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

\(ĐKXĐ:x\ne1;x\ne0\)

\(MTC:x\left(x-1\right)\) 

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

\(\Rightarrow150x-140\left(x-1\right)=5x\left(x-1\right)\)

\(\Leftrightarrow150x-140x+140=5x^2-5x\)

\(\Leftrightarrow150x-140x+140-5x^2+5x=0\)

\(\Leftrightarrow-5x^2+15x+140=0\)

\(\Leftrightarrow-5x^2-20x+35x+140=0\)

\(\Leftrightarrow\left(-5x^2+35x\right)+\left(-20x+140\right)=0\)

\(\Leftrightarrow-5x\left(x-7\right)-20\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(-5x-20\right)=0\)

HOẶC \(x-7=0\Leftrightarrow x=7\)(nhận)

HOẶC\(-5x-20=0\Leftrightarrow x=-4\)(nhận)

VẬY TẬP NGHIỆM CỦA PT LÀ \(S=\left\{7;-4\right\}\)

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

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

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

\(\Leftrightarrow\frac{150x-140x+140}{x\left(x-1\right)}=5\)

\(\Leftrightarrow\frac{10x+140}{x\left(x-1\right)}=5\)

\(\Leftrightarrow10x+140=5x\left(x-1\right)\)

\(\Leftrightarrow5\left(2x+28\right)=5x\left(x-1\right)\)

\(\Leftrightarrow2x+28=x\left(x-1\right)\)

\(\Leftrightarrow28=x\left(x-1\right)-2x\)

\(\Leftrightarrow28=x\left(x-1-2\right)\)

\(\Leftrightarrow28=x\left(x-3\right)\)

\(\Leftrightarrow x\left(x-3\right)=7.4=\left(-4\right)\left(-7\right)\)

\(\Leftrightarrow x\in\left\{7;-4\right\}\)

x=-4 hoặc x=7

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

\(\Leftrightarrow\frac{150}{x-1}.x\left(x-1\right)-\frac{140}{x}.x\left(x-1\right)=5.x\left(x-1\right)\)

\(\Leftrightarrow150x-140\left(x-1\right)=5x\left(x-1\right)\)

\(\Leftrightarrow10x+140=5x^2-5x\)

\(\Leftrightarrow5x^2-5x=10x+140\)

\(\Leftrightarrow5x^2-5x-140=10x+140-140\)

\(\Leftrightarrow5x^2-5x-140=10x\)

\(\Leftrightarrow5x^2-5x-140=10x-10\)

\(\Leftrightarrow5x^2-5x-140=0\)

\(\Rightarrow\hept{\begin{cases}x=7\\x=-4\end{cases}}\)

Không chắc nha

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

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

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

\(\Leftrightarrow x-1=2\)

\(\Leftrightarrow x=3\)

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.

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

sorry, em mới học lớp 6 thui

mình cũng giải câu này rồi nhưng  ko biết đúng ko

2)x+6/x-5 + x-5/x+6 = 2x²+23x+61/x²+x-30

dkxd:x khắc 5;x khác-6

mc:(x-5)(x+6)

2x²+2x+61 =2x²+23x+61

2x=23x

2x=0 suy ra x=0

23x=0 suyra x=0 

s={0}

3)6/x-5 + x+2/x-8 = 18/9(x-5)(8-x) - 1

dkxd: x khác 5 ; x khác -8

mc(x-5)(x-8) 

3x+x²-58 =36x-x²+264

3x-58=36x+264  

3x-58=0 suy ra x=58/3

36x+264=0 suy ra x=-22/8

s={58/3;-22/3}

bạn giúp mình giải 3 câu này nhé

ở câu a sai đề cho x+14 đúng là x-14 

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