Cho a+b+c = 4034 ; \(\frac{1}{c+b}+\frac{1}{a+c}+\frac{1}{a+b}=\frac{1}{2}\)
Tính \(\frac{a}{c+b}+\frac{b}{a+c}+\frac{c}{a+b}\)
Cho a+b+c=4034 vµ \(\dfrac{1}{c+b}+\dfrac{1}{a+c}+\dfrac{1}{a+b}=\dfrac{1}{2}\) .Tính \(\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\)
\(\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\)
\(=\left(\dfrac{a}{c+b}+1\right)+\left(\dfrac{b}{a+c}+1\right)+\left(\dfrac{c}{a+b}+1\right)-3\)
\(=\dfrac{a+c+b}{c+b}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}-3\)
\(=\left(a+b+c\right)\left(\dfrac{1}{c+b}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)-3\)
\(=4034.\dfrac{1}{2}-3=2014\)
Guể?
\(\dfrac{1}{c+b}+\dfrac{1}{a+c}+\dfrac{1}{a+b}=\dfrac{1}{2}\)
\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{c+a}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)=\dfrac{4034}{2}=2017\)
\(\Rightarrow1+\dfrac{a}{c+b}+1+\dfrac{b}{a+c}+1+\dfrac{c}{a+b}=2017\)
\(\Rightarrow\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=2014\)
Cho a + b + c = 4034 và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{2}\)
Tính \(S=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
cho a+b+c=4034 và \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{2}\)
tính S=\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
ai giải đc mình tick cho
\(S=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(\Rightarrow S=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)-3\)
\(\Rightarrow S=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3\)
\(\Rightarrow S=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)
\(\Rightarrow S=4034.\frac{1}{2}-3=2014\)
Vậy S=2014
key ban di chi gium n he khong k luon nhe
Cho A=1+1/2+1/3+...+1/4034, B=1+1/3+1/5+...+1/4033. So sánh A/B với 1+2017/2018
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
Tìm x, biết:
a) 8x(x - 2017) - 2x + 4034 = 0; b) x 2 + x 2 8 = 0;
c) 4 - x = 2 ( x - 4 ) 2 ; d) ( x 2 + 1)(x - 2) + 2x = 4.
tìm x
a,8x(x-2017)-2x+4034
b,4-x=2(x-4)^2
c,(x^2+1)(x-2)+2x=4
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 \(x^2+3\ge3>0\)
=> \(x-2=0\Rightarrow x=2\)
a) 4034 nhân x + 6,76 nhân x = 4,995
b) 15,6 : x - 4,3 = 0,9