51. will live

52. will have

53. will be

54. will live

55. won't be

56. will be

51 will live

52 will go

53 will be

54 will live

55 won't be

56 will be

Vì CK và CI lần lượt là phân giác trong và ngoài tại đỉnh C của ΔABC nên CK⊥CI

=>\(\hat{ICK}=90^0\)

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

=>\(\hat{ABC}+\hat{ACB}=180^0-70^0=110^0\)

=>\(2\left(\hat{IBC}+\hat{ICB}\right)=110^0\)

=>\(\hat{IBC}+\hat{ICB}=55^0\)

Xét ΔIBC có \(\hat{KIC}\) là góc ngoài tại đỉnh I

nên \(\hat{KIC}=\hat{IBC}+\hat{ICB}=55^0\)

ΔCIK vuông tại C

=>\(\hat{CIK}+\hat{CKI}=90^0\)

=>\(\hat{CKI}=90^0-55^0=35^0\)

Qua B, kẻ Bm//a//b(tia Bm nằm giữa hai tia BA và BC)

Bm//Aa

=>\(\widehat{mBA}=\widehat{aAB}=40^0\)

Ta có: Bm//Cb

=>\(\widehat{mBC}=\widehat{bCB}=180^0-130^0=50^0\)

\(\widehat{ABC}=\widehat{mBA}+\widehat{mBC}=40^0+50^0=90^0\)

Ta có: AB//DC

=>\(\widehat{A_1}=\widehat{D_4}\)(hai góc so le trong)

=>\(\widehat{D_4}=110^0\)
Ta có: \(\widehat{D_1}=\widehat{D_4}\)(hai góc đối đỉnh)

mà \(\widehat{D_4}=110^0\)

nên \(\widehat{D_1}=110^0\)

Ta có: AB//DC

=>\(\widehat{C_3}=\widehat{B_2}\)(hai góc so le trong)

=>\(\widehat{B_2}=135^0\)

Ta có: \(\widehat{B_1}+\widehat{B_2}=180^0\)(hai góc kề bù)

=>\(\widehat{B_1}=180^0-135^0=45^0\)

Bài 6:

a: \(\left|x+\dfrac{1}{2}\right|>=0\forall x;\left|y-\dfrac{3}{4}\right|>=0\forall y;\left|z-1\right|>=0\forall z\)

Do đó: \(\left|x+\dfrac{1}{2}\right|+\left|y-\dfrac{3}{4}\right|+\left|z-1\right|>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+\dfrac{1}{2}=0\\y-\dfrac{3}{4}=0\\z-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{3}{4}\\z=1\end{matrix}\right.\)

b: \(\left|x-\dfrac{3}{4}\right|>=0\forall x;\left|\dfrac{2}{5}-y\right|>=0\forall y;\left|x-y+z\right|>=0\forall x,y,z\)

Do đó: \(\left|x-\dfrac{3}{4}\right|+\left|\dfrac{2}{5}-y\right|+\left|x-y+z\right|>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{3}{4}=0\\\dfrac{2}{5}-y=0\\x-y+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=\dfrac{2}{5}\\z=-x+y=-\dfrac{3}{4}+\dfrac{2}{5}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=\dfrac{2}{5}\\z=-\dfrac{7}{20}\end{matrix}\right.\)

c: \(\left|x-\dfrac{2}{3}\right|>=0\forall x;\left|x+y+\dfrac{3}{4}\right|>=0\forall x,y;\left|y-z-\dfrac{5}{6}\right|>=0\forall y,z\)

Do đó: \(\left|x-\dfrac{2}{3}\right|+\left|x+y+\dfrac{3}{4}\right|+\left|y-z-\dfrac{5}{6}\right|>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{2}{3}=0\\x+y+\dfrac{3}{4}=0\\y-z-\dfrac{5}{6}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-x-\dfrac{3}{4}=-\dfrac{2}{3}-\dfrac{3}{4}\\z=y-\dfrac{5}{6}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{17}{12}\\z=-\dfrac{17}{12}-\dfrac{10}{12}=-\dfrac{27}{12}=-\dfrac{9}{4}\end{matrix}\right.\)

Bài 5:

a: \(\left|-\dfrac{3}{5}+\dfrac{1}{2}\right|-\left(\dfrac{3}{4}-\dfrac{5}{8}\right)+\left|-\dfrac{3}{2}\right|\)

\(=\left|-\dfrac{6}{10}+\dfrac{5}{10}\right|-\dfrac{1}{8}+\dfrac{3}{2}\)

\(=\dfrac{1}{10}-\dfrac{1}{8}+\dfrac{3}{2}=\dfrac{4}{40}-\dfrac{5}{40}+\dfrac{60}{40}=\dfrac{59}{40}\)

b: \(\dfrac{2}{3}-\left|-\dfrac{7}{3}+\dfrac{3}{4}\right|-\left|-\dfrac{5}{2}+1\right|\)

\(=\dfrac{2}{3}-\left|-\dfrac{28}{12}+\dfrac{9}{12}\right|-\left|-\dfrac{5}{2}+\dfrac{2}{2}\right|\)

\(=\dfrac{2}{3}-\dfrac{19}{12}-\dfrac{3}{2}=\dfrac{8}{12}-\dfrac{19}{12}-\dfrac{18}{12}\)

\(=-\dfrac{29}{12}\)

c: \(\dfrac{1}{5}-\left(\dfrac{3}{10}-\dfrac{-3}{5}\right)-\left|\dfrac{1}{4}-\dfrac{2}{5}\right|\)

\(=\dfrac{1}{5}-\dfrac{3}{10}-\dfrac{3}{5}-\left|\dfrac{5}{20}-\dfrac{8}{20}\right|\)

\(=-\dfrac{7}{10}-\left|\dfrac{-3}{20}\right|=-\dfrac{7}{10}-\dfrac{3}{20}=-\dfrac{17}{20}\)

d: \(\left|-\dfrac{5}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right|-\left(-\dfrac{3}{4}+\dfrac{-5}{3}\right)\)

\(=\left|-\dfrac{30}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right|+\dfrac{3}{4}+\dfrac{5}{3}\)

\(=\dfrac{25}{12}+\dfrac{9}{12}+\dfrac{20}{12}=\dfrac{54}{12}=\dfrac{9}{2}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)

Để A là số nguyên thì \(\sqrt{x}+2⋮\sqrt{x}-3\)

=>\(\sqrt{x}-3+5⋮\sqrt{x}-3\)

=>\(5⋮\sqrt{x}-3\)

=>\(\sqrt{x}-3\in\left\{1;-1;5;-5\right\}\)

=>\(\sqrt{x}\in\left\{4;2;8\right\}\)

=>\(x\in\left\{16;4;64\right\}\)

Thể tích phần bể chứa nước ban đầu là:

\(80\cdot50\cdot35=140000\left(cm^3\right)\)

Thể tích phần bể chứa nước lúc này sau khi thêm hòn đá là:

\(140000+20000=160000\left(cm^3\right)\)

Mực nước trong bể lúc này cao là:

\(160000:80:50=40\left(cm\right)\)

Thể tích ban đầu: 80 x 50 x 35 = 140.000 cm3

Sau khi thêm hòn đá: 140.000+20.000 =  160.000 cm3

=> Chiều cao mực nước = 160.000 / (80x50) = 40 cm

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