Bài 5

Ta có:

\(x^2-x-6=\left(x-3\right)\left(x+2\right)\) và đa thức chia bậc 2 nên dư là \(ax+b\)

Vậy \(f\left(x\right)=\left(x-3\right)\left(x+2\right)\left(x^2+4\right)+ax+b\)

Theo định lí Bezout, dư trong phép chia \(f\left(x\right)\) cho \(x-3\) là \(f\left(3\right)=21\) cho \(x+2\) là \(f\left(-2\right)=4\) nên ta có: \(\left\{{}\begin{matrix}3a+b=21\\-2a+b=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=5\\b=6\end{matrix}\right.\)

Đa thức cần tìm là \(\left(x+2\right)\left(x-3\right)\left(x^2+4\right)+5x+6=x^4-x^3-2x^2+x-18\)

Bài 4:

\(2n^2+6n-7⋮n-2\)

=>\(2n^2-4n+10n-20+13⋮n-2\)

=>\(13⋮n-2\)

=>\(n-2\in\left\{1;-1;13;-13\right\}\)

=>\(n\in\left\{3;1;15;-11\right\}\)

\(a.\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{11}{4}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{13}}:\dfrac{\dfrac{3}{5}-\dfrac{3}{8}+\dfrac{3}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\\ =\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}{11\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}:\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}\\ =\dfrac{3}{11}:\dfrac{3}{7}\\ =\dfrac{3}{11}\cdot\dfrac{7}{3}\\ =\dfrac{7}{11}\\ b.\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{19\cdot21}\\ =\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{19\cdot21}\right)\\ =\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\cdot\left(1-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\cdot\dfrac{20}{21}=\dfrac{10}{21}\)

a: ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(AC^2=BC^2-AB^2=20^2-12^2=256=16^2\)

=>AC=16(cm)

b: Xét ΔHID vuông tại H và ΔHBA vuông tại H có

HI=HB

HD=HA

Do đó: ΔHID=ΔHBA

=>DI=BA

ΔHID=ΔHBA

=>\(\widehat{HDI}=\widehat{HAB}\)

=>DI//AB

c: Ta có: DI//AB

AB\(\perp\)AC

Do đó: DI\(\perp\)AC

Xét ΔCAD có

DI,CH là các đường cao

DI cắt CH tại I

Do đó: I là trực tâm của ΔCAD

=>AI\(\perp\)CD
d: Xét ΔABC vuông tại A có AH là đường cao

nên \(AH\cdot BC=AB\cdot AC\)

=>\(AH\cdot20=12\cdot16=192\)

=>AH=192/20=9,6(cm)

ΔAHB vuông tại H

=>\(HA^2+HB^2=AB^2\)

=>\(HB=\sqrt{12^2-9,6^2}=7,2\left(cm\right)\)

=>AH>HB

mà AD=2AH và BI=2BH

nên AD>BI

1+1=3

\(0,\left(1\right)=\dfrac{1}{9}\\ 0,\left(01\right)=\dfrac{1}{99}\\ 1,\left(8\right)=\dfrac{17}{9}\\ 0,\left(27\right)=\dfrac{3}{11}\)

mình cảm ơn ạ

Lời giải:

Tháng thứ hai, sau khi bác bán được 5/7 số gà còn lại thì bác còn: $28-4=24$ (con gà) 

Tháng thứ hai, trước khi bán 5/7 số gà còn lại thì bác có: $24:(1-\frac{5}{7})=84$ (con) 

Tháng thứ nhất, sau khi bán 3/5 số gà thì bác còn: $84+6=90$ (con) 

Trang trại của bác có tất cả số gà là: $90:(1-\frac{3}{5})=225$ (con)

Đáp án A.

AB//CD

=>\(y=\widehat{BDC}\)(hai góc so le trong)

=>\(y=45^0\)

AB//CD

=>\(x+100^0=180^0\)

=>\(x=80^0\)

\(x-y=80^0-45^0=35^0\)

xx'//yy'

=>\(\widehat{xAB}+\widehat{yBz}=180^0\)(hai góc trong cùng phía)

=>\(\widehat{yBz}+70^0=180^0\)

=>\(\widehat{yBz}=110^0\)

xx'//yy'

=>\(\widehat{xAB}=\widehat{yBz'}\)(hai góc đồng vị)

=>\(\widehat{yBz'}=70^0\)

bài 2:

a: \(\dfrac{1}{9}-0,3\cdot\dfrac{5}{9}+\dfrac{1}{3}\)

\(=\dfrac{1}{9}-\dfrac{3}{10}\cdot\dfrac{5}{9}+\dfrac{1}{3}\)

\(=\dfrac{1}{9}-\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{2}{18}-\dfrac{3}{18}+\dfrac{6}{18}=\dfrac{5}{18}\)

b: \(\left(-\dfrac{2}{3}\right)^2+\dfrac{1}{6}-\left(-0,5\right)^3\)

\(=\dfrac{4}{9}+\dfrac{1}{6}-\left(-\dfrac{1}{2}\right)^3\)

\(=\dfrac{8}{18}+\dfrac{3}{18}-\dfrac{-1}{8}=\dfrac{11}{18}+\dfrac{1}{8}=\dfrac{44}{72}+\dfrac{9}{72}=\dfrac{53}{72}\)

c: \(0,3-\dfrac{8}{3}:\dfrac{4}{3}\cdot\dfrac{1}{5}+1\)

\(=1,3-\dfrac{8}{3}\cdot\dfrac{3}{4}\cdot\dfrac{1}{5}=1,3-\dfrac{2}{5}=1,3-0,4=0,9=\dfrac{9}{10}\)

d: \(-\dfrac{2}{3}-4\cdot\left(\dfrac{1}{2}+\dfrac{3}{4}\right)^2\)

\(=-\dfrac{2}{3}-4\cdot\left(\dfrac{2}{4}+\dfrac{3}{4}\right)^2\)

\(=-\dfrac{2}{3}-4\cdot\dfrac{25}{16}=-\dfrac{2}{3}-\dfrac{25}{4}\)

\(=-\dfrac{8}{12}-\dfrac{75}{12}=-\dfrac{83}{12}\)

e: \(2\cdot\left(-1\right)^6+\left(\dfrac{3}{4}\right)^2-\dfrac{3}{8}\)

\(=2\cdot1+\dfrac{9}{16}-\dfrac{3}{8}\)

\(=2+\dfrac{9}{16}-\dfrac{6}{16}=2+\dfrac{3}{16}=\dfrac{35}{16}\)

f: \(\left[\left(\dfrac{3}{5}-\dfrac{1}{3}\right)\cdot6+\dfrac{-1}{3}\right]\cdot5\)

\(=\left[\dfrac{9-5}{15}\cdot6+\dfrac{-1}{3}\right]\cdot5\)

\(=\left(\dfrac{4}{15}\cdot6-\dfrac{1}{3}\right)\cdot5=\dfrac{19}{15}\cdot5=\dfrac{19}{3}\)

g: \(\dfrac{3}{4}:\left(\dfrac{2}{3}-\dfrac{5}{9}\right)+\dfrac{9}{4}=\dfrac{3}{4}:\dfrac{6-5}{9}+\dfrac{9}{4}\)

\(=\dfrac{3}{4}\cdot9+\dfrac{9}{4}=\dfrac{27}{4}+\dfrac{9}{4}=\dfrac{36}{4}=9\)

h: \(0,8:\left\{0,2-8\cdot\left[\dfrac{7}{48}+\left(\dfrac{5}{24}-\dfrac{5}{16}\right)\right]\right\}\)

\(=0,8:\left\{0.2-8\cdot\left[\dfrac{7}{48}+\dfrac{10}{48}-\dfrac{15}{48}\right]\right\}\)

\(=0,8:\left\{0.2-8\cdot\dfrac{2}{48}\right\}\)

\(=0,8:\left\{0.2-\dfrac{1}{3}\right\}=0,8:\left\{\dfrac{1}{5}-\dfrac{1}{3}\right\}=0,8:\dfrac{-2}{15}=0,8\cdot\dfrac{15}{-2}=\dfrac{12}{-2}=-6\)

Bài 1:

a: \(x+\dfrac{-2}{5}=-\dfrac{1}{3}\)

=>\(x=-\dfrac{1}{3}+\dfrac{2}{5}=-\dfrac{5}{15}+\dfrac{6}{15}=\dfrac{1}{15}\)

b: \(0,5-x=-\dfrac{5}{14}\)

=>\(\dfrac{1}{2}-x=-\dfrac{5}{14}\)

=>\(x=\dfrac{1}{2}-\left(-\dfrac{5}{14}\right)=\dfrac{1}{2}+\dfrac{5}{14}=\dfrac{7}{14}+\dfrac{5}{14}=\dfrac{12}{14}=\dfrac{6}{7}\)

c: \(3\dfrac{1}{5}-x=1,6+\dfrac{7}{10}\)

=>\(3,2-x=1,6+0,7=2,3\)

=>x=3,2-2,3=0,9

d: \(\dfrac{11}{12}-\left(x+\dfrac{2}{5}\right)=\dfrac{2}{3}\)

=>\(x+\dfrac{2}{5}=\dfrac{11}{12}-\dfrac{2}{3}=\dfrac{11}{12}-\dfrac{8}{12}=\dfrac{3}{12}=\dfrac{1}{4}\)

=>\(x=\dfrac{1}{4}-\dfrac{2}{5}=\dfrac{5}{20}-\dfrac{8}{20}=-\dfrac{3}{20}\)

e: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

=>\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=-\dfrac{7}{20}\)

=>\(x=-\dfrac{1}{4}:\dfrac{7}{20}=-\dfrac{1}{4}\cdot\dfrac{20}{7}=\dfrac{-5}{7}\)

f: \(\dfrac{3}{4}x-\dfrac{1}{2}=\dfrac{3}{7}\)

=>\(\dfrac{3}{4}x=\dfrac{1}{2}+\dfrac{3}{7}=\dfrac{7}{14}+\dfrac{6}{14}=\dfrac{13}{14}\)

=>\(x=\dfrac{13}{14}:\dfrac{3}{4}=\dfrac{13}{14}\cdot\dfrac{4}{3}=\dfrac{52}{42}=\dfrac{26}{21}\)

mình cảm ơn ạ

Ta có: \(\widehat{MAB}=\widehat{ABC}\)

mà hai góc này là hai góc ở vị trí so le trong

nên MA//BC

Ta có: \(\widehat{NAC}=\widehat{ACB}\)

mà hai góc này là hai góc ở vị trí so le trong

nên NA//BC

Ta có: MA//BC

NA//BC

MA,NA có điểm chung là A

Do đó: M,A,N thẳng hàng