\(\dfrac{2a-5b}{a-3b}=\dfrac{2\left(\dfrac{a}{b}\right)-5}{\left(\dfrac{a}{b}\right)-3}=\dfrac{2.\dfrac{3}{4}-5}{\dfrac{3}{4}-3}=\dfrac{14}{9}\)

`a/b=3/5=>a=3/5b`

Thay `a=3/5b` vào `[2a-5b]/[a-3b]` có:

     `[2. 3/5b-5b]/[3/5b-3b]`

`=[6/5b-5b]/[3/5b-3b]`

`=[-19/5b]/[-12/5b]`

`=[-19/5]/[-12/5]=19/12`

bài 1 mình làm ở câu trước rồi nhé

bạn chụp rõ hơn được ko chứ mấy dấu nhân hay trừ chả phân biệt được

Bài 2)

a)

\(\left(x-5\right).2=16\)

\(x-5=16:2\)

\(x-5=8\)

\(x=8+5\)

\(x=13\)

b) 

\(6x-6=24\)

\(6x=30\)

\(x=30:6\)

\(x=5\)

c)

\(23+x.2=37\)

\(x.2=37-23\)

\(x.2=14\)

\(x=14:2\)

\(x=7\)

d)

\(\left(x.7-7\right).\left(x.12-24\right)=0\)

\(=>\left[{}\begin{matrix}x.7-7=0\\x.12-24=0\end{matrix}\right.=>\left[{}\begin{matrix}x.7=7\\x.24=24\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\)

e)

\(140.10.x=120\)

\(10.x=120:140\)

\(10.x=\dfrac{6}{7}\)

\(x=\dfrac{6}{7}:10\)

\(x=\dfrac{3}{35}\)

g)

\(x.5-127=275\)

\(x.5=275+127\)

\(x.5=402\)

\(x=402:5\)

\(x=\dfrac{402}{5}=80.4\)

Bài 3: 

a: \(3^x=9\)

nên x=2

b: \(5^x=125\)

nên x=3

c: \(3^{x+1}=9\)

nên x+1=2

hay x=1

d: \(6^{x+1}=36\)

nên x+1=2

hay x=1

e: \(3^{2x+1}=27\)

=>2x+1=3

hay x=1

f: \(x^{50}=x\)

nên \(x^{50}-x=0\)

=>x(x-1)=0

=>x=0 hoặc x=1

baif 3

a)\(3^x=9\Leftrightarrow3^x=3^2\Rightarrow x=2\)

b)\(5^x=125\Leftrightarrow5^x=5^3\Rightarrow x=3\)

c)\(3^{x+1}=9\Leftrightarrow3^{x+1}=3^{1+1}\Rightarrow x=1\)

d)\(6^{x-1}=36\Leftrightarrow6^{x-1}=6^{3-1}\Rightarrow x=3\)

e)\(3^{x+1}=27\Leftrightarrow3^{x+1}=3^{2+1}\Rightarrow x=2\)

g)\(x^{50}=x\Rightarrow x^{50}:x=1\Rightarrow x^{49}=1\Rightarrow x=1\)

hoặc \(x^{50}-x=0\Rightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

Refer:
1,

a) {145 -[130 -(246 - 236)]: 2}. 5

= {145 -[130 - 10]: 2}. 5

= {145 - 120 : 2}. 5

= {145 - 60}. 5

= 85 .5 

=425

 b) 100 :{250 :[450 -(4 . 5³ - 2² . 25)]}

= 100 :{250 :[450 -(4 . 125 - 4 . 25)]}

= 100 :{250 :[450 -(4 .(125 - 25)]}

= 100 :{250 :[450 -(4 . 100)]}

= 100 :{250 :[450 - 400]}

= 100 :{250 : 50}

= 100 : 5

= 20

c) 325 - 5 .[4³ -(27 - 5²): 1^18]

= 325 - 5 .[64 -(27 - 25): 1]

= 325 - 5 .[64 - 2 : 1]

= 325 - 5 . 62

= 325 - 310

= 15

d) 17^0 -[5^13 : 5^11 +(135 - 130)³]

= 1 -[5^13-11 +(135 - 130)³]

= 1 -[5² + 5³]

= 1 -[25 + 125]

= 1 - 150

= -149

\(\Leftrightarrow\left|2x-1\right|=\dfrac{7}{2}:\dfrac{21}{22}=\dfrac{7}{2}\cdot\dfrac{22}{21}=\dfrac{11}{3}\)

=>2x-1=11/3 hoặc 2x-1=-11/3

=>2x=14/3 hoặc 2x=-8/3

=>x=7/3 hoặc x=-4/3

Nếu \(x>\dfrac{1}{2}\) , ta có:

\(3\dfrac{1}{2}:|2x-1|=\dfrac{21}{22}\Rightarrow\dfrac{7}{2}:\left(2x-1\right)=\dfrac{21}{22}\Rightarrow x=\dfrac{7}{3}\left(tm\right)\)

Nếu \(x< \dfrac{1}{2}\), ta có:

\(3\dfrac{1}{2}:|2x-1|=\dfrac{21}{22}\Rightarrow\dfrac{7}{2}:\left(1-2x\right)=\dfrac{21}{22}\Rightarrow-2x=\dfrac{8}{3}\Rightarrow x=-\dfrac{4}{3}\left(tm\right)\)

Vậy \(x=\dfrac{7}{3};x=\dfrac{4}{3}\)

\(=>\left|2x-1\right|=3\dfrac{1}{2}:\dfrac{21}{22}=\dfrac{7}{2}:\dfrac{21}{22}=\dfrac{11}{3}\)

\(=>\left[{}\begin{matrix}2x-1=\dfrac{11}{3}\\2x-1=-\dfrac{11}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{14}{3}\\2x=-\dfrac{8}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

C

C

C

\(A=\dfrac{4}{1.2}+\dfrac{4}{2.3}+...+\dfrac{4}{1010.1011}\\ =4\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{1010.1011}\right)\\ =4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{1010}-\dfrac{1}{1011}\right)\\ =4\left(1-\dfrac{1}{1011}\right)\\ =4.\dfrac{1010}{1011}\\ =\dfrac{4040}{1011}\)

\(A=4.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{1010.1011}\right)\)

\(A=4.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{1010}-\dfrac{1}{1011}\right)\)

\(A=4.\left(1-\dfrac{1}{1011}\right)\)

\(A=4.\dfrac{1010}{1011}\)

\(A=\dfrac{4040}{1010}\)

\(A=4\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{1010.1011}\right)\)

\(A=4\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1010}-\dfrac{1}{1011}\right)\)

\(A=4\left(1-\dfrac{1}{1010}\right)=4\left(\dfrac{1011}{1011}-\dfrac{1}{1011}\right)=\dfrac{4040}{1011}\)

\(f\left(x\right)=x^{10}+101x^9+101x^8-101x^7+...-101x+101\)

\(=x^{10}-100x^9-x^9+100x^8+x^8-100x^7-x^7+....-101x+101\)

\(=x^9.\left(x-100\right)-x^8\left(x-100\right)+x^7\left(x-100\right)-.....+x\left(x-100\right)-\left(x-101\right)\)

\(\Rightarrow f\left(100\right)=1\)

Ta có:

`101=100+1=x+1`

`⇒f(x)=x^10 - 101 x^9 +  ... -101x+101`

`⇒ f(100)=  x^10 - (x+1) x^9 + ... -(x+1).x+x+1`

              `=x^10 - x^10 - x^9 + ... -x^2 -x +x+1`

              `=1`

\(\left(\dfrac{2x}{3}-3\right):\left(-10\right)=\dfrac{2}{5}\)

\(\dfrac{2x}{3}-3=\dfrac{2}{5}\times\left(-10\right)\)

\(\dfrac{2x}{3}-3=-4\)

\(\dfrac{2x}{3}=-4+\left(-3\right)\)

\(\dfrac{2x}{3}=-7\)

\(\)\(2.x:3=-7\)

\(2x=-7\times3\)

\(2x=-21\)

\(x=-21:2\)

\(x=-\dfrac{21}{2}\)

\(=\dfrac{5^4\cdot5^4\cdot2^8}{5^{10}\cdot2^{10}}=\dfrac{1}{5^2\cdot2^2}=\dfrac{1}{100}\)

\(=\dfrac{5^4.5^4.2^8}{5^{10}.2^{10}}=\dfrac{5^8.2^8}{5^{10}.2^{10}}=\dfrac{1}{5^2.2^2}=\dfrac{1}{100}=0,01\)

\(=\dfrac{1}{15}+\dfrac{4}{9}:\dfrac{8}{3}-\dfrac{5}{6}\)

\(=\dfrac{2}{30}-\dfrac{25}{30}+\dfrac{4}{9}\cdot\dfrac{3}{8}\)

\(=\dfrac{-23}{30}+\dfrac{1}{6}=\dfrac{-23+5}{30}=\dfrac{-18}{30}=\dfrac{-3}{5}\)