\(2\cdot11^x=\left(3^2+2\right)^3:\left(5^3-2^5:2^3\right)\)

\(\Leftrightarrow11^x\cdot2=1331:121\)

\(\Leftrightarrow11^x\cdot2=11\)

=> Phương trình vô nghiệm

\(-x-2=\frac{5}{4}\)

\(\Rightarrow-x=\frac{5}{4}+2=\frac{13}{4}\)

\(\Rightarrow x=\frac{-13}{4}\)

\(-x-2=\frac{5}{4}\)

\(-x=\frac{5}{4}+2\)

\(-x=\frac{13}{4}\)

\(x=-\frac{13}{4}\)

-x - 2 = 5/4

=> -x = 5/4 + 2

=> -x = 13/4

=> x = -13/4

x⁹ : x⁴ = 243

<=> x^9-4 = 243

=>x⁵ = 243

mà 3⁵ = 243

=> x=3

Vậy

Có x⁹:x⁴=x⁵ suy ra x⁵=243=3⁵

Do đó x=5

\(x^9:x^4=243\)

\(x^5=243\)

\(x^5=3^5\)

\(\Rightarrow x=3\)

Vậy x = 3

Đề mik ko hiểu lắm

\(2x\left(x-4\right)-6x^2\left(4-x\right)=0\)

\(\Leftrightarrow6x^2\left(x-4\right)+2x\left(x-4\right)=0\)

\(\Leftrightarrow2x\left(x-4\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-\dfrac{1}{3}\end{matrix}\right.\)

\(2xy-x-y=2\\ \Rightarrow x\left(2y-1\right)-y=2\\ \Rightarrow2x\left(2y-1\right)-2y+1=4+1\\ \Rightarrow2x\left(2y-1\right)-\left(2y-1\right)=5\\ \Rightarrow\left(2x-1\right)\left(2y-1\right)=5\)

Ta có bảng:

Vậy \(\left(x,y\right)\in\left\{\left(-2;0\right);\left(0;-2\right);\left(1;3\right);\left(3;1\right)\right\}\)

\(2x^2+5x-3=0\)

\(\Leftrightarrow2x^2-x+6x-3=0\)

\(\Leftrightarrow x\left(2x-1\right)+3\left(2x-1\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{1}{2}\end{cases}}}\)

\(4x+2x=0\)

\(\Rightarrow6x=0\)

\(x=0:6\)

\(x=0\)

\(\dfrac{x}{15}\)+\(\dfrac{x}{12}\)=4/1+1/2=9/2

=>x(\(\dfrac{1}{15}\)+\(\dfrac{1}{12}\))=9/2

=>x\(\cdot\)\(\dfrac{3}{20}\)=9/2

=>x=9/2:3/20=30

Vậy x=30

\(\dfrac{x}{15}+\dfrac{x}{12}=\dfrac{9}{2}\Rightarrow\left(\dfrac{1}{15}+\dfrac{1}{12}\right)x=\dfrac{9}{2}\)

\(\Rightarrow\left(\dfrac{12+18}{180}\right)x=\dfrac{9}{2}\Rightarrow\dfrac{30}{180}x=\dfrac{9}{2}\Rightarrow\dfrac{1}{6}x=\dfrac{9}{2}\Rightarrow x=\dfrac{9}{2}.6=27\)

ai giúp mik vs