\(\dfrac{3x-40}{50}+\dfrac{3x-10+2x+60+4x-360}{40}=0\)

=> \(\dfrac{3x-40}{50}+\dfrac{9x-310}{40}=0\)

=> \(\dfrac{3x-40}{50}=\dfrac{-9x+310}{40}\)

=> \(40\left(3x-40\right)=50\left(-9x+310\right)\)

=> \(120x-1600=-450x+15500\)

=> \(120x+450x=15500+1600\)

Hay \(570x=17100\)

=>x = 30

Hơi dài nhé bạn

\(\dfrac{3x-40}{50}\)+\(\dfrac{3x-10+2x+60+4x-360}{40}\)=0

⇒\(\dfrac{3x-40}{50}\)+\(\dfrac{9x-310}{40}\)=0

⇒\(\dfrac{3x-40}{50}\)=\(\dfrac{9x-310}{40}\)

⇒40(3x -40) = 50(-9x+310)

⇒120x - 1600 = -450x + 15500

⇒120x + 450x = 15500 + 1600

Mặt khác: 570x = 17100

⇒x = 30

a) Để biểu thức vô nghĩa thì \(\dfrac{3x-2}{5}-\dfrac{x-4}{3}=0\)

\(\Leftrightarrow\dfrac{3x-2}{5}=\dfrac{x-4}{3}\)

\(\Leftrightarrow3\left(3x-2\right)=5\left(x-4\right)\)

\(\Leftrightarrow9x-6=5x-20\)

\(\Leftrightarrow9x-5x=-20+6\)

\(\Leftrightarrow4x=-14\)

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

Đk: \(x\ne5;x\ne-10\)

Pt: \(\Rightarrow\dfrac{\left(x-2\right)\left(x+5\right)}{x^2}-\dfrac{40}{\left(x-5\right)\left(x+10\right)}=0\)

     \(\Rightarrow\left(x-2\right)\left(x+5\right)\left(x-5\right)\left(x+10\right)-40x^2=0\)

     \(\Rightarrow\left(x^2-12x+20\right)\left(x^2-25\right)-40x^2=0\)

     \(\Rightarrow x^4-12x^3-45x^2+300x=500\)

     \(\Rightarrow\left\{{}\begin{matrix}x=5\left(loại\right)\\x=-5\left(tm\right)\end{matrix}\right.\)

a: \(\Leftrightarrow x^2+x-6+2x-6=10x-20+50\)

\(\Leftrightarrow x^2+3x-12-10x-30=0\)

\(\Leftrightarrow x^2-7x-42=0\)

\(\text{Δ}=\left(-7\right)^2-4\cdot1\cdot\left(-42\right)=217>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{217}}{2}\\x_2=\dfrac{7+\sqrt{217}}{2}\end{matrix}\right.\)

b: \(\Leftrightarrow x^2-3x+5=-x^2+4\)

\(\Leftrightarrow2x^2-3x+1=0\)

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

hay \(x\in\left\{\dfrac{1}{2};1\right\}\)

a: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\cdot\left(x^2-2x-3\right)=0\)

=>(7x+10)(x-3)=0

=>x=3 hoặc x=-10/7

b: \(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow13\left(x+3\right)+x^2-9-12x-42=0\)

\(\Leftrightarrow x^2-12x-51+13x+39=0\)

\(\Leftrightarrow x^2+x-12=0\)

=>(x+4)(x-3)=0

=>x=-4

\(a,=\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}=\dfrac{45}{10}=4,5\\ b,=\dfrac{4}{5}\times\left(\dfrac{3}{8}+\dfrac{5}{8}-\dfrac{7}{8}\right)\times2=\dfrac{8}{5}\times\dfrac{1}{8}=\dfrac{1}{5}=0,2\)

a) Rút gọn các phân số về tối giản, ta được:
\(\dfrac{1}{10}\)+\(\dfrac{2}{10}\)+\(\dfrac{3}{10}\)+\(\dfrac{4}{10}\)+\(\dfrac{5}{10}\)+\(\dfrac{6}{10}\)+\(\dfrac{7}{10}\)+\(\dfrac{8}{10}\)+\(\dfrac{9}{10}\)= kết quả là \(\dfrac{45}{10}\) ra số thập phân = \(4,5\)
b) \(\dfrac{4}{5}\) \(\times\) \(\left(\dfrac{3}{8}+\dfrac{5}{8}-\dfrac{7}{8}\right)\) = \(\dfrac{4}{5}\times\dfrac{1}{8}\) = \(\dfrac{4}{40}=\dfrac{1}{10}\)\(\div\dfrac{1}{2}\)
 = \(\dfrac{2}{10}=0,2\)

b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)

=>(7x+10)(x-3)=0

hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)

d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)

\(\Leftrightarrow x^2+14x+68=0\)

hay \(x\in\varnothing\)

x^2-3x+2=(x-1)(x-2)

dk x≠1;2

1+(x-5)(x-1)=3/10(x^2-3x+2)

10+10x^2-60x+50=3x^2-9x+6

7x^2-54x-54=0

x=(27±3√123)/7

\(\dfrac{1}{x^2-3x+2}-\dfrac{x-5}{2-x}=\dfrac{3}{10}\)

⇔ \(\dfrac{1}{x^2-x-2x+2}+\dfrac{x-5}{x-2}=\dfrac{3}{10}\)

⇔ \(\dfrac{10}{10\left(x-1\right)\left(x-2\right)}+\dfrac{10\left(x-5\right)\left(x-1\right)}{10\left(x-1\right)\left(x-2\right)}=\dfrac{3\left(x^2-3x+2\right)}{10\left(x-1\right)\left(x-2\right)}\)( x # 1 ; x # 2)

⇔ 10 + 10( x² - 6x + 5)= 3(x² - 3x + 2)

⇔ 10 + 10x² - 60x + 50 = 3x² - 9x + 6

⇔ 7x² - 51x - 54 = 0

Phân tích ra

\(\text{a) }\left(x^2-9\right)^2-9\left(x-3\right)^2=0\\ \Leftrightarrow\left(x+3\right)^2\left(x-3\right)^2-9\left(x-3\right)^2=0\\ \Leftrightarrow\left(x^2+6x+9-9\right)\left(x-3\right)^2=0\\ \Leftrightarrow\left(x^2+6x\right)\left(x-3\right)^2=0\\ \Leftrightarrow x\left(x+6\right)\left(x-3\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\\x=3\end{matrix}\right.\)

Vậy phương trình có tập nghiệm \(S=\left\{0;3;-6\right\}\)

\(\text{b) }\dfrac{3x^2+7x-10}{x}=0\\ ĐKXĐ:x\ne0\\ \Rightarrow3x^2+7x-10=0\\ \Leftrightarrow3x^2-3x+10x-10=0\\ \Leftrightarrow\left(3x^2-3x\right)+\left(10x-10\right)=0\\ \Leftrightarrow3x\left(x-1\right)+10\left(x-1\right)=0\\ \Leftrightarrow\left(3x+10\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x+10=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-10\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{10}{3}\\x=1\end{matrix}\right.\left(T/m\right)\)

Vậy phương trình có tập nghiệm \(S=\left\{-\dfrac{10}{3};1\right\}\)

\(\text{c) }x+\dfrac{2x+\dfrac{x-1}{5}}{3}=1-\dfrac{3x+\dfrac{1-2x}{3}}{5}\left(\text{Chữa đề}\right)\\ \Leftrightarrow15x+5\left(2x+\dfrac{x-1}{5}\right)=15-3\left(3x+\dfrac{1-2x}{3}\right)\\ \Leftrightarrow15x+10x+\left(x-1\right)=15-9x+\left(1-2x\right)\\ \Leftrightarrow15x+10x+x-1=15-9x+1-2x\\ \Leftrightarrow26x+11x=16+1\\ \Leftrightarrow37x=17\\ \Leftrightarrow x=\dfrac{17}{37}\\ \)

Vậy phương trình có nghiệm \(x=\dfrac{17}{37}\)