![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{150}{x-1}-\frac{140}{x}=5\left(ĐKXĐ:x\ne1,x\ne0\right)\\ \Leftrightarrow\frac{150x-140\left(x-1\right)}{x\left(x-1\right)}=\frac{5x\left(x-1\right)}{x\left(x-1\right)}\\ \Leftrightarrow150x-140x+140=5x^2-5x\\5x^2-5x-10x-140=0\\ \Leftrightarrow5x^2-15x-140=0\\ \Leftrightarrow5\left(x^2-3x-28\right)=0\\ \Leftrightarrow5\left[\left(x^2-3x+\frac{9}{4}\right)-28-\frac{9}{4}\right]=0\\ \Leftrightarrow5\left[\left(x-\frac{1}{2}\right)^2-\frac{121}{4}\right]=0\\ \Leftrightarrow5\left(x-\frac{1}{2}-\frac{11}{2}\right)\left(x-\frac{1}{2}+\frac{11}{2}\right)=0\\ \Leftrightarrow5\left(x-6\right)\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-6=0\\x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\\ Vậy...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{150}{x-1}-\frac{140}{x}=5\)
\(\Leftrightarrow\frac{150}{x-1}.x\left(x-1\right)-\frac{140}{x}.x\left(x-1\right)=5.x\left(x-1\right)\)
\(\Leftrightarrow150x-140\left(x-1\right)=5x\left(x-1\right)\)
\(\Leftrightarrow10x+140=5x^2-5x\)
\(\Leftrightarrow5x^2-5x=10x+140\)
\(\Leftrightarrow5x^2-5x-140=10x+140-140\)
\(\Leftrightarrow5x^2-5x-140=10x\)
\(\Leftrightarrow5x^2-5x-140=10x-10\)
\(\Leftrightarrow5x^2-5x-140=0\)
\(\Rightarrow\hept{\begin{cases}x=7\\x=-4\end{cases}}\)
Không chắc nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{150}{x-1}-\frac{140}{x-1}=5\left(ĐK:x\ne1\right)\)
\(\Leftrightarrow\frac{10}{x-1}=5\)
\(\Leftrightarrow x-1=2\)
\(\Leftrightarrow x=3\)
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.
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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}
![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(\frac{150}{x-1}-\frac{140}{x}=5\)
\(ĐKXĐ:x\ne1;x\ne0\)
\(MTC:x\left(x-1\right)\)
\(\Leftrightarrow\frac{150x}{x\left(x-1\right)}-\frac{140\left(x-1\right)}{x\left(x-1\right)}=\frac{5x\left(x-1\right)}{x\left(x-1\right)}\)
\(\Rightarrow150x-140\left(x-1\right)=5x\left(x-1\right)\)
\(\Leftrightarrow150x-140x+140=5x^2-5x\)
\(\Leftrightarrow150x-140x+140-5x^2+5x=0\)
\(\Leftrightarrow-5x^2+15x+140=0\)
\(\Leftrightarrow-5x^2-20x+35x+140=0\)
\(\Leftrightarrow\left(-5x^2+35x\right)+\left(-20x+140\right)=0\)
\(\Leftrightarrow-5x\left(x-7\right)-20\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(-5x-20\right)=0\)
HOẶC \(x-7=0\Leftrightarrow x=7\)(nhận)
HOẶC\(-5x-20=0\Leftrightarrow x=-4\)(nhận)
VẬY TẬP NGHIỆM CỦA PT LÀ \(S=\left\{7;-4\right\}\)
\(\frac{150}{x-1}-\frac{140}{x}=5\)
\(\Leftrightarrow\frac{150x}{x\left(x-1\right)}-\frac{140\left(x-1\right)}{x\left(x-1\right)}=5\)
\(\Leftrightarrow\frac{150x-140\left(x-1\right)}{x\left(x-1\right)}=5\)
\(\Leftrightarrow\frac{150x-140x+140}{x\left(x-1\right)}=5\)
\(\Leftrightarrow\frac{10x+140}{x\left(x-1\right)}=5\)
\(\Leftrightarrow10x+140=5x\left(x-1\right)\)
\(\Leftrightarrow5\left(2x+28\right)=5x\left(x-1\right)\)
\(\Leftrightarrow2x+28=x\left(x-1\right)\)
\(\Leftrightarrow28=x\left(x-1\right)-2x\)
\(\Leftrightarrow28=x\left(x-1-2\right)\)
\(\Leftrightarrow28=x\left(x-3\right)\)
\(\Leftrightarrow x\left(x-3\right)=7.4=\left(-4\right)\left(-7\right)\)
\(\Leftrightarrow x\in\left\{7;-4\right\}\)