\(\dfrac{3}{7}\times\dfrac{6}{13}\times\dfrac{13}{6}\times\dfrac{7}{3}.\\ =\left(\dfrac{6}{13}\times\dfrac{13}{6}\right)\times\left(\dfrac{3}{7}\times\dfrac{7}{3}\right).\\ =1\times1=1.\\ \dfrac{6}{15}\times\dfrac{3}{4}\times\dfrac{30}{12}\times\dfrac{8}{3}.\\ =\dfrac{2}{5}\times\dfrac{3}{4}\times\dfrac{5}{2}\times\dfrac{8}{3}.\\ =\left(\dfrac{2}{5}\times\dfrac{5}{2}\right)\times\left(\dfrac{3}{4}\times\dfrac{8}{3}\right).\\ =1\times2=2.\\ \dfrac{17}{35}\times\dfrac{21}{45}\times\dfrac{7}{17}\times\dfrac{9}{21}.\\ =\left(\dfrac{17}{35}\times\dfrac{7}{17}\right)\times\left(\dfrac{21}{45}\times\dfrac{9}{21}\right).\\ =\dfrac{1}{5}\times\dfrac{1}{5}=\dfrac{1}{25}.\)

\(a,\widehat{xOm}=\widehat{yOn}\) (Do đối đỉnh).

Mà \(\widehat{xOm}=40^o\left(gt\right).\)

\(\Rightarrow\widehat{yOn}=40^o.\)

\(b,\widehat{mOy}+\widehat{xOm}=180^o\) (Do đối đỉnh).

\(\Rightarrow\widehat{mOy}=180^o-\widehat{xOm}=180^o-40^o=140^o.\)

\(c,Oz\) là tia phân giác của \(\widehat{xOm}\left(gt\right).\)

\(\Rightarrow\widehat{zOm}=\dfrac{\widehat{xOm}}{2}=\dfrac{40^o}{2}=20^o.\\ \widehat{zOy}=\widehat{zOm}+\widehat{mOy}.\\ =20^o+140^o.\\ =160^o.\)

\(a,24:\dfrac{0,3x+2,5}{x}-12=18.\\ \Leftrightarrow24:\dfrac{0,3x+2,5}{x}=30.\\ \Leftrightarrow\dfrac{0,3x+2,5}{x}=\dfrac{4}{5}.\\ \Leftrightarrow0,3x+2,5=\dfrac{4}{5}x.\\ \Leftrightarrow2,5=\dfrac{4}{5}x-0,3x.\\ \Leftrightarrow2,5=\dfrac{1}{2}x.\\ \Leftrightarrow x=5.\)

\(b,5,6.\dfrac{x.3,4-2,5}{x}+1,6=3,14.\\ \Leftrightarrow5,6.\dfrac{x.3,4-2,5}{x}=1,54.\\ \Leftrightarrow\dfrac{3,4x-2,5}{x}=0,275.\\ \Leftrightarrow3,4x-2,5=0,275x.\\ \Leftrightarrow3,125x=2,5.\\ \Leftrightarrow x=0,8.\)

\(c,x:\dfrac{3}{7}+\dfrac{x+27}{9}+0,37=25,37.\\ \Leftrightarrow\dfrac{7}{3}x+\dfrac{x+27}{9}=25.\\ \Leftrightarrow\dfrac{21x}{9}+\dfrac{x}{9}+\dfrac{27}{9}=\dfrac{25}{9}.\\ \Leftrightarrow\dfrac{22x}{9}=-\dfrac{2}{9}.\\ \Leftrightarrow x=-\dfrac{1}{11}.\)

\(d,15+\dfrac{x}{7}-123=189.\\ \Leftrightarrow\dfrac{x}{7}=297.\\ \Leftrightarrow x=2079.\\ e,x.36+x=925.\\ \Leftrightarrow37x=925.\\ \Leftrightarrow x=25.\\ g,\dfrac{9}{4}.\dfrac{8}{x}=\dfrac{3}{5}.\\ \Leftrightarrow\dfrac{8}{x}=\dfrac{4}{15}.\\ \Leftrightarrow x=30.\)

\(1,\left(3x-1\right)\left(x+2\right).\\ =3x^2+6x-x-2.\\ =3x^2+5x-2.\\ 2,\left(4-x\right)\left(5+3x\right).\\ =20+12x-5x-3x^2.\\ =-3x^2+7x+20.\\ 3,\left(2x+3y\right)\left(2x-3y\right).\\ =4x^2-9y^2.\\ 4,3\left(x+2\right)\left(1-2x\right).\\ =\left(3x+6\right)\left(1-2x\right).\\ =3x-6x^2+6-12x.\\ =-6x^2-9x+6.\)

T gửi lại ở phía trên nhé

a) Xét \(\text{△ BHA}\) và \(\text{△ BHD}:\)

\(HA=HD\left(gt\right).\)

\(\widehat{AHB}=\widehat{DHB}\) (Đối đỉnh).

\(BHchung.\)

\(\Rightarrow\) \(\text{△ BHA}\) \(\sim\) \(\text{△ BHD}\) \(\left(c-g-c\right).\)

\(\Rightarrow BD=BA\) (2 cạnh tương ứng).

b) Xét \(\Delta ABC:\)

AH là trung tuyến (H là trung điểm của cạnh đáy BC).

\(\Rightarrow\) AH là đường cao (T/c tam giác cân).

\(\Rightarrow AH\perp HC\) hay \(AD\perp HC.\)

Trên tia đối tia HA lấy điểm D sao cho \(\text{HD = HA}\left(gt\right).\)

\(\Rightarrow\) H là trung điểm của AD.

Xét \(\Delta ACD:\) 

HC là trung tuyến (H là trung điểm của AD).

HC là đường cao \(\left(AD\perp HC\right).\)

\(\Rightarrow\) \(\Delta ACD\) cân tại C.

\(\Rightarrow\) HC là tia phân giác của \(\widehat{ACD}\) (T/c tam giác cân).

hay CB là tia phân giác của \(\widehat{ACD}\) (T/c tam giác cân).

\(a,\dfrac{26}{x+3}\in Z.\\ \Leftrightarrow x+3\inƯ\left(26\right)=\left\{1;-1;2;-2;26;-26\right\}.\)

\(TH1:x+3=1.\\ \Leftrightarrow x=-2\left(TM\right).\\ TH2:x+3=-1.\\ \Leftrightarrow x=-4\left(TM\right).\\ TH3:x+3=2.\\ \Leftrightarrow x=-1\left(TM\right).\\ TH4:x+3=-2.\\ \Leftrightarrow x=-5\left(TM\right).\\ TH5:x+3=26.\\ \Leftrightarrow x=23\left(TM\right).\\ \Leftrightarrow TH6:x+3=-26.\\ \Leftrightarrow x=-29\left(TM\right).\)

Vậy \(x\in\left\{-2;-4;-5;23;-29\right\}.\)

\(b,\dfrac{x+6}{x+1}=1+\dfrac{5}{x+1}\left(x\ne-1\right).\\ 1+\dfrac{5}{x+1}\in Z.\\ \Leftrightarrow x+1\inƯ\left(5\right)=\left\{1;-1;-5;5\right\}.\)

\(TH1:x+1=1.\\ \Leftrightarrow x=0\left(TM\right).\\ TH2:x+1=-1.\\ \Leftrightarrow x=-2\left(TM\right).\\ TH3:x+1=5.\\ \Leftrightarrow x=4\left(TM\right).\\ TH4:x+1=-5.\\ \Leftrightarrow x=-6\left(TM\right).\)

Vậy \(x\in\left\{0;-2;4;-6\right\}.\)