\(\frac{150}{x-1}-\frac{140}{x}=5\left(ĐKXĐ:x\ne1,x\ne0\right)\\ \Leftrightarrow\frac{150x-140\left(x-1\right)}{x\left(x-1\right)}=\frac{5x\left(x-1\right)}{x\left(x-1\right)}\\ \Leftrightarrow150x-140x+140=5x^2-5x\\5x^2-5x-10x-140=0\\ \Leftrightarrow5x^2-15x-140=0\\ \Leftrightarrow5\left(x^2-3x-28\right)=0\\ \Leftrightarrow5\left[\left(x^2-3x+\frac{9}{4}\right)-28-\frac{9}{4}\right]=0\\ \Leftrightarrow5\left[\left(x-\frac{1}{2}\right)^2-\frac{121}{4}\right]=0\\ \Leftrightarrow5\left(x-\frac{1}{2}-\frac{11}{2}\right)\left(x-\frac{1}{2}+\frac{11}{2}\right)=0\\ \Leftrightarrow5\left(x-6\right)\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-6=0\\x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\\ Vậy...\)

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

\(ĐKXĐ:x\ne1\)

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

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

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

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

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

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

\(\Leftrightarrow10\left(x+14\right)=5x\left(x-1\right)\)

\(\Leftrightarrow2\left(x+14\right)=x\left(x-1\right)\)

\(\Leftrightarrow2x+28=x^2-x\)

\(\Leftrightarrow3x+28=x^2\)

\(\Leftrightarrow3x-x^2=-28\)

\(\Leftrightarrow x\left(3-x\right)=-28\)

\(\Leftrightarrow x\left(3-x\right)=-4.7\)

\(\Leftrightarrow x=-4\)

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

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

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

\(\Leftrightarrow x-1=2\)

\(\Leftrightarrow x=3\)

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.

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

\(\frac{x}{140}=\frac{-18}{y}=\frac{z}{-21}=\frac{-135}{t}=\frac{2679}{6251}\)

\(\Rightarrow x=\frac{140.2679}{6251}=60\)

\(y=\frac{-18.6251}{2679}=-42\)

\(z=\frac{-21.2679}{6251}=-9\)

\(t=\frac{-135.6251}{2679}=315\)

Ta có: x + y = 140 => x = 180 - y

Thay x = 180 - y vào x - x/8 = y + 8/x ta đc:

\(180-y-\frac{180-y}{8}=y+\frac{8}{180-y}\)

\(\Rightarrow\left(180-y\right)\left(180-y\right).8-\left(180-y\right)\left(180-y\right)=8y\left(180-y\right)+8.8\)

\(\Rightarrow\left(180-y\right)^2.8-\left(180-y\right)^2=8y\left(180-y\right)+64\)

\(\Rightarrow\left(32400-360y+y^2\right).8-\left(32400-360y+y^2\right)=1440y-8y^2+64\)

\(\Rightarrow259200-2880y+8y^2-32400+360y-y^2-1440y+8y^2-64=0\)

\(\Rightarrow15y^2-3960y+226736=0\)

\(\Rightarrow y=180\) hoặc y = 84

Khi y = 180 => x = 0

Khi y = 84 => x = 96

hình như tớ lm lộn rồi :v , thui kệ đi ..... mệt

\(A=\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}\Leftrightarrow A=3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right) \)
                                                               \(\Leftrightarrow A=3.\left(\frac{1}{5}-\frac{1}{x+3}\right)\)
Không có gtri A=? ak bạn??

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