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

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

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

giả sử : x = 5k; y = 4k; z = 3k (k là N*)

ta có: \(P=\dfrac{5k+2\left(4k\right)-3\left(3k\right)}{5k-2\left(4k\right)+3\left(3k\right)}=\dfrac{5k+8k-9k}{5k-8k+9k}=\dfrac{4k}{6k}=\dfrac{4}{6}=\dfrac{2}{3}\)

vậy P = \(\dfrac{2}{3}\)

giúp mình với mình xin cảm ơn

vì góc A và góc B là 2 góc bù nhau nên

góc A + góc B = 180 độ (1)

mà góc A = 1/6B (2)

từ (1) (2) => 1/6B + B = 180

7/6B = 180

B = 154 độ

A = 180 - 154 = 26

vậy góc A = 26 độ

a: \(\dfrac{8}{9}=1-\dfrac{1}{9}\)

\(\dfrac{108}{109}=1-\dfrac{1}{109}\)

Vì 9<109 nên \(\dfrac{1}{9}>\dfrac{1}{109}\)

=>\(-\dfrac{1}{9}< -\dfrac{1}{109}\)

=>\(-\dfrac{1}{9}+1< -\dfrac{1}{109}+1\)

=>\(\dfrac{8}{9}< \dfrac{108}{109}\)

b: \(\dfrac{97}{100}=0,97;\dfrac{98}{99}=0,\left(98\right)\)

mà 0,97<0,(98)

nên \(\dfrac{97}{100}< \dfrac{98}{99}\)

c: \(\dfrac{19}{18}=1+\dfrac{1}{18}\)

\(\dfrac{2021}{2020}=1+\dfrac{1}{2020}\)

Vì 18<2020 nên \(\dfrac{1}{18}>\dfrac{1}{2020}\)

=>\(1+\dfrac{1}{18}>1+\dfrac{1}{2020}\)

=>\(\dfrac{19}{18}>\dfrac{2021}{2020}\)

d: \(\dfrac{131}{171}=\dfrac{130+1}{170+1}>\dfrac{130}{170}=\dfrac{13}{17}\)

a, \(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}\) và \(5z-3x-4y=50\) (1)

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

\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}\)

\(=\dfrac{\left(5z-25\right)-\left(3x-3\right)-\left(4y+12\right)}{30-6-16}\)

\(=\dfrac{\left(5z-3x-4y\right)-25+3-12}{8}\)

\(=\dfrac{50-34}{8}=\dfrac{16}{8}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=2.2=4\\y+3=2.4=8\\z-5=2.6=12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\y=5\\z=17\end{matrix}\right.\)

b, \(\dfrac{4}{3x-2y}=\dfrac{3}{2z-4x}=\dfrac{2}{4y-3z}\) và \(x+y+z=-10\)  (2)

(ĐK: \(x\ne\dfrac{2}{3}y;z\ne2x;y\ne\dfrac{3}{4}z\))

Áp dụng tính chất của dãy tỉ số bằng nhau và (2), ta được:

\(\dfrac{4}{3x-2y}=\dfrac{3}{2z-4x}=\dfrac{2}{4y-3z}=\dfrac{16}{12x-8y}=\dfrac{9}{6z-12x}=\dfrac{4}{8y-6z}\)

\(=\dfrac{16+9+4}{12x-8y+6z-12x+8y-6z}=\dfrac{29}{0}\) 

\(\Rightarrow x,y,z\in\varnothing\)

a) Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3\left(x-1\right)-4\left(y+3\right)+5\left(z-5\right)}{-3\cdot2+\left(-4\right)\cdot4+5\cdot6}\)

\(=\dfrac{\left(5x-3x-4y\right)+\left(3-12-25\right)}{-6-16+30}=\dfrac{50-34}{8}=2\)

\(\Rightarrow\dfrac{x-1}{2}=2\Rightarrow x-1=4\Rightarrow x=5\)

\(\Rightarrow\dfrac{y+3}{4}=2\Rightarrow y+3=8\Rightarrow y=5\)

\(\Rightarrow\dfrac{z-5}{6}=2\Rightarrow z-5=12\Rightarrow z=17\) 

Bài 3:

a) \(AB//CD\)

\(\Rightarrow\widehat{A}+\widehat{D}=180^o\)

\(\Rightarrow\widehat{D}=180^o-\widehat{A}=180^o-100^o=80^o\)

b) Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)

\(\Rightarrow\widehat{C}=180^o-\left(\widehat{B}+\widehat{A}\right)=180^o-\left(50^o+70^o\right)\)

\(\Rightarrow\widehat{C}=60^o\)

Mà: \(\widehat{C}+\widehat{C_n}=180^o\) (kề bù)

\(\Rightarrow\widehat{C_n}=180^o-60^o=120^o\)

Bài 4:

a: Xét ΔABD và ΔAED có

AD chung

\(\widehat{BAD}=\widehat{EAD}\)

AB=AE

Do đó: ΔABD=ΔAED

b: Ta có: ΔABD=ΔAED

=>DB=DE

=>D nằm trên đường trung trực của BE(1)

Ta có: AB=AE
=>A nằm trên đường trung trực của BE(2)

Từ (1),(2) suy ra AD là đường trung trực của BE

=>AD\(\perp\)BE