a: Ta có: \(\widehat{ABC}=\widehat{ACB}\)(ΔABC cân tại A)

\(\widehat{ACB}=\widehat{ECN}\)(hai góc đối đỉnh)

Do đó: \(\widehat{ABC}=\widehat{ECN}\)

Xét ΔMBD vuông tại D và ΔNCE vuông tại E có

BD=CE

\(\widehat{MBD}=\widehat{NCE}\)

Do đó: ΔMBD=ΔNCE

=>DM=EN

b: Ta có: DM\(\perp\)BC

EN\(\perp\)BC

Do đó: DM//EN

Xét ΔIDM vuông tại D và ΔIEN vuông tại E có

MD=EN

\(\widehat{MDI}=\widehat{ENC}\)(hai góc so le trong, DM//EN)

Do đó: ΔIDM=ΔIEN

=>IM=IN

=>I là trung điểm của MN

\(\frac{3^{20}+9^{20}}{9^{11}+9}=\frac{3^{20}+3^{40}}{3^{22}+3^2}\\ =\frac{3^{20}\left(1+3^{20}\right)}{3^2\left(3^{20}+1\right)}\\ =3^{18}\)

Bài 2: 

a: \(f\left(x\right)=-9x^3-2x^2+6x-3\)

\(G\left(x\right)=9x^3-6x+53\)

b: \(H\left(x\right)=9x^3-6x+53-9x^3-2x^2+6x-3=-2x^2+50\)

c: Đặt H(x)=0

=>2x²-50=0

=>x=5 hoặc x=-5

a) \(\Rightarrow\left|\dfrac{3}{4}+x\right|=0\Rightarrow\dfrac{3}{4}+x=0\Rightarrow x=-\dfrac{3}{4}\)

b) \(\Rightarrow x+0,4=\dfrac{4}{9}:\dfrac{2}{3}=\dfrac{2}{3}\Rightarrow x=\dfrac{2}{3}-0,4=\dfrac{4}{15}\)

Bài 5: 

d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y-z}{2+3-4}=\dfrac{-20}{1}=-20\)

Do đó: x=-40; y=-60; z=-80

Vì x là số dương nên ta Giả sử \(\hept{\begin{cases}x^2=a\\\frac{2}{x}=b\end{cases}}\) với a,b là hai số tự nhiên

Vậy \(x=\frac{2}{b}\Rightarrow x^2=\frac{4}{b^2}=a\Leftrightarrow4=ab^2\)

Do b là số tự nhiên nên \(\orbr{\begin{cases}b=1\Rightarrow a=4\\b=2\Rightarrow a=1\end{cases}}\) vậy \(\orbr{\begin{cases}x=1\\x=2\end{cases}}\)

Câu 1:

a: \(\frac37+\left(-\frac52\right)+\frac35\)

\(=\frac37-\frac52+\frac35\)

\(=\frac{30}{70}-\frac{175}{70}+\frac{42}{70}=\frac{30+42-175}{70}=-\frac{103}{70}\)

b: \(0,8-\left(-\frac27\right)-\frac{7}{10}=\frac45+\frac27-\frac{7}{10}\)

\(=\frac{56}{70}+\frac{20}{70}-\frac{49}{70}=\frac{56+20-49}{70}=\frac{56-29}{70}=\frac{27}{70}\)

c: \(\frac34-\left\lbrack\left(-\frac53\right)-\left(\frac{1}{12}+\frac29\right)\right\rbrack\)

\(=\frac34+\frac53+\frac{1}{12}+\frac29\)

\(=\frac{27}{36}+\frac{60}{36}+\frac{3}{36}+\frac{8}{36}=\frac{98}{36}=\frac{49}{18}\)

d: \(19\frac13\cdot\frac45-\frac45\cdot15\frac13\)

\(=\frac45\left(19+\frac13-15-\frac13\right)\)

\(=\frac45\cdot4=\frac{16}{5}\)

e: \(\frac38\cdot\left(-\frac{3}{11}\right)+\frac38\cdot\left(-\frac{8}{11}\right)+2\frac57\)

\(=\frac38\left(-\frac{3}{11}-\frac{8}{11}\right)+2+\frac57\)

\(=-\frac38+2+\frac57=\frac{13}{8}+\frac57=\frac{91}{56}+\frac{40}{56}=\frac{131}{56}\)

f: \(\left(-\frac{5}{11}:\frac{13}{8}-\frac{5}{11}:\frac{13}{3}\right)-\left|-\frac{1}{33}\right|\)

\(=-\frac{5}{11}\cdot\frac{8}{13}-\frac{5}{11}\cdot\frac{3}{13}-\frac{1}{33}\)

\(=-\frac{5}{11}\cdot\frac{11}{13}-\frac{1}{33}=-\frac{5}{13}-\frac{1}{33}=\frac{-165-13}{13\cdot33}=\frac{-178}{429}\)

g: \(\frac{4}{23}+\frac{5}{21}+\frac12-\frac{4}{23}+\frac{16}{21}\)

\(=\left(\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+\frac12\)

\(=1+\frac12=\frac32\)

h: \(\frac{13}{25}+\frac{6}{41}+\frac{-38}{25}+\frac{35}{41}-0,5\)

\(=\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-0,5\)

\(=-\frac{25}{25}+\frac{41}{41}-0,5=-0,5\)

i: \(\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-1\frac{15}{17}+\frac23\)

\(=\left(\frac{15}{34}+\frac{19}{34}\right)+\left(\frac13+\frac23\right)-1-\frac{15}{17}\)

\(=1+1-1-\frac{15}{17}=1-\frac{15}{17}=\frac{2}{17}\)

j: \(\left(\frac{2022}{2023}\right)^0-\left(-3\right)_{}^3+\frac12:\left(-\frac12\right)^3\)

\(=1-\left(-27\right)+\frac12:\frac{-1}{8}\)

=28-4

=24

k: \(\left(-2\right)^3+\frac12:\frac{-1}{4}-\left|15\right|\)

\(=-8+\frac12\cdot\left(-4\right)-15\)

=-8-2-15

=-25

l: \(3:\left(-\frac32\right)^2+\frac19\cdot\sqrt{36}\)

\(=3:\frac94+\frac19\cdot6=3\cdot\frac49+\frac69=\frac{12}{9}+\frac69=\frac{18}{9}\)

=2

m: \(\sqrt{81}\cdot\frac13-\left(-3\right)^3=9\cdot\frac13-\left(-27\right)\)

=3+27

=30

n: \(\sqrt{\frac{16}{25}}\cdot\sqrt{\frac{121}{64}}-1\frac{3}{10}\)

\(=\frac45\cdot\frac{11}{8}-\frac{13}{10}\)

\(=\frac{11}{10}-\frac{13}{10}=-\frac{2}{10}=-\frac15\)

o: \(\frac{3^2}{2^3}\cdot\sqrt{144}-50\%\)

\(=\frac98\cdot12-\frac12=9\cdot\frac32-\frac12=\frac{27}{2}-\frac12=\frac{26}{2}\)

=13

p: \(\frac{6^2+3\cdot6^2+3^2}{-13}\)

\(=\frac{36+3\cdot108+9}{-13}=\frac{45+324}{-13}=\frac{369}{-13}=-\frac{369}{13}\)

q: \(\frac{5^4\cdot20^4}{25^5\cdot4^5}=\frac{\left(5\cdot20\right)^4}{\left(25\cdot4\right)^5}=\frac{100^4}{100^5}=\frac{1}{100}\)

r: \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)

\(=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)

\(=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}=\frac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot5}=\frac23\cdot\frac65=\frac{12}{15}=\frac45\)

\(A=x^3+3xy\left(x+y\right)+y^3-xy\left(x+y\right)+x^2+y^2+xy+2\)

\(A=\left(x+y\right)^3-xy.\left(-1\right)+x^2+y^2+xy+2\)

\(A=\left(-1\right)^3+x^2+y^2+2xy+2\)

\(A=\left(x+y\right)^2+1\)

\(A=\left(-1\right)^2+1=2\)

\(\left|x-\frac{1}{3}\right|=\left|2-3x\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{3}=2-3x\\x-\frac{1}{3}=3x-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{7}{3}\\2x=-\frac{5}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{12}\\x=-\frac{5}{6}\end{matrix}\right.\)

Vậy : \(x\in\left\{\frac{7}{12},-\frac{5}{6}\right\}\)

@Akai Haruma cô xem hộ em có thiếu TH không ạ ?

@Akai Haruma em có bị thiếu không ạ ?