge3\sqrt[3]{\frac{7x^2.14.14}{4.x^2}}+3\sqrt[3]{\frac{y^2.1.1}{2.2y.2y}}+\frac{\left(x+y\right)^2}{4+2}\)

      \(=3.\sqrt[3]{\frac{7.14.14}{4}}+\frac{3}{\sqrt[3]{2^3}}+\frac{3^2}{6}=24\)

Dấu "=" khi x = 2 ; y = 1 

Bài toán easy!

\(P=\left(2x^2+8\right)+\left(y^2+1\right)+\frac{28}{x}+\frac{1}{y}-9\)

Áp dụng BĐT AM-GM,ta có:

\(P\ge8x+2y+\frac{28}{x}+\frac{1}{y}-9\)

\(=\left(7x+\frac{28}{x}\right)+\left(y+\frac{1}{y}\right)+\left(x+y\right)-9\)

\(\ge2\sqrt{7x.\frac{28}{x}}+2\sqrt{y.\frac{1}{y}}+\left(x+y\right)-9\)

\(\ge28+2+3-9=24\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}2x^2=8\\y^2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)

Vậy \(P_{min}=24\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)