Tổng 3 góc của 1 tam giác là 180^o

Hình 1.

90^o + 55^o + x = 180^o

145^o + x = 180^o

x = 180^o - 145^o

x = 35^o

Hình 2.

30^o + x + 40^o = 180^o

70^o + x = 180^o

x = 180^o - 70^o

x = 110^o

Hình 3.

50^o + x + x = 180^o

2x = 180^o - 50^o

2x = 130^o

x = 130^o : 2

x = 65^o

Lời giải:

$\frac{131}{171}=1-\frac{40}{171}> 1-\frac{40}{170}=1-\frac{4}{17}=\frac{13}{17}$
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$\frac{51}{61}=1-\frac{10}{61}=1-\frac{100}{610}$

$\frac{515}{616}=1-\frac{101}{616}$

Xét hiệu:

$\frac{100}{610}-\frac{101}{616}=\frac{100.616-101.610}{610.616}$

$=\frac{100(610+6)-101.610}{610.616}$

$=\frac{600-610}{610.616}<0$

$\Rightarrow \frac{100}{610}< \frac{101}{616}$

$\Rightarrow 1-\frac{100}{610}> 1-\frac{101}{616}$

$\Rightarrow \frac{51}{61}> \frac{515}{616}$ 

a: \(-\dfrac{4}{15}=\dfrac{5}{15}-\dfrac{9}{15}=\dfrac{1}{3}-\dfrac{3}{5}=\dfrac{1}{3}+\left(-\dfrac{3}{5}\right)\)

b: \(\dfrac{-4}{15}=\dfrac{-2\cdot2}{3\cdot5}=\dfrac{-2}{3}\cdot\dfrac{2}{5}\)

c: \(\dfrac{-4}{15}=\dfrac{-2}{3}\cdot\dfrac{2}{5}=\dfrac{-2}{3}:\dfrac{5}{2}\)

a) \(-\dfrac{4}{15}=\left(-1\right)+\dfrac{11}{15}\)

b) \(-\dfrac{4}{15}=\left(-\dfrac{2}{3}\right).\dfrac{2}{5}\)

c) \(-\dfrac{4}{15}=\left(-\dfrac{2}{3}\right):\dfrac{5}{2}\)

a) \(\dfrac{2}{3}< a-\dfrac{1}{6}< \dfrac{8}{9}\\ \Rightarrow\dfrac{2}{3}+\dfrac{1}{6}< a-\dfrac{1}{6}+\dfrac{1}{6}< \dfrac{8}{9}+\dfrac{1}{6}\\ \dfrac{5}{6}< a< \dfrac{19}{18}\)

Do a là số nguyên nên a=1

b) \(\dfrac{12}{9}< \dfrac{4}{a}< \dfrac{8}{3}\left(a\ne0\right)\\ \Rightarrow\dfrac{4}{3}< \dfrac{4}{a}< \dfrac{4}{\dfrac{3}{2}}\\ \Rightarrow3>a>1,5\)

Do a là số nguyên nên a=2

a: \(\dfrac{2}{3}< \dfrac{a-1}{6}< \dfrac{8}{9}\)

=>\(\dfrac{12}{18}< \dfrac{3\left(a-1\right)}{18}< \dfrac{16}{18}\)

=>12<3(a-1)<16

=>12<3a-3<16

=>15<3a<19

=>\(5< a< \dfrac{19}{3}\)

mà a nguyên

nên a=6

b: \(\dfrac{12}{9}< \dfrac{4}{a}< \dfrac{8}{3}\)

=>\(\dfrac{24}{18}< \dfrac{24}{6a}< \dfrac{24}{9}\)

=>9<6a<18

mà a nguyên

nên 6a=12

=>a=2

a: \(\dfrac{1}{2}< \dfrac{12}{a}< \dfrac{4}{3}\)

=>\(\dfrac{12}{24}< \dfrac{12}{a}< \dfrac{12}{9}\)

=>9<a<24

mà a nguyên

nên \(a\in\left\{10;11;...;23\right\}\)

b: \(\dfrac{7}{4}< \dfrac{a}{8}< 3\)

=>\(\dfrac{14}{8}< \dfrac{a}{8}< \dfrac{24}{8}\)

=>14<a<24

mà a nguyên

nên \(a\in\left\{15;16;...;23\right\}\)

\(\dfrac{7}{2x+2}=\dfrac{3}{2y-4}=\dfrac{5}{z+4}\) và \(x+y+z=17\) (1)

ĐK: \(x\ne-1;y\ne2;z\ne-4\)

Áp dụng tính chất của dãy tỉ số bằng nhau và (1), ta được:

\(\dfrac{7}{2x+2}=\dfrac{3}{2y-4}=\dfrac{5}{z+4}=\dfrac{10}{2z+8}\)

\(=\dfrac{7+3+10}{2x+2+2y-4+2z+8}\)

\(=\dfrac{20}{2\left(x+y+z\right)+6}=\dfrac{20}{2.17+6}=\dfrac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}2x+2=2.7=14\\2y-4=2.3=6\\z+4=5.2=10\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=12\\2y=10\\z=6\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=6\left(tm\right)\\y=5\left(tm\right)\\z=6\left(tm\right)\end{matrix}\right.\)

a: \(\dfrac{7}{5}>\dfrac{7}{9}\)

=>\(-\dfrac{7}{5}< -\dfrac{7}{9}\)

\(\dfrac{3}{2}=1,5;\dfrac{4}{5}=0,8;\dfrac{9}{11}=0,\left(9\right);-\dfrac{3}{-4}=0,75\)

mà 0<0,75<0,8<0,(9)<1,5

nên \(0< \dfrac{3}{4}< \dfrac{4}{5}< \dfrac{9}{11}< \dfrac{3}{2}\)

=>\(-\dfrac{7}{5}< -\dfrac{7}{9}< 0< \dfrac{-3}{-4}< \dfrac{4}{5}< \dfrac{9}{11}< \dfrac{3}{2}\)

b: \(-\dfrac{11}{12}=-1+\dfrac{1}{12};\dfrac{-3}{4}=-1+\dfrac{1}{4};\dfrac{-18}{19}=-1+\dfrac{1}{19};\dfrac{-4}{5}=-1+\dfrac{1}{5};-\dfrac{25}{26}=-1+\dfrac{1}{26}\) 

=>

Vì 4<5<12<19<26

nên \(\dfrac{1}{4}>\dfrac{1}{5}>\dfrac{1}{12}>\dfrac{1}{19}>\dfrac{1}{26}\)

=>\(\dfrac{1}{4}-1>\dfrac{1}{5}-1>\dfrac{1}{12}-1>\dfrac{1}{19}-1>\dfrac{1}{26}-1\)

=>\(\dfrac{-3}{4}>-\dfrac{4}{5}>\dfrac{-11}{12}>\dfrac{-18}{19}>\dfrac{-25}{26}\)

=>

\(\dfrac{-25}{26}< \dfrac{-18}{19}< \dfrac{-11}{12}< \dfrac{-4}{5}< -\dfrac{3}{4}\)

mà \(\dfrac{-3}{4}< 0< \dfrac{-4}{-5}\)

nên \(-\dfrac{25}{26}< -\dfrac{18}{19}< \dfrac{-11}{12}< -\dfrac{4}{5}< -\dfrac{3}{4}< \dfrac{-4}{-5}\)

a) \(\dfrac{1}{2};0;-\dfrac{2}{9};-\dfrac{4}{9};-\dfrac{5}{9};-\dfrac{7}{9};-\dfrac{10}{9}\)

b) \(\dfrac{7}{15};\dfrac{3}{10};0;\dfrac{2}{-5};-\dfrac{3}{4};-\dfrac{5}{6}\)

Giải thích:

b) \(\dfrac{7}{15}=\dfrac{14}{30}>\dfrac{3}{10}=\dfrac{9}{30}\)

\(\dfrac{2}{-5}=-\dfrac{4}{10}=-0,4>-\dfrac{3}{4}=-\dfrac{75}{100}=-0,75\)

\(-\dfrac{3}{4}=-\dfrac{9}{12}>-\dfrac{5}{6}=-\dfrac{10}{12}\)

?

a/

$\frac{97}{100}< \frac{98}{100}< \frac{98}{99}$

b/

$\frac{19}{18}=1+\frac{1}{18}> 1+\frac{1}{2020}=\frac{2021}{2020}$
c/

$\frac{131}{171}=1-\frac{40}{171}> 1-\frac{40}{170}=1-\frac{4}{17}=\frac{13}{17}$
d/

$\frac{51}{61}=1-\frac{10}{61}=1-\frac{100}{610}$

$\frac{515}{616}=1-\frac{101}{616}$

Xét hiệu:

$\frac{100}{610}-\frac{101}{616}=\frac{100.616-101.610}{610.616}$

$=\frac{100(610+6)-101.610}{610.616}$

$=\frac{600-610}{610.616}<0$

$\Rightarrow \frac{100}{610}< \frac{101}{616}$

$\Rightarrow 1-\frac{100}{610}> 1-\frac{101}{616}$

$\Rightarrow \frac{51}{61}> \frac{515}{616}$ 

Lời giải:

ĐKĐB $\Rightarrow \frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}$

Áp dụng TCDTSBN:

$\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}$

$=\frac{4(3x-2y)}{16}=\frac{3(2z-4x)}{9}=\frac{2(4y-3z)}{4}$

$=\frac{4(3x-2y)+3(2z-4x)+2(4y-3z)}{16+9+4}$

$=\frac{0}{29}=0$

$\Rightarrow 3x-2y=2z-4x=4y-3z=0$

$\Rightarrow 3x=2y; 4y=3z\Rightarrow \frac{x}{2}=\frac{y}{3}=\frac{z}{4}$

Áp dụng TCDTSBN:

$\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x+y-z}{2+3-4}=\frac{-10}{1}=-10$

$\Rightarrow x=(-10).2=-20; y=3(-10)=-30; z=4(-10)=-40$