1) Ta có: \(5\left(x-2\right)=3x+10\)

\(\Leftrightarrow5x-10-3x-10=0\)

\(\Leftrightarrow2x-20=0\)

\(\Leftrightarrow2\left(x-10\right)=0\)

Vì 2>0

nên x-10=0

hay x=10

Vậy: x=10

2) Ta có: \(x^2\left(x-5\right)-4x+20=0\)

\(\Leftrightarrow x^2\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\\x=-2\end{matrix}\right.\)

Vậy: x∈{-2;2;5}

3) Ta có: \(\frac{3x+1}{4}+\frac{8x-21}{20}=\frac{3\left(x+2\right)}{5}-2\)

\(\Leftrightarrow\frac{5\left(3x+1\right)}{20}+\frac{8x-21}{20}-\frac{12\left(x+2\right)}{20}+\frac{40}{20}=0\)

\(\Leftrightarrow15x+5+8x-21-12\left(x+2\right)+40=0\)

\(\Leftrightarrow15x+5-8x-21-12x-24+40=0\)

\(\Leftrightarrow-5x=0\)

hay x=0

Vậy: x=0

4) ĐKXĐ: x≠5; x≠-5

Ta có: \(\frac{3}{4x-20}+\frac{7}{6x+30}=\frac{15}{2x^2-50}\)

\(\Leftrightarrow\frac{3}{4\left(x-5\right)}+\frac{7}{6\left(x+5\right)}-\frac{15}{2\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\frac{9\left(x+5\right)}{12\left(x-5\right)\left(x+5\right)}+\frac{14\left(x-5\right)}{12\left(x+5\right)\left(x-5\right)}-\frac{180}{12\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow9x+45+14x-70-180=0\)

\(\Leftrightarrow23x-205=0\)

\(\Leftrightarrow23x=205\)

hay \(x=\frac{205}{23}\)(tm)

Vậy: \(x=\frac{205}{23}\)

Giải:

a) $\frac{}{}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}$⇔⇔ 9x² + 12x + 4 - 18x + 12 = 9x² ⇔ 9x² + 12x + 4 - 18x + 12 - 9x² = 0

⇔ 16 + 6x = 0 ⇔ 2(8 + 3x) = 0 ⇔ 8 + 3x = 0 ⇔ x = \(\frac{-8}{3}\)

Vậy nghiệm của phương trình là x = \(\frac{-8}{3}\) .

b) \(\frac{3}{5x-1}+\frac{3}{3-5x}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\text{⇔ }\frac{-3}{1-5x}+\frac{-3}{5x-3}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\)

⇔ \(\frac{9-15x}{\left(1-5x\right)\left(5x-3\right)}+\frac{15x-3}{\left(1-5x\right)\left(5x-3\right)}=\frac{4}{\left(1-5x\right)\left(5x-3\right)}\) ⇔ 9 - 15x + 15x - 3 = 4

⇔ 8 = 4 ( vô lí)

Vậy phương trình trên vô nghiệm.

Mình chỉ làm 2 câu a, b thôi nhé! Các bài tập này cách làm giống nhau, bạn tự hoàn thành những bài còn lại nhé!

a) A=\(\frac{x+1}{6x^3-6x^2}-\frac{x-2}{8x^3-8x}=\frac{x+1}{6x^2\left(x-1\right)}-\frac{x-2}{8x\left(x-1\right)\left(x+1\right)}=\frac{4\left(x+1\right)^2-3x\left(x-2\right)}{24x^2\left(x-1\right)\left(x+1\right)}=\frac{4x^2+8x+4-3x^2+6x}{24x^2\left(x-1\right)\left(x+1\right)}=\frac{x^2+14x+10}{24x^2\left(x-1\right)\left(x+1\right)}\)

d: =>4x+6=15x-12

=>4x-15x=-12-6=-18

=>-11x=-18

hay x=18/11

e: =>\(45x+27=12+24x\)

=>21x=-15

hay x=-5/7

f: =>35x-5=96-6x

=>41x=101

hay x=101/41

g: =>3(x-3)=90-5(1-2x)

=>3x-9=90-5+10x

=>3x-9=10x+85

=>-7x=94

hay x=-94/7

\(1,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2x+6}-\frac{x-6}{x\left(2x-6\right)}=\frac{3x-x+6}{x\left(2x-6\right)}=\frac{2x+6}{x\left(2x-6\right)}\)

\(2,\frac{1}{1-x}+\frac{2x}{x^2-1}=\frac{-1\left(x+1\right)+2x}{x^2-1}=\frac{x-1}{x^2-1}=\frac{1}{x+1}\)

\(3,\frac{1}{xy-x^2}-\frac{1}{y^2-xy}=\frac{1}{x\left(y-x\right)}-\frac{1}{y\left(y-x\right)}=\frac{y-x}{xy\left(y-x\right)}=\frac{1}{xy}\)

\(4,\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}=\frac{-5}{2}\)

\(5,\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}=\frac{3\left(1+2x\right)}{2x\left(x+4\right)}\)

\(6,\frac{12x}{5y^3}.\frac{15y^4}{8x^3}=\frac{9y}{2x^2}\)

a)

Ta có

\(\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{3x}{6}=\frac{y}{5}\)

Áp dụng tc của dãy tỉ só bằng nhau

\(\Rightarrow\frac{3x}{6}=\frac{y}{5}=\frac{3x-y}{6-5}=\frac{10}{1}=10\)

=> x=2.10=20

    y=5.10=50

Ta có

\(\frac{x}{2}=\frac{y}{5}\Rightarrow\frac{x^2}{4}=\frac{y^2}{25}=\frac{xy}{10}=\frac{30}{10}=3\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\sqrt{12}\\x=-\sqrt{12}\end{array}\right.\)

     \(\left[\begin{array}{nghiempt}y=\sqrt{75}\\y=-\sqrt{75}\end{array}\right.\)

Mà 2;5 cùng dấu

=> x; y cùng dấu

Vậy \(\left(x;y\right)=\left(\sqrt{12};\sqrt{75}\right);\left(-\sqrt{12};-\sqrt{75}\right)\)

a) Ta có: \(\frac{x}{2}\) = \(\frac{y}{5}\) và 3x-y = 10

=> \(\frac{3x}{6}\) = \(\frac{y}{5}\) và 3x-y = 10

Áp dụng tính chất của dãy tỉ số bằng nhau ta có: 

\(\frac{3x}{6}\) = \(\frac{y}{5}\) = \(\frac{3x-y}{6-5}\) = \(\frac{11}{1}\) = 11

=> x= \(\frac{11.6}{3}\) = 22

=> y= 11.5= 55

Vậy x= 22

       y= 55