Theo đề bài ta có:

\(\left(x+y+z\right)\cdot\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\right)=2017\cdot\dfrac{1}{672}\)

\(\Rightarrow\dfrac{x+y+z}{x+y}+\dfrac{x+y+z}{y+z}+\dfrac{x+y+z}{z+x}=\dfrac{2017}{672}\)

\(\Rightarrow1+\dfrac{z}{x+y}+1+\dfrac{x}{y+z}+1+\dfrac{y}{z+x}=\dfrac{2017}{672}\)

\(\Rightarrow C=\dfrac{2017}{672}-3=\dfrac{1}{672}\)

a,

Vì Â₁ + Â₂ = 180^o [kề bù] và Â₁ = 90^o

=> Â₂ = 90^o

Xét ∆ABC và ∆ADE, ta có:

- AB = AD [gt]

- Â₁ = Â₂ = 90^o [cmt]

- AC = AE [gt]

=> ∆ABC = ∆ADE [c-g-c]

=> Ê₁ = \(\widehat{C_1}\)

Mà \(\widehat{D_1}=\widehat{D_2}\left(đ^2\right)\)

=> \(\widehat{F_1}\) = Â₂ = 90^o

=> DE vuông góc BC

b,

\(4\widehat{B}=5\widehat{C}\Leftrightarrow\dfrac{\widehat{B}}{5}=\dfrac{\widehat{C}}{4}\)

Đặt k = \(\dfrac{\widehat{B}}{5}=\dfrac{\widehat{C}}{4}\)

\(\Rightarrow\left\{{}\begin{matrix}\widehat{B}=5k\\\widehat{C}=4k\end{matrix}\right.\)

\(\Rightarrow90^o+4k+5k=180^o\Leftrightarrow9k=90^o\Leftrightarrow k=10^o\)

\(\Rightarrow\left\{{}\begin{matrix}\widehat{B}=50^o\\\widehat{C}=40^o\end{matrix}\right.\)

Mà ∆ABC = ∆ADE [cmt]

=> góc C = góc E = 40^o

Vậy AÊD = 40^o

AM = MC

∆MAC cân tại M

=> \(\widehat{A}=\widehat{ACM}\)

BM = MC

=> ∆MAB cân tại M

=> \(\widehat{B}=\widehat{MCB}\)

=> \(\widehat{A}+\widehat{B}+\widehat{C}=\widehat{A}+\widehat{ACM}+\widehat{BCM}+\widehat{B}=2\widehat{A}+2\widehat{B}=180^o\Leftrightarrow\widehat{A}+\widehat{B}=90^o\Leftrightarrow\widehat{C}=90^o\)

1x1x1x1:2=1:2=0,5

Cố gắng học giỏi nha ^_^

GIẢI CHI TIẾT GIÚP MÌNH NHÉ!

1995 x 1997 > 1996 x 1996

CM A chia hết cho 7 và 11.
* 36 mod 7 = 1 nên 36³⁸ mod 7 = 1; 41 mod 7 = -1 nên 41³³ mod 7 = (-1)³³ = -1
suy ra A mod 7 = 0 hay A chia hết cho 7.
* 36 mod 11 = 3, 41 mod 11 =-3 nên A mod 11 = 3³⁸ - 3³³ =3³³ (3⁵ - 1) =3³³. 242
Vì 242 chia hết cho 11 nên A mod 11 = 0.
Vậy A chia hết cho 7.11 =77

Vì 90 = 9 x 10

hay ta tách 90 = 1 x 2 x 5 x 9

Vậy ta được số thập phân nhỉ nhất có tích = 90 là 1,259

a,

\(\left(\dfrac{3}{5}x-\dfrac{2}{3}x-x\right)\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)

\(\Rightarrow\dfrac{-16}{15}x\cdot\dfrac{1}{7}=-\dfrac{5}{21}\)

\(\Rightarrow\dfrac{-16}{15}x=\dfrac{-\dfrac{5}{21}}{\dfrac{1}{7}}=-\dfrac{5}{3}\)

\(\Rightarrow x=\dfrac{-\dfrac{5}{3}}{-\dfrac{16}{15}}=\dfrac{25}{16}\)

b,

\(\left(5x-1\right)\left(2x+\dfrac{1}{3}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\2x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{6}\end{matrix}\right.\)

c,

\(\dfrac{5\left|x+1\right|}{2}=\dfrac{90}{\left|x+1\right|}\)

\(\Rightarrow5\left|x+1\right|^2=180\)

\(\Rightarrow\left|x+1\right|^2=36\)

Mà \(\left|x+1\right|\ge0\)

=> x + 1 = 6 <=> x = 7