\(\frac{150}{x-1}-\frac{140}{x-1}=5\left(ĐK:x\ne1\right)\)

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

\(\Leftrightarrow x-1=2\)

\(\Leftrightarrow x=3\)

ĐKXĐ: x-1\(\ne\)0=> x\(\ne\)1

=> \(\frac{150-140}{x-1}\)=5

=> \(\frac{10}{x-1}\)=5

=> 10= 5(x-1)=> x-1=2=> x=1(ko thỏa mã ĐKXĐ x\(\ne\)1)

phương trình này vô nghiệm.

\(\frac{150}{x-1}-\frac{140}{x-1}\)= 5

\(\frac{150-140}{x-1}=5\)

\(\frac{10}{x-1}=5\)

\(\frac{10}{x-1}=\frac{5x-5}{x-1}\)

10 = 5x - 5 

-5x = -10 - 5

-5x = -15

x = -15 : -5

x = 3

a/ Chia làm 2 trường hợp :

+) x - 1 = 2x => -x = 1 => x = -1

+) x - 1 = -2x => 3x = 1 => x = 1/3 

Vậy x = -1 ; x = 1/3

b/ \(\Rightarrow x=x-5+\left(x+5\right)\left(1-x\right)\)

\(\Rightarrow x=x-5+x-x^2+5-5x\)

\(\Rightarrow x^2+4x=0\Rightarrow x\left(x+4\right)=0\)

\(\Rightarrow x=0\) hoặc \(x+4=0\Rightarrow x=-4\)

Vậy x = 0 ; x = -4

Đặt \(A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)

=>  \(\frac{1}{5}.A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}+\frac{1}{5^{100}}\)

=> \(A-\frac{1}{5}A=\frac{4}{5}.A=1-\frac{1}{5^{100}}\Rightarrow\frac{4}{5}.A=\frac{5^{100}-1}{5^{100}}\Rightarrow A=\frac{5^{100}-1}{4.5^{99}}\)

Tính \(\frac{1}{50}+\frac{1}{150}+\frac{1}{300}+...+\frac{1}{9500}=\frac{1}{25}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{380}\right)\)

\(=\frac{1}{25}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\right)=\frac{1}{25}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)\)\(=\frac{1}{25}.\left(1-\frac{1}{20}\right)=\frac{19}{20.25}=\frac{19}{4.5^3}\)

vậy phương trình đã cho trở thành:

\(\frac{5^{100}-1}{4.5^{99}}.x+\frac{1}{4.5^{99}.x}=\frac{19}{4.5^3}\Rightarrow\left(5^{100}-1\right)x^2+1=19.5^{96}.x\)

\(\left(5^{100}-1\right)x^2-19.5^{96}.x+1=0\)

bạn kiểm tra lại đề lần nữa, phương trình này có nghiệm  rất lẻ , nghiệm lớn

xét x âm dương rồi nhân 2 vế thêm x

\(\Leftrightarrow\left(x-\frac{1}{x}\right)+\sqrt{x-\frac{1}{x}}=\left(2x-\frac{5}{x}\right)+\sqrt{2x-\frac{5}{x}}\)

\(a=\sqrt{x-\frac{1}{x}};\text{ }b=\sqrt{2x-\frac{5}{2}};\text{ }a,\text{ }b>0\)

\(a^2+a=b^2+b\Leftrightarrow\left(a-b\right)\left(a+b\right)+\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)

\(\Leftrightarrow a=b\text{ }\left(do\text{ }a+b+1\ge1>0\right)\)

\(\Leftrightarrow x-\frac{1}{x}=2x-\frac{5}{x}\Leftrightarrow x-\frac{4}{x}=0\Leftrightarrow x^2-4=0\Leftrightarrow x=\pm2\)

Cộng cả tử và mẫu với 1=> Kết quả là -66

Bạn cộng cả tử và mẫu vs 1 là xong 

-66                                    

Cần gì phải thế.

Đặt \(\sqrt{x-\frac{1}{x}}=a\ge0;\sqrt{2x-\frac{5}{x}}=b\ge0\Rightarrow x-\frac{4}{x}=b^2-a^2\)

\(\Rightarrow a=b^2-a^2+b\Leftrightarrow\left(a-b\right)\left(a+b+1\right)=0\)

Đến đây tự làm tiếp

a)\(pt\Leftrightarrow-\frac{x}{2x^2-5}-\frac{25}{2x^2-50}+\frac{x}{x^2-5}+\frac{5}{x^2-5}=\frac{x}{2x^2+10x}-\frac{5}{2x^2+10x}\)

=>\(-\frac{x}{2x^2+10x}+\frac{5}{2x^2+10x}-\frac{x}{2x^2-50}-\frac{25}{2x^2-50}+\frac{x}{x^2-5}+\frac{5}{x^2-5}=0\)

\(\Leftrightarrow-\frac{5\left(x^2+8x-5\right)}{2\left(x-5\right)x\left(x^2-5\right)}=0\)

\(\Rightarrow\frac{1}{x-5}=0\Leftrightarrow\frac{1}{x}=0\Rightarrow\frac{1}{x^2-5}=0\)

=>x²+8x-5=0

=>8²-(-4(1.5))=84

=>x₁=(-8)+8:2=\(\sqrt{21}-4\)

=>x₂=(-8)+8:2=\(-\sqrt{21}-4\)

=>x=±\(\sqrt{21}-4\)

b)\(\Leftrightarrow-\frac{x}{x+1}+\frac{1}{x+1}+\frac{x}{x-1}+\frac{1}{x-1}=\frac{16}{x^2-1}\)

\(\Rightarrow-\frac{16}{x^2-1}-\frac{x}{x+1}+\frac{1}{x+1}+\frac{x}{x-1}+\frac{1}{x-1}=0\)

\(\Rightarrow\frac{4\left(x-4\right)}{\left(x-1\right)\left(x+1\right)}=0\Leftrightarrow\frac{1}{x-1}=0\Rightarrow\frac{1}{x+1}=0\)

=>x=4

c)\(\Leftrightarrow-\frac{x^2}{x+1}-\frac{x}{x+1}+\frac{2}{x+1}+x+2=\frac{x}{x+1}-\frac{1}{x+1}+\frac{x}{x-1}+\frac{1}{x-1}\)

\(\Rightarrow-\frac{x^2}{x+1}-\frac{2x}{x+1}+\frac{3}{x+1}-\frac{x}{x-1}+x-\frac{1}{x-1}+2=0\)

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

=>x=3

\(\frac{3x-1}{x-1}-\frac{2x-5}{x+3}+\frac{4}{x^2+2x-3}=1\)

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

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

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

ĐKXĐ: x \(\ne\) 1 và x \(\ne\) - 3

\(\left(3x-1\right)\left(x+3\right)-\left(2x-5\right)\left(x-1\right)+4=\left(x+3\right)\left(x-1\right)\)

3x² + 9x - x - 3 - 2x² + 2x + 5x - 5 + 4 = x² - x + 3x - 3

3x² + 9x - x - 3 - 2x² + 2x + 5x - 5 + 4 - x² + x - 3x + 3 = 0

13x - 1 = 0

x = \(\frac{1}{13}\)

chính là 1/13 

nếu đúng thì