Xét ΔABC có \(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^0\)

=>\(2\left(\widehat{IBC}+\widehat{ICB}\right)+50^0=180^0\)

=>\(\widehat{IBC}+\widehat{ICB}=\dfrac{180^0-50^0}{2}=65^0\)

Xét ΔIBC có \(\widehat{IBC}+\widehat{ICB}+\widehat{BIC}=180^0\)

=>\(\widehat{BIC}+65^0=180^0\)

=>\(\widehat{BIC}=180^0-65^0=115^0\)

đề làm sao ấy bn nhỉ ?

a: Ta có: \(\widehat{HIA}+\widehat{HAI}=90^0\)(ΔHAI vuông tại H)

\(\widehat{KIB}+\widehat{KBI}=90^0\)(ΔKIB vuông tại K)

mà \(\widehat{HIA}=\widehat{KIB}\)(hai góc đối đỉnh)

nên \(\widehat{HAI}=\widehat{KBI}\)

=>\(x=40^0\)

b: Xét tứ giác BEDC có \(\widehat{BEC}=\widehat{BDC}=90^0\)

nên BEDC là tứ giác nội tiếp

=>\(x=\widehat{EBD}=\widehat{ECD}=35^0\)

c: Ta có: \(\widehat{IMP}+\widehat{IPM}=90^0\)(ΔMIP vuông tại I)

\(\widehat{MPN}+\widehat{MNP}=90^0\)(ΔMNP vuông tại M)

Do đó: \(x=\widehat{IMP}=\widehat{N}=60^0\)

\(\left|\dfrac{4}{3}x-\dfrac{1}{4}\right|>=0\forall x\)

=>\(A=\left|\dfrac{4}{3}x-\dfrac{1}{4}\right|-\dfrac{2}{11}>=-\dfrac{2}{11}\forall x\)

Dấu '=' xảy ra khi \(\dfrac{4}{3}x-\dfrac{1}{4}=0\)

=>\(\dfrac{4}{3}x=\dfrac{1}{4}\)

=>\(x=\dfrac{1}{4}:\dfrac{4}{3}=\dfrac{3}{16}\)

\(\left|6x+22\right|>=0\forall x;\left(y-21\right)^2>=0\forall y\)

Do đó: \(\left|6x+22\right|+\left(y-21\right)^2>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}6x+22=0\\y-21=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-\dfrac{11}{3}\\y=21\end{matrix}\right.\)

a: \(\left|-\dfrac{1}{3}\right|-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)

\(=\dfrac{1}{3}-1+\dfrac{1}{4}:2=-\dfrac{2}{3}+\dfrac{1}{8}=\dfrac{-16}{24}+\dfrac{3}{24}=-\dfrac{13}{24}\)

b: \(\left(\dfrac{2}{3}\right)^3+\sqrt{\dfrac{49}{81}}-\left|-\dfrac{7}{3}\right|:3\)

\(=\dfrac{8}{27}+\dfrac{7}{9}-\dfrac{7}{3}\cdot\dfrac{1}{3}\)

\(=\dfrac{8}{27}+\dfrac{7}{9}-\dfrac{7}{9}=\dfrac{8}{27}\)

c: \(\sqrt{\dfrac{25}{49}}+\left(5555\right)^0+\left|-\dfrac{2}{7}\right|\)

\(=\dfrac{5}{7}+1+\dfrac{2}{7}\)

=1+1=2

d: \(\left|-5-\sqrt{2}\right|=5+\sqrt{2}\)

c: \(\left|4+\sqrt{3}\right|=4+\sqrt{3}\)

d: \(\left|-\dfrac{4}{15}\right|=\dfrac{4}{15}\)

a: \(\left|3,02\right|=3,02\)

a: \(\sqrt{50}>\sqrt{49}\)

mà \(\sqrt{49}=7\)

nên \(\sqrt{50}>7\)

b: \(\sqrt{27}>\sqrt{25}=5\)

=>\(\dfrac{4}{\sqrt{27}}< \dfrac{4}{5}\)

c: \(\dfrac{3}{\sqrt{7}}>1;\dfrac{\sqrt{7}}{3}< 1\)

Do đó: \(\dfrac{3}{\sqrt{7}}>\dfrac{\sqrt{7}}{3}\)

Bài 2:

a:

\(-4,4\left(9\right)-5,8\left(1\right)\simeq-4,5-5,8=-10,3\)

 \(-4,4\left(9\right)-5,8\left(1\right)\)

\(=-\dfrac{9}{2}-\dfrac{-523}{90}=-\dfrac{9}{2}+\dfrac{523}{90}=\dfrac{118}{90}=\dfrac{59}{45}\)

b:

\(-12,\left(7\right)\cdot3,\left(12\right)\simeq-12,8\cdot3,1\simeq-40\)

 \(-12,\left(7\right)\cdot3,\left(12\right)\)

\(=-\dfrac{115}{9}\cdot\dfrac{103}{33}=\dfrac{11845}{297}\)

c: \(9,\left(49\right):\left[-5,\left(09\right)\right]\simeq9,5:\left(-5,1\right)\simeq-1,9\)

\(9,\left(49\right):\left[-5,\left(09\right)\right]\)

\(=\dfrac{940}{99}:\dfrac{-56}{11}=\dfrac{940}{99}\cdot\dfrac{11}{-56}\)

\(=\dfrac{940}{-56}\cdot\dfrac{1}{9}=-\dfrac{235}{14\cdot9}=-\dfrac{235}{126}\)

Bài 1:

a: \(9,4\simeq9\)

b: \(3,51\simeq4\)

c: \(-7,505\simeq-8\)

d: \(-1.199\simeq-1\)