mình dịch hộ bạn thôi nha

\(7,\\ a,ĐK:a>0;a\ne1\\ b,K=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}-1+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\\ K=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\dfrac{a-1}{\sqrt{a}}\\ c,a=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\\ \Leftrightarrow K=\dfrac{3+2\sqrt{2}-1}{\sqrt{2}+1}=\dfrac{2+2\sqrt{2}}{\sqrt{2}+1}=\dfrac{2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=2\\ d,K< 0\Leftrightarrow a-1< 0\left(\sqrt{a}>0\right)\Leftrightarrow0< a< 1\)

Bài 2:

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

\(\frac{x}{5}=\frac{y}{7}=\frac{z}{17}=\frac{2x-3y-4z}{2\cdot5-3\cdot7-4\cdot17}=\frac{-237}{-79}=3\)

=>\(\begin{cases}x=3\cdot5=15\\ y=3\cdot7=21\\ z=3\cdot17=51\end{cases}\)

b: 2x=3y=5z

=>\(\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\)

=>\(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}\)

mà x+y-z=76

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

\(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{76}{25-6}=\frac{76}{19}=4\)

=>\(\begin{cases}x=4\cdot15=60\\ y=4\cdot10=40\\ z=4\cdot6=24\end{cases}\)

c: 2x=3y

=>\(\frac{x}{3}=\frac{y}{2}\)

=>\(\frac{x}{21}=\frac{y}{14}\left(1\right)\)

5y=7z

=>\(\frac{y}{7}=\frac{z}{5}\)

=>\(\frac{y}{14}=\frac{z}{10}\) (2)

Từ (1),(2) suy ra \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)

mà x-y+z=85

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

\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{x-y+z}{21-14+10}=\frac{85}{17}=5\)

=>\(\begin{cases}x=5\cdot21=105\\ y=5\cdot14=70\\ z=5\cdot10=50\end{cases}\)

e: \(\frac{x}{y}=\frac74\)

=>\(\frac{x}{7}=\frac{y}{4}\)

=>\(\frac{x}{21}=\frac{y}{12}\left(3\right)\)

\(\frac{y}{z}=\frac{12}{5}\)

=>\(\frac{y}{12}=\frac{z}{5}\) (4)

Từ (3),(4) suy ra \(\frac{x}{21}=\frac{y}{12}=\frac{z}{5}\)

mà x-2y+z=16

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

\(\frac{x}{21}=\frac{y}{12}=\frac{z}{5}=\frac{x-2y+z}{21-2\cdot12+5}=\frac{16}{21-24+5}=\frac{16}{2}=8\)

=>\(\begin{cases}x=8\cdot21=168\\ y=8\cdot12=96\\ z=8\cdot5=40\end{cases}\)

Bài 1:

a: \(\frac{-4}{79}=\frac{-4\cdot5}{79\cdot5}=\frac{-20}{395}\)

\(\frac{5}{-88}=\frac{-5}{88}=\frac{-5\cdot4}{88\cdot4}=\frac{-20}{352}\)

Ta có: 395>352

=>\(\frac{20}{395}<\frac{20}{352}\)

=>\(-\frac{20}{395}>-\frac{20}{352}\)

=>\(\frac{-4}{79}>\frac{-5}{88}\)

b: \(\frac{796}{1013}=\frac{1809-1013}{1013}=\frac{1809}{1013}-1\)

\(\frac{798}{1011}=\frac{1809-1011}{1011}=\frac{1809}{1011}-1\)

Ta có: 1013>1011

=>\(\frac{1809}{1013}<\frac{1809}{1011}\)

=>\(\frac{1809}{1013}-1<\frac{1809}{1011}-1\)

=>\(\frac{796}{1013}<\frac{798}{1011}\)

=>\(\frac{-796}{1013}>\frac{-798}{1011}\)

mà \(\frac{-798}{1011}>\frac{-799}{1011}\)

nên \(\frac{-796}{1013}>-\frac{799}{1011}\)

c: \(\frac{57}{169}>\frac{57}{171}=\frac13\)

\(\frac13=\frac{67}{201}>\frac{67}{203}\)

Do đó: \(\frac{57}{169}>\frac{67}{203}\)

=>\(-\frac{57}{169}<-\frac{67}{203}\)

d: \(\frac{-237}{327}>\frac{-327}{327}=-1;-1=\frac{-723}{723}>\frac{-732}{723}\)

Do đó: \(-\frac{237}{327}>\frac{-732}{723}\)

f: \(\frac{-83}{17}=\frac{-85+2}{17}=-5+\frac{2}{17}\)

\(\frac{-277}{55}=\frac{-275-2}{55}=-5-\frac{2}{55}\)

mà \(\frac{2}{17}>0>-\frac{2}{55}\)

nên \(-\frac{83}{17}>-\frac{277}{55}\)

g: \(\frac{2021}{-2020}=\frac{-2021}{2020}=\frac{-2020-1}{2020}=-1-\frac{1}{2020}\)

\(\frac{-2022}{2021}=\frac{-2021-1}{2021}=-1-\frac{1}{2021}\)

Ta có: 2020<2021

=>\(\frac{1}{2020}>\frac{1}{2021}\)

=>\(-\frac{1}{2020}<-\frac{1}{2021}\)

=>\(-\frac{1}{2021}-1<-\frac{1}{2021}-1\)

=>\(\frac{2021}{-2020}<\frac{2022}{-2021}\)

Bài 2: 

a: =>168x+20=6x-21

=>162x=-41

hay x=-41/162

b: \(\Leftrightarrow2\left(3x-8\right)=3\left(5-x\right)\)

=>6x-16=15-3x

=>9x=31

hay x=31/9

c: \(\Leftrightarrow4\left(x^2+8x-20\right)-\left(x+4\right)\left(x+10\right)=3\left(x^2+2x-8\right)\)

\(\Leftrightarrow4x^2+32x-80-x^2-14x-40-3x^2-6x+24=0\)

=>12x-96=0

hay x=8

bn ơi bài 7 mk chẳng thấy tia Cz nằm đâu cả

Bài 4: 

a: Xét tứ giác OBAC có 

\(\widehat{OBA}+\widehat{OCA}=180^0\)

Do đó: OBAC là tứ giác nội tiếp

hay O,B,A,C cùng thuộc 1 đường tròn

Bài 5:

\(\sqrt{x+2021}-y^3=\sqrt{y+2021}-x^3\\ \Leftrightarrow\left(\sqrt{x+2021}-\sqrt{y+2021}\right)+\left(x^3-y^3\right)=0\\ \Leftrightarrow\dfrac{x-y}{\sqrt{x+2021}+\sqrt{y+2021}}+\left(x-y\right)\left(x^2+xy+y^2\right)=0\\ \Leftrightarrow\left(x-y\right)\left(\dfrac{1}{\sqrt{x+2021}+\sqrt{y+2021}}+x^2+xy+y^2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-y=0\\\dfrac{1}{\sqrt{x+2021}+\sqrt{y+2021}}+x^2+xy+y^2=0\left(1\right)\end{matrix}\right.\)

Dễ thấy \(\left(1\right)>0\) với mọi x,y

Do đó \(x-y=0\) hay \(x=y\)

\(\Leftrightarrow M=x^2+2x^2-2x^2+2x+2022=x^2+2x+1+2021\\ \Leftrightarrow M=\left(x+1\right)^2+2021\ge2021\)

Dấu \("="\Leftrightarrow x=y=-1\)

P = (10.10².10³.10⁴...10⁹) : (10⁵.10¹⁰.10²⁵)

= 10¹⁺²⁺³⁺⁴⁺⁵⁺⁶⁺⁷⁺⁸⁺⁹ 10⁵⁺¹⁰⁺²⁵

= 10⁴⁵ : 10⁴⁰

= 10⁴⁵⁻⁴⁰

= 10⁵

= 100000

\(P=\left(10.10^2.10^3.10^4.....10^9\right):\left(10^5.10^{10}.10^{25}\right)\)

\(P=10^{45}:10^{40}\)

\(P=10^5\)

đáp án "E" nha,k cho mik nha
 

b: \(\sqrt{8^2+6^2}-\sqrt{16}+\dfrac{1}{2}\sqrt{\dfrac{4}{25}}\)

\(=10-4+\dfrac{1}{2}\cdot\dfrac{2}{5}=6+\dfrac{1}{5}=\dfrac{31}{5}\)

4:

a: căn x-1=3

=>căn x=4

=>x=16

b: căn x-3=2

=>căn x=5

=>x=25

c: 5/12*căn x-1/6=1/3

=>5/12*căn x=1/2

=>căn x=1/2:5/12=1/2*12/5=12/10=6/5

=>x=36/25

d: căn x-1-3=0

=>căn x-1=3

=>x-1=9

=>x=10

e: =>căn x+1=1

=>x+1=1

=>x=0

f: =>|3/5*căn x-1/20|=1/5+3/4=19/20

=>3/5*căn x-1/20=-19/20 hoặc 3/5*căn x-1/20=19/20

=>3/5*căn x=1

=>căn x=5/3

=>x=25/9

:))