\(\dfrac{x+3}{x+1}\)-2=\(\dfrac{1-x}{x}\)
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CM
7 tháng 12 2017
\(\frac{2\frac{1}{2}x-1}{\frac{2}{3}}=\frac{\frac{-2}{3}}{1-2\frac{1}{2}x}\) ĐKXĐ \(x\ne\frac{2}{5}\)
\(\Leftrightarrow\)\(\frac{\frac{5}{2}x-1}{\frac{2}{3}}=\frac{\frac{2}{3}}{\frac{5}{2}x-1}\)\(\Leftrightarrow\)\(\left(\frac{5}{2}x-1\right)^2=\frac{4}{9}\)\(\Leftrightarrow\)\(\frac{25}{4}x^2-5x+1=\frac{4}{9}\)
\(\Leftrightarrow\)\(\frac{25}{4}x^2-5x+\frac{5}{9}=0\)\(\Leftrightarrow\)\(\frac{25}{4}x^2-\frac{25}{6}x-\frac{5}{6}x+\frac{5}{9}=0\)
\(\Leftrightarrow\)\(\left(\frac{25}{4}x^2-\frac{25}{6}x\right)-\left(\frac{5}{6}x-\frac{5}{9}\right)=0\)\(\Leftrightarrow\)\(\frac{25}{2}x\left(\frac{1}{2}x-\frac{1}{3}\right)-\frac{5}{3}\left(\frac{1}{2}x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\)\(\left(\frac{25}{2}x-\frac{5}{3}\right)\left(\frac{1}{2}x-\frac{1}{3}\right)=0\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{2}{15}\end{cases}}\)
7 tháng 12 2017
Tích chéo ta có:
\(-(2\frac{1}{2}x-1) ^2=-(\frac{2}{3})^2 \)
<=>\(2\frac{1}{2}x -1=\frac{2}{3} \)
<=>\(2\frac{1}{2}x =\frac{5}{3} \)
<=>\(\frac{5}{2}x=\frac{5}{3} \)
<=>\(x=\frac{5}{3}:\frac{5}{2} \)
<=>\(x=\frac{2}{3} \)
LM
7 tháng 12 2017
bạn giải thích rõ chỗ :\(\left(2\dfrac{1}{2}x-1\right)\times\left(1-2\dfrac{1}{2}x\right)=-\left(2\dfrac{1}{2}x-1\right)^2\)hộ mik với
AH
Akai Haruma
Giáo viên
8 tháng 5 2023
Lời giải:
PT $\Leftrightarrow (\frac{x+1}{2022}+1)+(\frac{x+2}{2021}+1)+...+(\frac{x+23}{2000}+1)=0$
$\Leftrightarrow \frac{x+2023}{2022}+\frac{x+2023}{2021}+...+\frac{x+2023}{2000}=0$
$\Leftrightarrow (x+2023)(\frac{1}{2022}+\frac{1}{2021}+...+\frac{1}{2000})=0$
Dễ thấy tổng trong () luôn dương
$\Rightarrow x+2023=0$
$\Leftrightarrow x=-2023$
AH
Akai Haruma
Giáo viên
30 tháng 7 2023
Lời giải:
a. $3(x-\frac{1}{2})-3(x-\frac{1}{3})=x$
$3[(x-\frac{1}{2})-(x-\frac{1}{3})]=x$
$3.\frac{-1}{6}=x$
$\Rightarrow x=\frac{-1}{2}$
b.
$\frac{1}{2}(x+2)-4(x-\frac{1}{4})=\frac{1}{2}x$
$\frac{1}{2}x+1-4(x-\frac{1}{4})=\frac{1}{2}x$
$1-4(x-\frac{1}{4})=0$
$x-\frac{1}{4}=\frac{1}{4}$
$x=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$
13 tháng 11 2017
a) \(\dfrac{x^2-y^2}{x^2-y^2+xz-yz}=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)+z\left(x-y\right)}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x-y\right)\left(x+y+z\right)}=\dfrac{x+y}{x+y+z}\)
b) \(\dfrac{x^2+y^2-z^2+2xy}{x^2+z^2-y^2-2xz}=\dfrac{\left(x+y\right)^2-z^2}{\left(x-z\right)^2-y^2}=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{\left(x-y-z\right)\left(x-z+y\right)}\)\(=\dfrac{x+y+z}{x-y-z}\)
c) \(\dfrac{x^2\left(x-3\right)-\left(x-3\right)}{x\left(x-3\right)}=\dfrac{\left(x-3\right)\left(x^2-1\right)}{x\left(x-3\right)}=\dfrac{x^2-1}{x}\)
d) \(\dfrac{4x^2\left(x-2\right)+3\left(x-2\right)}{4x^2\left(3x+1\right)+3\left(3x+1\right)}=\dfrac{\left(x-2\right)\left(4x^2+3\right)}{\left(3x+1\right)\left(4x^2+3\right)}=\dfrac{x-2}{3x+1}\)
12 tháng 9 2021
\(a,M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\sqrt{x}+x}\right)\left(x>0;x\ne1\right)\\ M=\dfrac{x+\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2\sqrt{x}+2-2+x}{x\left(\sqrt{x}+1\right)}\\ M=\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\\ M=\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(b,M=-\dfrac{1}{2}\Leftrightarrow\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=-\dfrac{1}{2}\\ \Leftrightarrow-4x=x+\sqrt{x}-2\\ \Leftrightarrow5x+\sqrt{x}-2=0\)
Đặt \(\sqrt{x}=t\)
\(\Leftrightarrow5t^2+t-2=0\\ \Delta=1^2-4\cdot5\left(-2\right)=41\\ \Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-1-\sqrt{41}}{10}\\t=\dfrac{-1+\sqrt{41}}{10}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\left(1+\sqrt{41}\right)^2}{100}=\dfrac{-42-2\sqrt{41}}{100}\\x=\dfrac{\left(\sqrt{41}-1\right)^2}{100}=\dfrac{42-2\sqrt{41}}{100}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-21-\sqrt{41}}{50}\left(L\right)\\x=\dfrac{21-\sqrt{41}}{50}\left(N\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{21-\sqrt{41}}{50}\)
12 tháng 9 2021
a: Ta có: \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{2}{x}+\dfrac{x-2}{x\sqrt{x}+x}\right)\)
\(=\dfrac{x+\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2\sqrt{x}+2+x-2}{x\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x}{\sqrt{x}-1}\cdot\dfrac{x}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2x\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
LH
2
14 tháng 6 2018
\(\left(5x-1\right)\left(2x-\dfrac{1}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy: ...............
KH
14 tháng 6 2018
* Trả lời:
\(\left(5x-1\right).\left(2x-\dfrac{1}{3}\right)=0\)
\(\Rightarrow5x-1=0\) hoặc \(2x-\dfrac{1}{3}=0\)
TH1: \(5x-1=0\)
\(\Rightarrow5x=1\)
\(\Rightarrow x=\dfrac{1}{5}\)
TH2: \(2x-\dfrac{1}{3}=0\)
\(\Rightarrow2x=\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{1}{6}\)
Vậy nghiệm của đa thức \(\left(5x-1\right).\left(2x-\dfrac{1}{3}\right)\) là \(x=\dfrac{1}{6};x=\dfrac{1}{5}\)
TH
19 tháng 12 2018
a) \(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{36}{x^2-9}\)
\(\Rightarrow\dfrac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow\dfrac{\left(x+3\right)^2-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow\left(x+3\right)^2-\left(x-3\right)^2=36\)
\(\Rightarrow\left(x^2+6x+9\right)-\left(x^2-6x+9\right)=36\)
\(\Rightarrow x^2+6x+9-x^2+6x-9=36\)
\(\Rightarrow12x=36\)
\(\Rightarrow x=\dfrac{36}{12}\)
Vậy x = 3
b) \(x^2-x-6=0\)
\(\Rightarrow x^2-3x+2x-6=0\)
\(\Rightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
c) \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+17}{15}\)
\(\Rightarrow\dfrac{3\left(2x-1\right)}{15}-\dfrac{5\left(x-2\right)}{15}=\dfrac{x+17}{15}\)
\(\Rightarrow\dfrac{3\left(2x-1\right)-5\left(x-2\right)}{15}=\dfrac{x+17}{15}\)
\(\Rightarrow\dfrac{6x-3-5x+10}{15}=\dfrac{x+17}{15}\)
... Phần còn lại cũng tương tự như vậy thôi
Đề yêu cầu gì em?
Lời giải:
Điều kiện: $x\neq 0; -1$
$\frac{x+3}{x+1}-2=\frac{1-x}{x}$
$1+\frac{2}{x+1}-2=\frac{1}{x}-1$
$\frac{2}{x+1}-1=\frac{1}{x}-1$
$\frac{2}{x+1}=\frac{1}{x}$
$\Rightarrow 2x=x+1$
$\Leftrightarrow x=1$ (thỏa mãn)