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Theo bài ra , ta có :

\(\dfrac{x+5}{25}+1+\dfrac{x+6}{24}+1+\dfrac{x+7}{23}=0\)

\(\Leftrightarrow\dfrac{x+5+25}{25}+\dfrac{x+6+24}{24}+\dfrac{x+7+23}{23}=0\)

\(\Leftrightarrow\dfrac{x+30}{25}+\dfrac{x+30}{24}+\dfrac{x+30}{23}=0\)

\(\Leftrightarrow\left(x+30\right)\left(\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}\right)=0\)

Vì \(\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}\ne0\)

\(\Leftrightarrow x+30=0\)

\(\Leftrightarrow x=-30\)

Vậy S={-30}

Chúc bạn học tốt =))

Ta có :

\(\dfrac{x+5}{25}+\dfrac{x+6}{24}+\dfrac{x+7}{23}=-3\)

=> \(\left(x+5\right).\dfrac{1}{25}+\left(x+5+1\right).\dfrac{1}{24}+\left(x+5+2\right).\dfrac{1}{23}=-3\)

=>\(\left(x+5\right).\dfrac{1}{25}+\left(x+5\right).\dfrac{1}{24}+\dfrac{1}{24}+\left(x+5\right).\dfrac{1}{23}+2.\dfrac{1}{23}\)= -3

=> (x + 5).\(\left(\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}\right)\) + \(\dfrac{1}{24}+\dfrac{2}{23}\) = -3

=> (x + 5). \(\dfrac{1727}{13800}\) + \(\dfrac{71}{552}\) = -3

=> (x + 5). \(\dfrac{1727}{13800}\) = -3 - \(\dfrac{71}{552}\)

=> (x + 5). \(\dfrac{1727}{13800}\) = \(\dfrac{-1727}{552}\)

=> x + 5 = -25

=> x = -25-5

=> x = -30

Vậy x = -30

\(x^4+2014x^2+2013x+2014\)

\(=x^4+2014x^2+2014x-x+2014\)

\(=\left(x^4-x\right)+\left(2014x^2+2014x+2014\right)\)

\(=x\left(x^3-1\right)+2014\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2014\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2014\right)\)

b)\(x^8+7x^4+6\)

\(=x^8+x^4+6x^4+6\)

\(=x^4\left(x^4+1\right)+6\left(x^4+1\right)\)

\(=\left(x^4+1\right)\left(x^4+6\right)\)

b) \(x^8+7x^4+16\)

\(=\left(x^8+8x^4+16\right)-x^4\)

\(=\left[\left(x^4\right)^2+2.x^4.4+4^2\right]-x^4\)

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

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

Vì AB=2BC nên góc đối diện với cạnh AB bằng 30^o hay \(\widehat{C}=30^o\)

Ta có:

\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)

hay \(2\widehat{A}+\widehat{A}=150^o\)

\(\Rightarrow\widehat{A}=50^o\)

\(\Rightarrow\widehat{B}=100^o\)

A B C D I K O

\(1,\hept{\begin{cases}OI//AB\Rightarrow\frac{OI}{AB}=\frac{OD}{BD}\\OI//CD\Rightarrow\frac{OI}{CD}=\frac{OA}{AC}\\AB//CD\Rightarrow\frac{OA}{AC}=\frac{OB}{BD}\end{cases}}\Rightarrow\frac{OI}{AB}+\frac{OI}{CD}=\frac{OD}{BD}+\frac{OA}{AC}=\frac{OD}{BD}+\frac{OB}{BD}=\frac{BD}{BD}=1\)

\(\hept{\begin{cases}OK//AB\Rightarrow\frac{OC}{AC}=\frac{OK}{AB}\\OK//CD\Rightarrow\frac{OK}{CD}=\frac{OB}{BD}\\\frac{CB}{BD}=\frac{OA}{AC}\end{cases}}\Rightarrow\frac{OK}{AB}+\frac{OK}{CD}=\frac{OC}{AC}+\frac{OB}{BD}=\frac{OC}{AC}+\frac{OA}{AC}=\frac{AC}{AC}=1\)

\(2,\hept{\begin{cases}\frac{OI}{AB}+\frac{OI}{CD}=1\\\frac{OK}{AB}+\frac{OK}{CD}=1\end{cases}}\Rightarrow\frac{OI}{AB}+\frac{OI}{CD}+\frac{OK}{AB}+\frac{OK}{CD}=2\)

\(\Leftrightarrow\frac{OI+OK}{AB}+\frac{OI+OK}{CD}=2\)

\(\Leftrightarrow\frac{IK}{AB}+\frac{IK}{CD}=2\)

\(\Leftrightarrow\frac{1}{AB}+\frac{1}{CD}=\frac{2}{IK}\left(đpcm\right)\)

Giúp mik bài này với: https://olm.vn/hoi-dap/detail/244594379058.html

y+1=(4bc)/[(b+c)^2-a^2] (1)

x+1=[(b+c)^2-a^2]/(2bc) (2)

Nhân với nhau (1)*(2): (x+1)(y+1)=2

(x+1)(y+1)=xy+x+y+1

Vậy P=1

câu hỏi hay chưa không biết cách trả lời không tệ:

đáp án có thể sai do cộng trừ chưa đúng