a.

\(2^x.4=16\)

\(2^x=16:4\)

\(2^x=4\)

\(2^x=2^2\)

\(x=2\)

b.

\(3^x.3=243\)

\(3^x=243:3\)

\(3^x=81\)

\(3^x=3^4\)

\(x=4\)

c.

\(64.4^x=16^8\)

\(4^3.4^x=\left(4^2\right)^8\)

\(4^{x+3}=4^{2.8}\)

\(4^{x+3}=4^{16}\)

\(x+3=16\)

\(x=13\)

d.

\(2^x.16^x=1024\)

\(2^x.\left(2^4\right)^x=2^{10}\)

\(2^x.2^{4x}=2^{10}\)

\(2^{x+4x}=2^{10}\)

\(2^{5x}=2^{10}\)

\(5x=10\)

\(x=2\)

a.

\(=\left(x^2+2xy+y^2\right)-9\)

\(=\left(x+y\right)^2-3^2\)

\(=\left(x+y-3\right)\left(x+y+3\right)\)

b.

\(=\left(x^2+4xy+4y^2\right)-4\)

\(=\left(x+2y\right)^2-2^2\)

\(=\left(x+2y-2\right)\left(x+2y+2\right)\)

c.

\(=\left(x^2+6xy+9y^2\right)-16\)

\(=\left(x+3y\right)^2-4^2\)

\(=\left(x+3y-4\right)\left(x+3y+4\right)\)

d.

\(=\left(x^2+8xy+16y^2\right)-9\)

\(=\left(x+4y\right)^2-3^2\)

\(=\left(x+4y-3\right)\left(x+4y+3\right)\)

em c.on ạ

\(\dfrac{x-5}{3}=\dfrac{x+7}{2}\\ =>2\left(x-5\right)=3\left(x+7\right)\\ =>2x-10=3x+21\\ =>3x-2x=-10-21\\ =>x=-31\)

Vậy: ... 

TH1: \(-\dfrac{1}{2}\le x\le\dfrac{5}{3}\)

\(\left(2x+1\right)+\left(5-3x\right)=6\\ =>2x+1+5-3x=6\\ =>\left(2x-3x\right)+6=6\\ =>x=0\left(tm\right)\)

TH2: \(x>\dfrac{5}{3}\)

\(\left(2x+1\right)-\left(5-3x\right)=6\\ =>2x+1-5+3x=6\\ =>2x+3x=6-1+5\\ =>5x=10\\ =>x=\dfrac{10}{5}=2\left(tm\right)\)

TH3: \(x< -\dfrac{1}{2}\)

\(-\left(2x+1\right)+\left(5-3x\right)=6\\ =>-2x-1+5-3x=6\\ =>-2x-3x+4=6\\ =>-5x=6-4=2\\ =>x=-\dfrac{2}{5}\left(ktm\right)\)

\(A=\dfrac{1}{4}:\left(1,5-1\dfrac{5}{6}\right)+1\\ =\dfrac{1}{4}:\left(\dfrac{3}{2}-\dfrac{11}{6}\right)+1\\ =\dfrac{1}{4}:\dfrac{9-11}{6}+1\\ =\dfrac{1}{4}:\dfrac{-2}{6}+1\\ =\dfrac{1}{4}\cdot\left(-3\right)+1\\ =\dfrac{-3}{4}+1\\ =\dfrac{1}{4}\)

Ta có: 

\(y^3+2y^2+y+4=\left(y+2\right)^3-\left(4y^2+11y+4\right)\)

Do y là số tự nhiên \(\Rightarrow4y^2+11y+4>0\Rightarrow\left(y+2\right)^3>y^3+2y^2+y+4\)

Đồng thời cũng do y tự nhiên \(\Rightarrow2y^2+y+4>0\Rightarrow y^3+2y^2+y+4>y^3\)

\(\Rightarrow y^3< y^3+2y^2+y+4< \left(y+2\right)^3\)

\(\Rightarrow y^3+2y^2+y+4\) là lập phương của 1 số tự nhiên khi và chỉ khi:

\(y^3+2y^2+y+4=\left(y+1\right)^3\)

\(\Leftrightarrow y^2+2y-3=0\Rightarrow y=1\)

\(\Rightarrow x^3=8\Rightarrow x=2\)

Ta có:

`(2x-5)^2022>=0` với mọi x 

`(3y-4)^2024>=0` với mọi y

`=>(2x-5)^2022+(3y-4)^2024>=0` với mọi x,y 

Mặt khác: `(2x-5)^2022+(3y-4)^2024<=0` 

`=>2x-5=0` và `3y-4=0`

`=>x=5/2` và `y=4/3` 

\(P+\left(5\cdot\dfrac{5}{2}-2\cdot\dfrac{4}{3}\right)=6\cdot\left(\dfrac{5}{2}\right)^2+9\cdot\dfrac{5}{2}\cdot\dfrac{4}{3}-\left(\dfrac{4}{3}\right)^2\\ =>P+\dfrac{59}{6}=\dfrac{1183}{18}\\ =>P=\dfrac{1183}{18}-\dfrac{59}{6}\\ =>P=\dfrac{503}{9}\)

\(\dfrac{1}{99\cdot97}-\dfrac{1}{97\cdot95}-...-\dfrac{1}{5\cdot3}-\dfrac{1}{3\cdot1}\\ =\dfrac{1}{99\cdot97}-\left(\dfrac{1}{97\cdot95}+...+\dfrac{1}{5\cdot3}+\dfrac{1}{3\cdot1}\right)\\ =\dfrac{1}{99\cdot97}-\left(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+...+\dfrac{1}{3}-\dfrac{1}{5}+1-\dfrac{1}{3}\right)\\ =\dfrac{1}{99\cdot97}-\left(-\dfrac{1}{97}+1\right)\\ =\dfrac{1}{99}-\dfrac{1}{97}+\dfrac{1}{97}-1\\ =\dfrac{1}{99}-1\\ =-\dfrac{98}{99}\)

\(\dfrac{11}{48}=\dfrac{11.3}{48.3}=\dfrac{33}{144}\)

\(\dfrac{17}{36}=\dfrac{17.4}{36.4}=\dfrac{68}{144}\)

Do \(68>33\Rightarrow\dfrac{68}{144}>\dfrac{33}{144}\Rightarrow\dfrac{17}{36}>\dfrac{11}{48}\)