a) \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}=\frac{x-4}{5}+\frac{x-5}{6}\)

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

\(\frac{x-1}{2}+\frac{x-1}{3}+\frac{x-1}{4}=\frac{x-1}{5}+\frac{x-1}{6}\)

\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0

\(x-1=0\)

\(x=1\)

Ai làm đc câu nào thì làm giúp mình với ạ, cảm ơn trc:(((

\(1,3x-5x+5=-8\)

\(\Leftrightarrow-2x+5+8=0\)

\(\Leftrightarrow-2x=-13\)

\(\Leftrightarrow x=\frac{13}{2}\)

\(Đkxđ:x\ge0\)

Ta có: Bất phương trình tương đương với:

\(\left(1+\sqrt{x}\right)\left(\frac{1}{\sqrt{x+3}}+\frac{1}{\sqrt{3x+1}}\right)=2\)

Áp dụng BĐT Cô - si ta có:

\(\frac{1}{\sqrt{3x+1}}=\sqrt{\frac{1}{x+1}.\frac{x+1}{3x+1}}\le\frac{1}{2}\left(\frac{1}{x+1}+\frac{x+1}{3x+1}\right)\)

\(\sqrt{\frac{x}{3x+1}}=\sqrt{\frac{1}{2}.\frac{2x}{3x+1}}\le\frac{1}{2}\left(\frac{1}{2}+\frac{2x}{3x+1}\right)\)

\(\Rightarrow\frac{1+\sqrt{x}}{\sqrt{3x+1}}\le\frac{1}{2}\left(\frac{1}{x+1}+\frac{1}{2}+1\right)=\frac{1}{2}\left(\frac{1}{x+1}+\frac{3}{2}\right)\left(1\right)\)

\(\frac{1}{\sqrt{x+3}}=\sqrt{\frac{1}{2}.\frac{2}{x+3}}\le\frac{1}{2}\left(\frac{1}{2}+\frac{2}{x+3}\right)\)

\(\frac{\sqrt{x}}{\sqrt{x+3}}=\sqrt{\frac{x}{x+1}.\frac{x+1}{x+3}}\le\frac{1}{2}\left(\frac{x}{x+1}+\frac{x+1}{x+3}\right)\)

\(\Rightarrow\frac{1+\sqrt{x}}{\sqrt{x+3}}\le\frac{1}{2}\left(\frac{x}{x+1}+\frac{3}{2}\right)\left(2\right)\)

Từ: \(\left(1\right)\left(2\right)\Rightarrow\left(1+\sqrt{x}\right)\left(\frac{1}{\sqrt{x+3}}+\frac{1}{\sqrt{3x+1}}\right)\le\frac{1}{2}\left(\frac{1}{x+1}+\frac{x}{x+1}+3\right)=2\)

Đẳng thức xảy ra \(\Leftrightarrow x=1\)

Vậy nghiệm của pt là \(x=1\)

Theo đề bài ta có: \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}-\frac{x-4}{5}-\frac{x-5}{6}>0\)

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

<=> \(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}>1\)

<=>\(\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)>1\)

<=> \(\left(x+1\right)\cdot\frac{43}{60}>1\)

<=>\(x+1>\frac{60}{43}\)

<=> x>\(\frac{17}{43}\)

Vậy x>17/43

\(a)\) \(3-2x>4x+5\)

\(\Leftrightarrow\)\(3-2x+2x>4x+2x+5\)

\(\Leftrightarrow\)\(6x+5< 3\)

\(\Leftrightarrow\)\(6x+5-5< 3-5\)

\(\Leftrightarrow\)\(6x< -2\)

\(\Leftrightarrow\)\(\frac{6x}{6}< \frac{-2}{6}\)

\(\Leftrightarrow\)\(x< \frac{-1}{3}\)

Vậy \(x< \frac{-1}{3}\)

Chúc bạn học tốt ~ 

1)

a) \(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}< =>\frac{2\left(x+5\right)}{2\left(3x-6\right)}-\frac{3x-6}{2\left(3x-6\right)}=\frac{3\left(2x-3\right)}{3\left(2x-4\right)}.\)

(đk:x khác \(\frac{1}{2}\))

\(\frac{2x+10}{6x-12}-\frac{3x-6}{6x-12}=\frac{6x-9}{6x-12}< =>2x+10-3x+6=6x-9< =>x=\frac{25}{7}\)

Vậy x=\(\frac{25}{7}\)

b) /7-2x/=x-3 \(x\ge\frac{7}{2}\)

(đk \(x\ge3,\frac{7}{2}< =>x\ge\frac{7}{2}\))

\(\Rightarrow\orbr{\begin{cases}7-2x=x-3\\7-2x=-\left(x-3\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\left(< \frac{7}{2}\Rightarrow l\right)\\x=4\left(tm\right)\end{cases}}}\)

Vậy x=4

2)

\(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}>\frac{x-4}{5}+\frac{x-5}{6}\)

\(\Leftrightarrow\frac{30\left(x-1\right)}{60}+\frac{20\left(x-2\right)}{60}+\frac{15\left(x-3\right)}{60}-\frac{12\left(x-4\right)}{60}-\frac{10\left(x-5\right)}{60}>0\)

\(\Leftrightarrow30x-30+20x-40+15x-45-12x+48-10x+50>0\Leftrightarrow43x-17>0\Leftrightarrow x>\frac{17}{43}\)