Đkxđ: ...

Đặt x²+ 15=a (a>0)

Pt ban đầu trở thành:

(a-10x)/(a-6x)=4x/(a-12x)

<=>a²-26ax+144x²=0

<=>(a-12x)(a-10x)=0

Xét th:a=10x

pt có nghiệm \(X=5\pm\sqrt{10}\)

Xét th:a=12x

Pt có nghiệm \(X=6\pm\sqrt{21}\)

 làm như thế này nha:blablablablablablablabla hiểu hơm

\(\Leftrightarrow\dfrac{x+1}{\left(x+1\right)^2-1}+\dfrac{x+6}{\left(x+6\right)^2-1}=\dfrac{x+2}{\left(x+2\right)^2-1}=\dfrac{x+5}{\left(x+5\right)^2-1}\)

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)^2-x-1+\left(x+6\right)\left(x+1\right)^2-x-6=\left(x+2\right)\left(x+5\right)^2-x-2+\left(x+5\right)\left(x+2\right)^2-x-5\)

=>(x+1)(x+6)^2+(x+6)(x+1)^2=(x+2)(x+5)^2+(x+2)^2(x+5)

=>(x+1)(x+6)(x+6+x+1)=(x+2)(x+5)(x+5+x+2)

=>(2x+7)[x^2+7x+6-x^2-7x-10]=0

=>(2x+7)=0

=>x=-7/2

=>5(4x-1)-2+x<=3(10x-3)

=>20x-5+x-2<=30x-9

=>21x-7<=30x-9

=>-9x<=-2

=>x>=2/9

\(\Leftrightarrow\dfrac{2}{\left(x+1\right)\left(x+3\right)}+\dfrac{2}{\left(x+3\right)\left(x+5\right)}+\dfrac{2}{\left(x+5\right)\left(x+7\right)}+\dfrac{2}{\left(x+7\right)\left(x+9\right)}=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+9}=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+9}=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{x+9-x-1}{\left(x+1\right)\left(x+9\right)}=\dfrac{2}{5}\)

=>2(x+1)(x+9)=5*8=40

=>x^2+9x+9=20

=>x^2+9x-11=0

hay \(x=\dfrac{-9\pm5\sqrt{5}}{2}\)

=>x^2+9x

a. \(\dfrac{6x+5}{2}-\dfrac{10x+3}{4}=2x+\dfrac{2x+1}{2}\)

\(\Leftrightarrow2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)

\(\Leftrightarrow12x+10-10x-3=8x+4x+2\)

\(\Leftrightarrow12x-10x-8x-4x=2-10+3\)

\(\Leftrightarrow-10x=-5\Leftrightarrow x=\dfrac{1}{2}\)

b. \(\left(x+1\right)^3-\left(x-1\right)^3=6\left(x^2+x+1\right)\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=6x^2+6x+6\)

\(\Leftrightarrow6x^2+2=6x^2+6x+6\)

\(\Leftrightarrow6x^2-6x^2-6x=6-2\Leftrightarrow-6x=4\)

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

c. \(\dfrac{x+2}{13}+\dfrac{2x+45}{15}=\dfrac{3x+8}{37}+\dfrac{4x+69}{9}\)

\(\Leftrightarrow\left(\dfrac{x+2}{13}+1\right)+\left(\dfrac{2x+45}{15}-1\right)=\left(\dfrac{3x+8}{37}+1\right)+\left(\dfrac{4x+69}{9}-1\right)\)

\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2\left(x+15\right)}{15}-\dfrac{3\left(x+15\right)}{37}-\dfrac{4\left(x+15\right)}{9}=0\)

\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)=0\)

Vì \(\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)>0\)

\(\Leftrightarrow x+15=0\Leftrightarrow x=-15\)

a) \(15 - 4x = x - 5\)

\( - 4x - x =  - 5 - 15\) (chuyển vế)

\( - 5x =  - 20\)

\(x = \left( { - 20} \right):\left( { - 5} \right)\) (chia cho một số)

\(x = 4\)

Vậy phương trình có nghiệm \(x = 4\).

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

\(\dfrac{{\left( {5x + 2} \right).3}}{{4.3}} + \dfrac{{\left( {3x - 2} \right).4}}{{3.4}} = \dfrac{{3.6}}{{2.6}}\) (quy đồng mẫu số)

\(\dfrac{{15x + 6}}{{12}} + \dfrac{{12x - 8}}{{12}} = \dfrac{{18}}{{12}}\)

\(15x + 6 + 12x - 8 = 18\) (chia cả hai vế cho một số)

\(15x + 12x = 18 - 6 + 8\) (chuyển vế)

\(27x = 20\) (rút gọn)

\(x = 20:27\) (chia cả hai vế co một số)

\(x = \dfrac{{20}}{{27}}\)

Vậy phương trình có nghiệm \(x = \dfrac{{20}}{{27}}\).