\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\)

Dễ thấy: \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\Rightarrow x+2004=0\Leftrightarrow x=-2014\)

x = -2014

ti-ck nha

.........

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\left(x+1\right)=0\)

\(\Rightarrow x+1=0\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\right)\)

\(x=-1\)

Vậy \(x=-1\)

a)\(-\left(\frac{2}{5}+x\right)=\frac{2}{3}-\frac{11}{12}\)

\(-\left(\frac{2}{5}+x\right)=\frac{-1}{4}\)

\(\frac{-2}{5}-x=\frac{-1}{4}\)

\(-x=\frac{-1}{4}+\frac{2}{5}\)

\(-x=\frac{3}{20}\)

\(x=\frac{-3}{20}\)

Vậy...

b)\(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}\)

\(\frac{1}{4}:x=\frac{-7}{20}\)

\(x=\frac{1}{4}:\left(\frac{-7}{20}\right)\)

\(x=\frac{-5}{7}\)

Vậy...

tk mk nhoaa bn

a) \(4\sqrt{x}+\frac{2}{\sqrt{x}}< 2x+\frac{1}{2x}+2\)

hay \(2\sqrt{x}+\frac{1}{\sqrt{x}}< x+\frac{1}{4x}+1\)

\(\Leftrightarrow0< x+\frac{1}{4x}+1-2\sqrt{x}-\frac{1}{\sqrt{x}}\)

\(\Leftrightarrow0< \left(\sqrt{x}\right)^2-2\sqrt{x}-2\sqrt{x}\cdot1+1+\frac{1}{\left(2\sqrt{x}\right)^2}-2\cdot\frac{1}{2\sqrt{x}}\)

\(\Leftrightarrow1< \left(\sqrt{x}-1\right)^2+\left(\frac{1}{2\sqrt{x}}-1\right)^2\)

\(\Rightarrow\hept{\begin{cases}x>0\\\sqrt{x}>1\\2\sqrt{x}>1\end{cases}\Rightarrow\hept{\begin{cases}x>1\\x>\frac{1}{4}\end{cases}\Rightarrow}x>1}\)

b) \(\frac{1}{1-x^2}>\frac{3}{\sqrt{1-x^2}}-1\left(1\right)\left(ĐK:-1< x< 1\right)\)

Ta có (1) <=> \(\frac{1}{1-x^2}-1-\frac{3x}{\sqrt{1-x^2}}+2>0\)\(\Leftrightarrow\frac{x^2}{1-x^2}-\frac{3x}{\sqrt{1-x^2}}+2>0\)

Đặt \(t=\frac{x}{\sqrt{1-x^2}}\)ta được

\(t^2-3t+2>0\Leftrightarrow\orbr{\begin{cases}\frac{x}{\sqrt{1-x^2}}< 1\\\frac{x}{\sqrt{1-x^2}}>2\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{1-x^2}>x\left(a\right)\\2\sqrt{1-x^2}< x\left(b\right)\end{cases}}}\)

(a) <=> \(\hept{\begin{cases}x< 0\\1-x^2>0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge0\\1-x^2>x^2\end{cases}}}\)

\(\Leftrightarrow-1< x< 0\)hoặc \(\hept{\begin{cases}x\ge0\\x^2< \frac{1}{2}\end{cases}}\)

\(\Leftrightarrow-1< x< 0\)hoặc \(0\le x\le\frac{\sqrt{2}}{2}\Leftrightarrow-1< x< \frac{\sqrt{2}}{2}\)

(b) \(\Leftrightarrow\hept{\begin{cases}1-x^2>0\\x>0\\4\left(1-x^2\right)< x^2\end{cases}\Leftrightarrow\hept{\begin{cases}0< x< 1\\x^2>\frac{4}{5}\end{cases}\Leftrightarrow}\frac{2}{\sqrt{5}}< x< 1}\)

ok đợi nấu ăn xong r làm cho

a) điều kiện x>0

khi đó

\(\left(a\right)\Leftrightarrow4\left(\sqrt{4}+\frac{1}{2\sqrt{x}}\right)< 2\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)^2\)

\(\Leftrightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}>2\Leftrightarrow2x-4\sqrt{x}+1>0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}< \frac{2-\sqrt{2}}{2}\\\sqrt{x}>\frac{2+\sqrt{2}}{2}\end{cases}}\)

a, x=-5/33

b, x=-30

c, (Nếu dấu của bạn là giá trị tuyệt đối)      x=1/3 hoặc x=-1/3

Nếu cần mik giải rõ ràng cho

a)Ta có: \(\frac{-2}{5}+\frac{6}{5}.\left(y-\frac{2}{3}\right)=\frac{-4}{15}\)

\(\Rightarrow\frac{6}{5}.\left(y-\frac{2}{3}\right)=\frac{-4}{15}-\frac{-2}{15}\)

\(\Rightarrow\frac{6}{5}.\left(y-\frac{2}{3}=\right)\frac{-2}{5}\)

\(\Rightarrow y-\frac{2}{3}=\frac{-2}{5}:\frac{6}{5}=\frac{-1}{3}\)

\(\Rightarrow y=\frac{-1}{3}+\frac{2}{3}=\frac{1}{3}\)

Vậy x = \(\frac{1}{3}\)

b) Ta có: \(\frac{-2}{5}+\frac{2}{3}x+\frac{1}{6}x=\frac{-4}{15}\)

        \(\Rightarrow\frac{-2}{5}+x.\left(\frac{2}{3}+\frac{1}{6}\right)=\frac{-4}{15}\)

        \(\Rightarrow x.\frac{5}{6}=\frac{-4}{15}-\frac{-2}{15}\)

         \(x.\frac{5}{6}=\frac{-2}{15}\)

\(\Rightarrow x=\frac{-2}{15}:\frac{5}{6}=\frac{-4}{25}\)

Vậy x = \(\frac{-4}{25}\)

c) Ta có: \(\frac{3}{2}x+\frac{-2}{5}-\frac{2}{3}.x=\frac{-4}{15}\)

\(\Rightarrow\frac{3}{2}x-\frac{2}{3}x+\frac{-2}{5}=\frac{-4}{15}\)

\(\Rightarrow x.\left(\frac{3}{2}-\frac{2}{4}\right)=\frac{-4}{15}-\frac{-2}{15}\)

\(\Rightarrow x.\frac{5}{6}=\frac{-2}{15}\)

\(\Rightarrow x=\frac{-2}{15}:\frac{5}{6}=\frac{-4}{25}\)

Vậy x = \(\frac{-4}{25}\)

Ủng hộ tớ nha m.n

a) \(ĐKXĐ:\hept{\begin{cases}x\ne3\\x\ne\pm2\end{cases}}\)

b) \(D=\left(\frac{2+x}{2-x}-\frac{2-x}{2+x}-\frac{4x^2}{x^2-4}\right)\div\left(\frac{x-3}{2-x}\right)\)

\(\Leftrightarrow D=\frac{\left(2+x\right)^2-\left(2-x\right)^2+4x^2}{\left(2-x\right)\left(2+x\right)}\cdot\frac{2-x}{x-3}\)

\(\Leftrightarrow D=\frac{4+4x+x^2-4+4x-x^2+4x^2}{\left(2+x\right)\left(x-3\right)}\)

\(\Leftrightarrow D=\frac{4x^2+8x}{\left(x+2\right)\left(x-3\right)}\)

\(\Leftrightarrow D=\frac{4x}{x-3}\)

c) Để D = 0

\(\Leftrightarrow\frac{4x}{x-3}=0\)

\(\Leftrightarrow4x=0\)

\(\Leftrightarrow x=0\)

Vậy để D = 0 \(\Leftrightarrow\)x = 0

d) Khi \(\left|2x-1\right|=5\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=5\\1-2x=5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=6\\2x=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\left(ktm\right)\\x=-2\left(ktm\right)\end{cases}}\)

Vậy khi \(\left|2x-1\right|=5\Leftrightarrow D\in\varnothing\)

a)\(-\frac{2}{5}+\frac{2}{3}x+\frac{1}{6}x=-\frac{4}{5}\Leftrightarrow\frac{5}{6}x=-\frac{2}{5}\Leftrightarrow x=-\frac{12}{25}\)
Vậy nghiệm là x = -12/25

b)\(\frac{3}{2}x-\frac{2}{5}-\frac{2}{3}x=-\frac{4}{15}\Leftrightarrow\frac{5}{6}x=\frac{2}{15}\Leftrightarrow x=\frac{4}{25}\)
Vậy nghiệm là x = 4/25

c)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\right)\)\(\Leftrightarrow x=-1\)
Vậy nghiệm là x = -1
 

Cảm ơn bạnh nha. Chúc bạn buổi tối ấm =)))) <3