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\(\frac{x-5}{x-1}+\frac{2}{x-3}=1\)(ĐKXĐ: x khác 1;3)
\(\Leftrightarrow\frac{\left(x-5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}+\frac{2\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}=1\)
\(\Leftrightarrow\frac{x^2-8x+15+2x-2}{x^2-4x+3}=1\)
\(\Leftrightarrow\frac{x^2-6x+13}{x^2-4x+3}=1\)\(\Rightarrow x^2-4x+3=x^2-6x+13\)
\(\Leftrightarrow x^2-4x+3-x^2+6x-13=0\)
\(\Leftrightarrow2x-10=0\Leftrightarrow2x=10\Leftrightarrow x=5\)(t/m ĐKXĐ)
Vậy nghiệm của pt là x=5.
ĐKXĐ: x khác 1, 3
\(\frac{x-5}{x-1}+\frac{2}{x-3}-1=0\Leftrightarrow\frac{\left(x-5\right).\left(x-3\right)}{\left(x-1\right).\left(x-3\right)}+\frac{2\left(x-1\right)}{\left(x-1\right).\left(x-3\right)}-\frac{\left(x-1\right).\left(x-3\right)}{\left(x-1\right).\left(x-3\right)}=0\\ \)
\(\Leftrightarrow\frac{\left(x^2-8x+15\right)+\left(2x-2\right)-\left(x^2-4x+3\right)}{\left(x-1\right).\left(x-3\right)}=0\)
\(\Leftrightarrow\left(x^2-8x+15\right)+2x-2-\left(x^2-4x+3\right)=0\)
\(\Leftrightarrow x^2-8x+15+2x-2-x^2+4x-3=0\)
\(\Leftrightarrow-2x+10=0\Leftrightarrow x=5\)
\(ĐKXĐ:x\ne-1;x\ne2\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Rightarrow\frac{x-2+5x+5}{\left(x+1\right)\left(2-x\right)}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Rightarrow x-2+5x+5=15\)
\(\Rightarrow6x+3=15\Leftrightarrow6x=12\Leftrightarrow x=2\)
Vậy x = 2
\(\frac{2}{x-3}+\frac{x-5}{x-1}=1\)
\(\Leftrightarrow\frac{2\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}+\frac{\left(x-3\right)\left(x-5\right)}{\left(x-1\right)\left(x-3\right)}=1\)
\(\Leftrightarrow\frac{2x-2+x^2-8x+15}{\left(x-3\right)\left(x-1\right)}=1\)
\(\Leftrightarrow\frac{x^2-6x+13}{x^2-4x+3}=1\)
\(\Leftrightarrow x^2-6x+13=x^2-4x+3\)
\(\Leftrightarrow2x=10\)
\(\Leftrightarrow x=5\)
Ta có :
\(\frac{2}{x-3}+\frac{x-5}{x-1}=1\)
\(\Leftrightarrow\frac{2\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}+\frac{\left(x-5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\frac{\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{2x-2}{\left(x-1\right)\left(x-3\right)}+\frac{x^2-8x+15}{\left(x-1\right)\left(x-3\right)}=\frac{x^2-4x+3}{\left(x-1\right)\left(x-3\right)}\)
\(\Rightarrow2x-2+x^2-8x+15=x^2-4x+3\)
\(\Leftrightarrow x^2-x^2+2x+4x-8x=3+2-15\)
\(\Leftrightarrow-2x=-10\Leftrightarrow x=5\)
Vậy x = 5 là ngiệm của PT.
\(\frac{3\text{x}-1}{x-1}-\frac{2\text{x}+5}{x+3}=1-\)\(\frac{4}{x^2+2\text{x}-3}\) \(\left(\text{Đ}K\text{X}\text{Đ}:x\ne1;x\ne-3\right)\)
\(\Leftrightarrow\frac{\left(3\text{x}-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2\text{x}+5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}=\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}\)
\(\Rightarrow\left(3\text{x}-1\right)\left(x+3\right)-\left(2\text{x}+5\right)\left(x-1\right)=\left(x-1\right)\left(x+3\right)-4\)
\(\Leftrightarrow3\text{x}^2+8\text{x}-3-2\text{x}^2-3\text{x}+5=x^2+2\text{x}-3-4\)
\(\Leftrightarrow3\text{x}^2-2\text{x}^2-x^2+8\text{x}-3\text{x}-2\text{x}=-3-4+3-5\Leftrightarrow3\text{x}=-9\Leftrightarrow x=-3\)(không thỏa mãn ĐKXĐ)
Vậy pt vô nghiệm
Huyền Subi x2 + 2x - 15 - (x2 - 1) + 8 = 2x - 6 chứ, sao lại là 2x + 6 được, bạn xem lại xem!
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
a.\(\Leftrightarrow\left(x+3\right)\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(-x^2+2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\\x=-1\end{matrix}\right.\)
(x-2)(x+1)(x+3)=(x+3)(x+1)(2x-58)
\(x^3+2x^2-5x-6\)=\(2x^3+3x^2-14x-15\)
\(-x^3-x^2+9x+9=0\)
\(-x^2\left(x+1\right)+9\left(x+1\right)=0\)
\(\left(x+1\right)\left(9-x^2\right)\)=0
(x+1)(3-x)(3+x)=0
*x+1=0 =>x=-1
*3-x=0=>x=3
*3+x=0=>x=-3
=> x + 1 = 1/5
<=> x = -4/5
1/x+1 = 5
x+1=1/5
x=-4/5