\(\frac{150}{x-1}-\frac{140}{x}=5\)
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14 tháng 2 2019
\(\frac{150}{x-1}-\frac{140}{x}=5\)
\(ĐKXĐ:x\ne1;x\ne0\)
\(MTC:x\left(x-1\right)\)
\(\Leftrightarrow\frac{150x}{x\left(x-1\right)}-\frac{140\left(x-1\right)}{x\left(x-1\right)}=\frac{5x\left(x-1\right)}{x\left(x-1\right)}\)
\(\Rightarrow150x-140\left(x-1\right)=5x\left(x-1\right)\)
\(\Leftrightarrow150x-140x+140=5x^2-5x\)
\(\Leftrightarrow150x-140x+140-5x^2+5x=0\)
\(\Leftrightarrow-5x^2+15x+140=0\)
\(\Leftrightarrow-5x^2-20x+35x+140=0\)
\(\Leftrightarrow\left(-5x^2+35x\right)+\left(-20x+140\right)=0\)
\(\Leftrightarrow-5x\left(x-7\right)-20\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(-5x-20\right)=0\)
HOẶC \(x-7=0\Leftrightarrow x=7\)(nhận)
HOẶC\(-5x-20=0\Leftrightarrow x=-4\)(nhận)
VẬY TẬP NGHIỆM CỦA PT LÀ \(S=\left\{7;-4\right\}\)
KN
14 tháng 2 2019
\(\frac{150}{x-1}-\frac{140}{x}=5\)
\(\Leftrightarrow\frac{150x}{x\left(x-1\right)}-\frac{140\left(x-1\right)}{x\left(x-1\right)}=5\)
\(\Leftrightarrow\frac{150x-140\left(x-1\right)}{x\left(x-1\right)}=5\)
\(\Leftrightarrow\frac{150x-140x+140}{x\left(x-1\right)}=5\)
\(\Leftrightarrow\frac{10x+140}{x\left(x-1\right)}=5\)
\(\Leftrightarrow10x+140=5x\left(x-1\right)\)
\(\Leftrightarrow5\left(2x+28\right)=5x\left(x-1\right)\)
\(\Leftrightarrow2x+28=x\left(x-1\right)\)
\(\Leftrightarrow28=x\left(x-1\right)-2x\)
\(\Leftrightarrow28=x\left(x-1-2\right)\)
\(\Leftrightarrow28=x\left(x-3\right)\)
\(\Leftrightarrow x\left(x-3\right)=7.4=\left(-4\right)\left(-7\right)\)
\(\Leftrightarrow x\in\left\{7;-4\right\}\)
18 tháng 1 2019
\(\frac{150}{x-1}-\frac{140}{x}=5\)
\(\Leftrightarrow\frac{150}{x-1}.x\left(x-1\right)-\frac{140}{x}.x\left(x-1\right)=5.x\left(x-1\right)\)
\(\Leftrightarrow150x-140\left(x-1\right)=5x\left(x-1\right)\)
\(\Leftrightarrow10x+140=5x^2-5x\)
\(\Leftrightarrow5x^2-5x=10x+140\)
\(\Leftrightarrow5x^2-5x-140=10x+140-140\)
\(\Leftrightarrow5x^2-5x-140=10x\)
\(\Leftrightarrow5x^2-5x-140=10x-10\)
\(\Leftrightarrow5x^2-5x-140=0\)
\(\Rightarrow\hept{\begin{cases}x=7\\x=-4\end{cases}}\)
Không chắc nha
LN
3
18 tháng 1 2019
\(\frac{150}{x-1}-\frac{140}{x-1}=5\left(ĐK:x\ne1\right)\)
\(\Leftrightarrow\frac{10}{x-1}=5\)
\(\Leftrightarrow x-1=2\)
\(\Leftrightarrow x=3\)
18 tháng 1 2019
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.
LD
Tìm số x,y,z,t biết : \(\frac{x}{140}=\frac{-18}{y}=\frac{z}{-21}=-\frac{135}{t}=\frac{2679}{6251}\)
1
MT
29 tháng 7 2015
\(\frac{x}{140}=\frac{-18}{y}=\frac{z}{-21}=\frac{-135}{t}=\frac{2679}{6251}\)
\(\Rightarrow x=\frac{140.2679}{6251}=60\)
\(y=\frac{-18.6251}{2679}=-42\)
\(z=\frac{-21.2679}{6251}=-9\)
\(t=\frac{-135.6251}{2679}=315\)
LN
2
31 tháng 1 2016
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
HH
22 tháng 3 2016
\(A=\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}\Leftrightarrow A=3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)
\)
\(\Leftrightarrow A=3.\left(\frac{1}{5}-\frac{1}{x+3}\right)\)
Không có gtri A=? ak bạn??
TN
4 tháng 3 2016
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
\(\frac{150}{x-1}-\frac{140}{x}=5\left(ĐKXĐ:x\ne1,x\ne0\right)\\ \Leftrightarrow\frac{150x-140\left(x-1\right)}{x\left(x-1\right)}=\frac{5x\left(x-1\right)}{x\left(x-1\right)}\\ \Leftrightarrow150x-140x+140=5x^2-5x\\5x^2-5x-10x-140=0\\ \Leftrightarrow5x^2-15x-140=0\\ \Leftrightarrow5\left(x^2-3x-28\right)=0\\ \Leftrightarrow5\left[\left(x^2-3x+\frac{9}{4}\right)-28-\frac{9}{4}\right]=0\\ \Leftrightarrow5\left[\left(x-\frac{1}{2}\right)^2-\frac{121}{4}\right]=0\\ \Leftrightarrow5\left(x-\frac{1}{2}-\frac{11}{2}\right)\left(x-\frac{1}{2}+\frac{11}{2}\right)=0\\ \Leftrightarrow5\left(x-6\right)\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-6=0\\x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\\ Vậy...\)