ĐKXĐ : \(2014x+1\ne0;2015x+2\ne0;2016x+3\ne0;2017x+4\ne0\)

=> \(x\ne-\frac{1}{2014};x\ne-\frac{2}{2015};x\ne-\frac{3}{2016};x\ne-\frac{4}{2017}\)

Ta có \(\frac{1}{2014x+1}-\frac{1}{2015x+2}=\frac{1}{2016x+3}-\frac{1}{2017x+4}\)

<=> \(\frac{x+1}{\left(2014x+1\right)\left(2015x+2\right)}=\frac{x+1}{\left(2016x+3\right)\left(2017x+4\right)}\)

<=> \(\left(x+1\right)\left(\frac{1}{\left(2014x+1\right)\left(2015x+2\right)}-\frac{1}{\left(2016x+3\right)\left(2017x+4\right)}\right)=0\)

<=> \(\left(x+1\right)\left(\frac{8062x^2+8072x+10}{\left(2014x+1\right)\left(2015x+2\right)\left(2016x+3\right)\left(2017x+4\right)}\right)=0\)

<=> \(\left(x+1\right)\frac{\left(x+1\right)\left(8062x+10\right)}{\left(2014x+1\right)\left(2015x+2\right)\left(2016x+3\right)\left(2017x+4\right)}=0\)

<=> (x + 1)²(8062x + 10) = 0

<=> \(\orbr{\begin{cases}x+1=0\\8062x+10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\left(tm\right)\\x=-\frac{5}{4031}\left(tm\right)\end{cases}}\)

Vậy \(x\in\left\{-1;-\frac{5}{4031}\right\}\)là nghiệm phương trình

Có x = 3

<=> -4.3 - 10 = 5.3 + a

<=> -22 = 15 + a

<=> a = -37

Vậy a = -37

xin lỗi mình nhẩm sai :v 

\(\Leftrightarrow-12-10=15+a\Leftrightarrow-22=15+a\Leftrightarrow a=-37\)

Vậy a = -37 nếu x = 3

3y³ - 7y² - 7y + 3 = 0

<=> 3y³ + 3y² - 10y² - 10y + 3y + 3 = 0

<=> 3y²( y + 1 ) - 10y( y + 1 ) + 3( y + 1 ) = 0

<=> ( y + 1 )( 3y² - 10y + 3 ) = 0

<=> ( y + 1 )( 3y² - 9y - y + 3 ) = 0

<=> ( y + 1 )[ 3y( y - 3 ) - ( y - 3 ) ] = 0

<=> ( y + 1 )( y - 3 )( 3y - 1 ) = 0

<=> y = -1 hoặc y = 3 hoặc y = 1/3

Vậy ...

2y⁴ - 9y³ + 14y² - 9y + 2 = 0

<=> 2y⁴ - 4y³ - 5y³ + 10y² + 4y² - 8y - y + 2 = 0

<=> 2y³( y - 2 ) - 5y²( y - 2 ) + 4y( y - 2 ) - ( y - 2 ) = 0

<=> ( y - 2 )( 2y³ - 5y² + 4y - 1 ) = 0

<=> ( y - 2 )( 2y³ - 2y² - 3y² + 3y + y - 1 ) = 0

<=> ( y - 2 )[ 2y²( y - 1 ) - 3y( y - 1 ) + ( y - 1 ) ] = 0

<=> ( y - 2 )( y - 1 )( 2y² - 3y + 1 ) = 0

<=> ( y - 2 )( y - 1 )( 2y² - 2y - y + 1 ) = 0

<=> ( y - 2 )( y - 1 )[ 2y( y - 1 ) - ( y - 1 ) ] = 0

<=> ( y - 2 )( y - 1 )²( 2y - 1 ) = 0

<=> y = 2 hoặc y = 1 hoặc y = 1/2

Vậy ...

\(ĐK:a,b,c\ne0\)

Ta có: \(\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}+\frac{a^2+b^2-c^2}{2ab}=1\)\(\Leftrightarrow\left(\frac{b^2+c^2-a^2}{2bc}+1\right)+\left(\frac{c^2+a^2-b^2}{2ca}-1\right)+\left(\frac{a^2+b^2-c^2}{2ab}-1\right)=0\)\(\Leftrightarrow\frac{\left(b+c\right)^2-a^2}{2bc}+\frac{\left(c-a\right)^2-b^2}{2ca}+\frac{\left(a-b\right)^2-c^2}{2ab}=0\)\(\Leftrightarrow\frac{\left(b+c-a\right)\left(b+c+a\right)}{2bc}+\frac{\left(c-a-b\right)\left(b+c-a\right)}{2ca}-\frac{\left(a-b+c\right)\left(b+c-a\right)}{2ab}=0\)\(\Leftrightarrow\left(b+c-a\right)\frac{a\left(a+b+c\right)+b\left(c-a-b\right)-c\left(a-b+c\right)}{2abc}=0\)\(\Leftrightarrow\frac{\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)}{2abc}=0\)

Trường hợp 1: \(b+c-a=0\)thì

+) \(\frac{\left(b+c\right)^2-a^2}{2bc}=\frac{\left(b+c-a\right)\left(a+b+c\right)}{2bc}=0\Rightarrow\frac{b^2+c^2-a^2}{2bc}=-1\)

+) \(\frac{\left(a-b\right)^2-c^2}{2ab}=\frac{\left(a-b-c\right)\left(a+c-b\right)}{2ab}=0\Rightarrow\frac{a^2+b^2-c^2}{2ab}=1\)

\(\Rightarrow\frac{c^2+a^2-b^2}{2ca}=1\)

Điều này chứng tỏ có hai phân thức có giá trị là 1 và một phân thức có giá trị -1

Trường hợp 2: \(c+a-b=0\) thì 

+) \(\frac{\left(a-b\right)^2-c^2}{2ab}=\frac{\left(a-b-c\right)\left(a+c-b\right)}{2ab}=0\Rightarrow\frac{a^2+b^2-c^2}{2ab}=1\)

+) \(\frac{\left(c+a\right)^2-b^2}{2ca}=\frac{\left(c+a-b\right)\left(c+a+b\right)}{2ca}=0\Rightarrow\frac{c^2+a^2-b^2}{2ca}=-1\)

\(\Rightarrow\frac{b^2+c^2-a^2}{2bc}=1\)

Điều này cũng chứng tỏ có hai phân thức có giá trị là 1 và một phân thức có giá trị -1

Trường hợp 3: \(a+b-c=0\)

+) \(\frac{\left(c-a\right)^2-b^2}{2ca}=\frac{\left(c-a-b\right)\left(c-a+b\right)}{2ca}=0\Rightarrow\frac{c^2+a^2-b^2}{2ca}=1\)

+) \(\frac{\left(a+b\right)^2-c^2}{2ab}=\frac{\left(a+b-c\right)\left(a+b+c\right)}{2ab}=0\Rightarrow\frac{a^2+b^2-c^2}{2ab}=-1\)

\(\Rightarrow\frac{b^2+c^2-a^2}{2bc}=1\)

Điều này cũng chứng tỏ có hai phân thức có giá trị là 1 và một phân thức có giá trị -1 (đpcm)

cho mình hỏi tại sao từ

\(\left(b+c-a\right)\cdot\frac{a\left(a+b+c\right)+b\left(c-a-b\right)-c\left(a-b+c\right)}{2abc}=0\)

lại có thể suy ra được

\(\frac{\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)}{2abc}=0\) vậy ?

\(S=3-2+3^2-2^2+3^3-2^3+...+3^{2019}-2^{2019}\)

\(=\left(3+3^2+3^3+...+3^{2019}\right)-\left(2+2^2+2^3+...+2^{2019}\right)\)

\(=B-C\)

\(B=3+3^2+3^3+...+3^{2019}\)

\(3B=3^2+3^3+3^4+...+3^{2020}\)

\(3B-B=\left(3^2+3^3+3^4+...+3^{2020}\right)-\left(3+3^2+3^3+...+3^{2019}\right)\)

\(2B=3^{2020}-3\)

\(B=\frac{3^{2020}-3}{2}\)

\(C=2+2^2+2^3+...+2^{2019}\)

\(2C=2^2+2^3+2^4+...+2^{2020}\)

\(2C-C=\left(2^2+2^3+2^4+...+2^{2020}\right)-\left(2+2^2+2^3+...+2^{2019}\right)\)

\(C=2^{2020}-2\)

\(\Rightarrow S=B-C=\frac{3^{2020}-3}{2}-\left(2^{2020}-2\right)\)

\(=\frac{3^{2020}-3}{2}-\frac{2.\left(2^{2020}-2\right)}{2}\)

\(=\frac{3^{2020}-3-2^{2021}+4}{2}\)

\(=\frac{3^{2020}-2^{2021}+1}{2}\)

Vậy \(S=\frac{3^{2020}-2^{2021}+1}{2}\)

ta có 

\(A=\frac{3-4x}{x^2+1}\Leftrightarrow Ax^2+4x+A-3=0\)

\(\Leftrightarrow\Delta'=4-A.\left(A-3\right)\ge0\Leftrightarrow A\in\left[-1;4\right]\)

Do đó giá trị nhỏ nhất của A là -1 khi x=2

*nháp

Ta có: \(A=\frac{3-4x}{x^2+1}\Leftrightarrow Ax^2+A=3-4x\Leftrightarrow Ax^2+4x+\left(A-3\right)=0\)

\(\Delta=4^2-4A\left(A-3\right)=-4A^2+12A+16\ge0\)

\(\Leftrightarrow A^2-3A-4\le0\Leftrightarrow\left(A^2+A\right)-\left(4A+4\right)\le0\)

\(\Leftrightarrow\left(A+1\right)\left(A-4\right)\le0\Rightarrow4\ge A\ge-1\)

Khi đó Min(A) = -1

Bài làm:

Ta có: \(A=\frac{3-4x}{x^2+1}=\frac{\left(x^2-4x+4\right)-x^2-1}{x^2+1}=\frac{\left(x-2\right)^2}{x^2+1}-1\ge-1\left(\forall x\right)\)

Dấu "=" xảy ra khi: x = 2

Vậy Min(A) = -1 khi x = 2

ta có 

\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\Leftrightarrow\frac{1}{1+a^2}-\frac{1}{1+ab}+\frac{1}{1+b^2}-\frac{1}{1+ab}\ge0\)

\(\Leftrightarrow\frac{a\left(b-a\right)}{\left(1+a^2\right)\left(1+ab\right)}+\frac{b\left(a-b\right)}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2\left(ab-1\right)}{\left(1+a^2\right)\left(1+ab\right)\left(1+b^2\right)}\ge0\) luôn đúng do ab>= 1

Bất đẳng thức cần CM tương đương:

\(\frac{2+a^2+b^2}{\left(1+a^2\right)\left(1+b^2\right)}\ge\frac{2}{1+ab}\)

\(\Leftrightarrow\left(2+a^2+b^2\right)\left(1+ab\right)\ge2\left(1+a^2+b^2+a^2b^2\right)\)

\(\Leftrightarrow ab\left(a^2+b^2\right)+2ab+2+a^2+b^2\ge2+2a^2+2b^2+2a^2b^2\)

\(\Leftrightarrow\left[ab\left(a^2+b^2\right)-2a^2b^2\right]-\left(a^2-2ab+b^2\right)\ge0\)

\(\Leftrightarrow ab\left(a^2-2ab+b^2\right)-\left(a^2-2ab+b^2\right)\ge0\)

\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\) (luôn đúng)

Dấu "=" xảy ra khi: a = b = 1

Ta có: \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Leftrightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)

\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{zx}{ca}\right)=1\)

\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\cdot\frac{xyc+yza+zxb}{abc}=1\)

Mà \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Leftrightarrow\frac{yza+zxb+xyc}{xyz}=0\)

\(\Rightarrow yza+zxb+xyc=0\)

\(\Rightarrow A=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\)