\(\frac{x-1}{99}-1+\frac{x-2}{49}-2+\frac{x-7}{31}-3+\frac{x-8}{23}-4=0\)

\(\frac{x-100}{99}+\frac{x-100}{49}+\frac{x-100}{31}+\frac{x-100}{23}=0\)

\(\left(x-100\right)\left(\frac{1}{99}+\frac{1}{49}+\frac{1}{31}+\frac{1}{23}\right)=0\)

x-100=0 ( vi 1/99+1/49+1/31+1/23 khác 0)

x=100

Thực ra cũng EZ thôi :

\(\frac{6}{x^2-9}-1+\frac{4}{x^2-11}-1-\frac{7}{x^2-8}+1-\frac{3}{x^2-12}+1=0=>\)

\(\frac{15-x^2}{x^2-9}+\frac{15-x^2}{x^2-11}-\frac{15-x^2}{x^2-8}-\frac{15-x^2}{x^2-12}=0\)

=> \(\left(15-x^2\right)\left(\frac{1}{x^2-9}+\frac{1}{x^2-11}-\frac{1}{x^2-8}-\frac{1}{x^2-12}\right)=0\)

=>\(15-x^2=0=>x=\pm\sqrt{15}\)

Hình như còn nghiệm , any body help me ?

<=> (x-1)/99-1 + (x-2)/49-2 + (x-7)/31-3 +(x-8)/23-4=0

<=> (x-100)/99 + (x-100)/49 + (x-100)/31 + (x-100)/23=0

<=> (x-100)(1/99 + 1/49 + 1/31 + 1/23)=0

<=> x-100=0(vì 1/99 + 1/49 + 1/31 +1/23)

<=> x=100

Vậy PT có TN S={100}

ĐKXĐ: \(\left\{{}\begin{matrix}x^2\ne9\\x^2\ne11\\x^2\ne8\\x^2\ne12\end{matrix}\right.\Leftrightarrow x\notin\left\{3;-3;\sqrt{11};-\sqrt{11};2\sqrt{2};-2\sqrt{2};2\sqrt{3};-2\sqrt{3}\right\}\)

Đặt \(x^2-11=a\)(Điều kiện: \(a\notin\left\{-2;0;-3;1\right\}\))

PT\(\Leftrightarrow\frac{6}{a+2}+\frac{4}{a}-\frac{7}{a+3}-\frac{3}{a-1}=0\)

\(\Leftrightarrow\frac{6}{a+2}-1+\frac{4}{a}-1+\frac{-7}{a+3}+1+\frac{-3}{a-1}+1=0\)

\(\Leftrightarrow\frac{6-a-2}{a+2}+\frac{4-a}{a}+\frac{-7+a+3}{a+3}+\frac{-3+a-1}{a-1}=0\)

\(\Leftrightarrow-\frac{a-4}{a+2}-\frac{a-4}{a}+\frac{a-4}{a+3}+\frac{a-4}{a-1}=0\)

\(\Leftrightarrow\left(a-4\right)\left(-\frac{1}{a+2}-\frac{1}{a}+\frac{1}{a+3}+\frac{1}{a-1}\right)=0\)

\(\Leftrightarrow a-4=0\)

hay a=4

\(\Leftrightarrow x^2-11=4\)

\(\Leftrightarrow x^2=15\)

hay \(x=\pm\sqrt{15}\)

\(\frac{7}{8}x-5\left(x-9\right)=\frac{20x-1,5}{6}\)

\(\Leftrightarrow\frac{7}{8}x-5x+45=\frac{10x}{3}-\frac{1}{4}\)

\(\Leftrightarrow\frac{-33}{8}x+45=\frac{10x}{3}-\frac{1}{4}\)

\(\Leftrightarrow\frac{-33}{8}x-\frac{10}{3}x=-\frac{1}{4}-45\)

\(\Leftrightarrow\frac{-179}{24}x=-\frac{181}{4}\)

\(\Leftrightarrow x=\frac{1086}{179}\)

\(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)

\(\Rightarrow\frac{7}{8}x-5x+45=\frac{20x}{6}+\frac{1}{4}\)

\(\Rightarrow\frac{7}{8}x-\frac{40}{8}x+45=\frac{10x}{3}+\frac{1}{4}\)

\(\Rightarrow\frac{-33}{8}x+45=\frac{10x}{3}+\frac{1}{4}\)

\(\Rightarrow\frac{-33}{8}x-\frac{10x}{3}=\frac{1}{4}-45\)

\(\Rightarrow\frac{-179}{24}x=\frac{-179}{4}\)

\(\Rightarrow x=6\)

Vậy phương trình có 1 nghiệm là 6

Câu 1 :

a, Ta có : \(3\left(x-1\right)-2\left(x+3\right)=-15\)

=> \(3x-3-2x-6=-15\)

=> \(3x-3-2x-6+15=0\)

=> \(x=-6\)

Vậy phương trình có nghiệm là x = -6 .

b, Ta có : \(3\left(x-1\right)+2=3x-1\)

=> \(3x-3+2=3x-1\)

=> \(3x-3+2-3x+1=0\)

=> \(0=0\)

Vậy phương trình có vô số nghiệm .

c, Ta có : \(7\left(2-5x\right)-5=4\left(4-6x\right)\)

=> \(14-35x-5=16-24x\)

=> \(14-35x-5-16+24x=0\)

=> \(-35x+24x=7\)

=> \(x=\frac{-7}{11}\)

Vậy phương trình có nghiệm là \(x=\frac{-7}{11}\) .

Bài 2 :

a, Ta có : \(\frac{x}{30}+\frac{5x-1}{10}=\frac{x-8}{15}-\frac{2x+3}{6}\)

=> \(\frac{x}{30}+\frac{3\left(5x-1\right)}{30}=\frac{2\left(x-8\right)}{30}-\frac{5\left(2x+3\right)}{30}\)

=> \(x+3\left(5x-1\right)=2\left(x-8\right)-5\left(2x+3\right)\)

=> \(x+15x-3=2x-16-10x-15\)

=> \(x+15x-3-2x+16+10x+15=0\)

=> \(24x+28=0\)

=> \(x=\frac{-28}{24}=\frac{-7}{6}\)

Vậy phương trình có nghiệm là \(x=\frac{-7}{6}\) .

b, Ta có : \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)

=> \(\frac{6\left(x+4\right)}{30}-\frac{30x}{30}+\frac{120}{30}=\frac{10x}{30}-\frac{15\left(x-2\right)}{30}\)

=> \(6\left(x+4\right)-30x+120=10x-15\left(x-2\right)\)

=> \(6x+24-30x+120=10x-15x+30\)

=> \(6x+24-30x+120-10x+15x-30=0\)

=> \(-19x+114=0\)

=> \(x=\frac{-114}{-19}=6\)

Vậy phương trình có nghiệm là x = 6 .

a/ \(\Rightarrow x^2+9x=7\left(x+3\right)^2\)

\(\Rightarrow x^2+9x=7x^2+42x+63\).

\(\Rightarrow6x^2+33x+63=0\)

Có denta = 33² - 4.6.63 = -423 < 0 

=> pt vô nghiệm 

Vậy k có giá trị nào của x thỏa mãn biểu thức => \(x\in\phi\)

b) ĐK : ........

 PT đã cho tương đương với :

\(\frac{3}{x-4+\frac{1}{x}}+\frac{2}{x+1+\frac{1}{x}}=\frac{8}{3}\)

Đặt  x + 1/x + 1 = a 

pt <=> \(\frac{3}{a-5}+\frac{2}{a}=\frac{8}{3}\)

giải pt với ẩn a 

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge16\\y\ge9\end{matrix}\right.\)

Từ pt thứ nhất của hệ:

\(\frac{8xy}{x^2+y^2+6xy}+\frac{17}{8}\left(\frac{x}{y}+\frac{y}{x}\right)=\frac{21}{4}\)

\(\Leftrightarrow\frac{8}{\frac{x}{y}+\frac{y}{x}+6}+\frac{17}{8}\left(\frac{x}{y}+\frac{y}{x}\right)=\frac{21}{4}\)

Đặt \(\frac{x}{y}+\frac{y}{x}=t\ge2\)

\(\Rightarrow\frac{8}{6+t}+\frac{17}{8}t=\frac{21}{4}\)

\(\Leftrightarrow\frac{17}{8}t^2+\frac{15}{2}t-\frac{47}{2}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-\frac{94}{17}< 0\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\frac{x}{y}+\frac{y}{x}=2\Leftrightarrow x^2+y^2=2xy\)

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

Thay xuống pt dưới:

\(\sqrt{x-16}+\sqrt{x-9}=7\)

\(\Leftrightarrow\sqrt{x-16}-3+\sqrt{x-9}-4=0\)

\(\Leftrightarrow\frac{x-25}{\sqrt{x-16}+3}+\frac{x-25}{\sqrt{x-9}+4}=0\)

\(\Leftrightarrow...\)