\(\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\)

\(=\left(\dfrac{a}{c+b}+1\right)+\left(\dfrac{b}{a+c}+1\right)+\left(\dfrac{c}{a+b}+1\right)-3\)

\(=\dfrac{a+c+b}{c+b}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}-3\)

\(=\left(a+b+c\right)\left(\dfrac{1}{c+b}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)-3\)

\(=4034.\dfrac{1}{2}-3=2014\)

Guể?

\(\dfrac{1}{c+b}+\dfrac{1}{a+c}+\dfrac{1}{a+b}=\dfrac{1}{2}\)

\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{c+a}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)=\dfrac{4034}{2}=2017\)

\(\Rightarrow1+\dfrac{a}{c+b}+1+\dfrac{b}{a+c}+1+\dfrac{c}{a+b}=2017\)

\(\Rightarrow\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=2014\)

\(S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)

\(\Rightarrow S=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)

\(\Rightarrow S=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3\)

\(\Rightarrow S=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)

\(\Rightarrow S=4034.\frac{1}{2}-3=2014\)

Vậy S=2014

key ban di chi gium n he khong k luon nhe

mình ko hiểu

TUI ĐOÁN BỪA : CHỌN C

Một hình lăng trụ có 36 cạnh . Hỏi hình lăng trụ đó có bao nhiêu mặt A. 2017 B. 6051 C. 4034 D. 6045.

Bài này đáp án là 14 mặt nha

a) Thiếu VP

b) 4 - x = 2( x - 4 )²

<=> 4 - x = 2( x² - 8x + 16 )

<=> 4 - x = 2x² - 16x + 32

<=> 2x² - 16x + 32 - 4 + x = 0

<=> 2x² - 15x + 28 = 0

<=> 2x² - 8x - 7x + 28 = 0

<=> 2x( x - 4 ) - 7( x - 4 ) = 0

<=> ( x - 4 )( 2x - 7 ) = 0

<=> \(\orbr{\begin{cases}x-4=0\\2x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{7}{2}\end{cases}}\)

c) ( x² + 1 )( x - 2 ) + 2x = 4

<=> x³ - 2x² + 3x - 2 - 4 = 0

<=> x³ - 2x² + 3x - 6 = 0

<=> x²( x - 2 ) + 3( x - 2 ) = 0

<=> ( x - 2 )( x² + 3 ) = 0

<=> x = 2 ( vì x² + 3 ≥ 3 > 0 ∀ x )

a, thiếu 

b, \(4-x=2\left(x-4\right)^2\Leftrightarrow4-x=2\left(x^2-8x+16\right)\)

\(\Leftrightarrow4-x=2x^2-16x+32\Leftrightarrow2x^2-15x+28=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x-7\right)=0\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{7}{2}\end{cases}}\)

c, \(\left(x^2+1\right)\left(x-2\right)+2x=4\Leftrightarrow x^3-2x^2+3x-6=0\Leftrightarrow x_1=2;x_2=\sqrt{3}i\)

a) Ta có: \(8x\left(x-2017\right)-2x=4034\) (hẳn đề là ntn-.-)

\(\Leftrightarrow8x^2-16136x-2x-4034=0\)

\(\Leftrightarrow8x^2-16138x-4034=0\)

Bấm nghiệm ra: \(x_1=2017,499938\) ; \(x_2=-0,24993...\)

b) Ta có: \(4-x=2\left(x-4\right)^2\)

\(\Leftrightarrow2\left(x^2-4x+4\right)+x-4=0\)

\(\Leftrightarrow2x^2-7x+4=0\)

\(\Leftrightarrow\left(x^2-\frac{7}{2}x+\frac{49}{16}\right)-\frac{17}{16}=0\)

\(\Leftrightarrow\left(x-\frac{7}{4}\right)^2-\left(\frac{\sqrt{17}}{4}\right)^2=0\)

\(\Leftrightarrow\left(x-\frac{7+\sqrt{17}}{4}\right)\left(x-\frac{7-\sqrt{17}}{4}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7+\sqrt{17}}{4}\\x=\frac{7-\sqrt{17}}{4}\end{cases}}\)

c) \(\left(x^2+1\right)\left(x-2\right)+2x=4\)

\(\Leftrightarrow\left(x^2+1\right)\left(x-2\right)+2\left(x-2\right)=0\)

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

Mà \(x^2+3\ge3>0\)

=> \(x-2=0\Rightarrow x=2\)