5x^x+1+5x=150

=>5x(x^x+1)=150

=>x^x+1=30/x

=>x^x=x+30/x

Đề sai à bạn

\(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}\)

\(\Rightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\)

\(15x^2+10x+3x+2=15x^2-5x+21x-7\)

\(15x^2+10x+3x-15x^2+5x-21x=-7-2\)

\(-3x=-9\)

\(x=3\)

Nhân chéo      

a ) \(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}\)

\(\Leftrightarrow\left(2x+3\right).\left(10x+2\right)=\left(5x+2\right).\left(4x+5\right)\)

\(\Leftrightarrow20x^2+4x+30x+6=20x^2+25x+8x+10\)

\(\Leftrightarrow4x+30x-25x-8x=10-6\)

\(\Leftrightarrow1x=4\)

\(\Leftrightarrow x=4:1\)

\(\Leftrightarrow x=4\)

b ) \(\frac{3x-1}{40-5x}=\frac{25-3x}{5x-34}\)

\(\Leftrightarrow\left(3x-1\right).\left(5x-34\right)=\left(40-5x\right).\left(25-3x\right)\)

\(\Leftrightarrow15x^2-102x-5x+34=1000-120x-125x+15x^2\)

\(\Leftrightarrow-102x-5x+120x+125x=1000-34\)

\(\Leftrightarrow138x=966\)

\(\Leftrightarrow x=966:138\)

\(\Leftrightarrow x=7\)

a. \(\left(2x+3\right)\left(10x+2\right)=\left(5x+2\right)\left(4x+5\right)\)

\(20x^2+34x+5=20x^2+33x+10\)

rút gọn \(x=5\)

mình gửi câu a tí nữa mình gửi câu b

Ta có : \(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)

\(\Leftrightarrow\dfrac{5x-150}{50}-1+\dfrac{5x-102}{49}-2+\dfrac{5x-56}{48}-3+\dfrac{5x-12}{47}-4+\dfrac{5x-660}{46}+10=0\)

\(\Leftrightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)

\(\Leftrightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)

\(\Leftrightarrow5x-200=0\)

\(\Leftrightarrow x=40\)

Vậy ...

Ta có: \(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)

\(\Leftrightarrow\dfrac{5x-150}{50}-1+\dfrac{5x-102}{49}-2+\dfrac{5x-56}{48}-3+\dfrac{5x-12}{47}-4+\dfrac{5x-660}{46}+10=0\)

\(\Leftrightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)

\(\Leftrightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)

mà \(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}>0\)

nên 5x-200=0

\(\Leftrightarrow5x=200\)

hay x=40

Vậy: S={40}

a/ \(=5x^2\left(x-y\right)-10x\left(x-y\right)=\left(x-y\right)\left(5x^2-10x\right)=5x\left(x-2\right)\left(x-y\right)\)

b/ \(=2x^2-7x+2x-7=x\left(2x-7\right)+\left(2x-7\right)=\left(x+1\right)\left(2x-7\right)\)

2819+289=

\(x=\frac{1}{3}\)