Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)

\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)

CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)

Cộng vế theo vế:

\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)

Dấu \("="\Leftrightarrow a=b=c=2\)

a: \(\Leftrightarrow-\dfrac{2}{3}\cdot\dfrac{4-6-9}{12}\ge x\ge-\dfrac{13}{3}\cdot\dfrac{3-1}{6}\)

\(\Leftrightarrow-\dfrac{2}{3}\cdot\dfrac{-11}{12}\ge x\ge\dfrac{-13}{3}\cdot\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{22}{36}\ge x\ge\dfrac{-13}{9}\)

mà x là số nguyên

nên \(x\in\left\{0;-1\right\}\)

b: \(\Leftrightarrow\dfrac{21}{100}+\dfrac{75}{100}-\dfrac{220}{100}>=2x-1>=-3-\dfrac{1}{2}+3+\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{-124}{100}\ge2x-1\ge\dfrac{-3}{10}\)

\(\Leftrightarrow-\dfrac{124}{100}+1\ge2x>=\dfrac{-3}{10}+1\)

\(\Leftrightarrow\dfrac{-3}{25}\ge2x\ge\dfrac{7}{10}\)(vô lý)

=>x không có giá trị

c: \(\Leftrightarrow43+\dfrac{1}{2}-39-\dfrac{1}{5}\le-3x+4\le9+\dfrac{1}{5}+50+\dfrac{1}{7}\)

\(\Leftrightarrow3+\dfrac{3}{10}\le-3x+4\le59+\dfrac{12}{35}\)

\(\Leftrightarrow\dfrac{33}{10}-4\le-3x\le59+\dfrac{12}{35}-4\)

\(\Leftrightarrow\dfrac{-7}{10}\le-3x\le\dfrac{1937}{35}\)

\(\Leftrightarrow\dfrac{7}{30}\ge x\ge-\dfrac{1937}{105}\)

mà x là số nguyên

nên \(x\in\left\{0;-1;-2;...;-18\right\}\)

a.\(\dfrac{1}{4}\left(x-1\right)\ge\dfrac{x-4}{6}\)

\(\Leftrightarrow\) \(\dfrac{1}{4}\left(x-1\right)12\ge\dfrac{x-4}{6}12\)

\(\Leftrightarrow3\left(x-1\right)\ge2\left(x-4\right)\)

\(\Leftrightarrow3x-3\ge2x-8\)

\(\Leftrightarrow3x-2x\ge-8+3\)

\(\Leftrightarrow x\ge-5\)

b.\(\dfrac{x-2}{4}\ge\dfrac{1}{3}\left(x-3\right)\)

\(\Leftrightarrow\dfrac{x-2}{4}12\ge\dfrac{1}{3}\left(x-3\right)12\)

\(\Leftrightarrow3\left(x-2\right)\ge4\left(x-3\right)\)

\(\Leftrightarrow3x-6\ge4x-12\)

\(\Leftrightarrow3x-4x\ge-12+6\)

\(\Leftrightarrow x\ge-6\)

`a, 2/3 +3/4 = (8+9)/12=17/12.`

`1 1/3+4/5 = 4/3 + 4/5 = (20+12)/15=32/15`.

`=> x=2.`

`b, 5/6-1/4=(20-6)/24=7/12`.

`2 1/3-2/5= 7/3-2/5 = (35-6)/15=29/15`.

`=> x=1`.

a) \(\dfrac{2}{3}+\dfrac{3}{4}=\dfrac{8+9}{12}=\dfrac{17}{12}\)

-> 1 1/3 + 4/5 = 4/3 + 4/5 =  20+12/15 = 32/15

vậy x có thể = 14/14 = 1 (x thuộc N)

a: \(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)

\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+1}\cdot\dfrac{2}{\sqrt{x}+3}=-\dfrac{6}{\sqrt{x}+3}\)

b: P>=-1/2

=>P+1/2>=0

=>\(\dfrac{-6}{\sqrt{x}+3}+\dfrac{1}{2}>=0\)

=>\(\dfrac{-12+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}>=0\)

=>căn x-9>=0

=>x>=81

c: căn x+3>=3

=>6/căn x+3<=6/3=2

=>-6/căn x+3>=-2

Dấu = xảy ra khi x=0

`@` `\text {Ans}`

`\downarrow`

`a)`

\(\dfrac{3}{2}\times\dfrac{4}{5}-x=\dfrac{2}{3}\)

\(\dfrac{6}{5}-x=\dfrac{2}{3}\)

\(x=\dfrac{6}{5}-\dfrac{2}{3}\)

\(x=\dfrac{18}{15}-\dfrac{10}{15}\)

\(x=\dfrac{8}{15}\)

Vậy, `x =`\(\dfrac{8}{15}\)

`b)`

\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)

\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)

\(x=\dfrac{40}{51}\div\dfrac{10}{3}\)

\(x=\dfrac{4}{17}\)

Vậy, \(x=\dfrac{4}{17}\)

`c)`

\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)

\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)

\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)

\(x=\dfrac{34}{7}\)

Vậy, `x = `\(\dfrac{34}{7}\)

a) \(\dfrac{3}{2}x\dfrac{4}{5}-x=\dfrac{2}{3}\Rightarrow\dfrac{6}{5}-x=\dfrac{2}{3}\Rightarrow x=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{18}{15}-\dfrac{10}{15}=\dfrac{8}{15}\)

b) \(x.3\dfrac{1}{3}=3\dfrac{1}{3}:4\dfrac{1}{4}\Rightarrow\dfrac{10}{3}.x=\dfrac{10}{3}:\dfrac{17}{4}\Rightarrow\dfrac{10}{3}.x=\dfrac{10}{3}.\dfrac{4}{17}\Rightarrow x=\dfrac{10}{3}.\dfrac{4}{17}:\dfrac{10}{3}=\dfrac{10}{3}.\dfrac{4}{17}.\dfrac{3}{10}=\dfrac{4}{17}\)

c) \(5\dfrac{2}{3}:x=3\dfrac{2}{3}-2\dfrac{1}{2}\Rightarrow\dfrac{17}{3}:x=\dfrac{11}{3}-\dfrac{5}{2}\Rightarrow\dfrac{17}{3}:x=\dfrac{22}{6}-\dfrac{15}{6}\Rightarrow\dfrac{17}{3}:x=\dfrac{7}{6}\Rightarrow x=\dfrac{17}{3}:\dfrac{7}{6}=\dfrac{17}{3}.\dfrac{7}{6}=\dfrac{119}{18}\)

\(5,\dfrac{4}{x-2}+\dfrac{x}{x+1}-\dfrac{x^2-2}{\left(x-2\right)\left(x+1\right)}=0\left(dkxd:x\ne2;-1\right)\)

\(\Rightarrow4\left(x+1\right)+x\left(x-2\right)-x^2-2=0\)

\(\Rightarrow4x+4+x^2-2x-x^2-2=0\)

\(\Rightarrow2x+2=0\)

\(\Rightarrow x=-1\left(loai\right)\)

Vậy \(S=\varnothing\)

\(4,\dfrac{x}{x-3}-\dfrac{1}{x+2}=0\left(dkxd:x\ne3;-2\right)\)

\(\Rightarrow x\left(x+2\right)-\left(x-3\right)=0\)

\(\Rightarrow x^2+3x-x+3=0\)

\(\Rightarrow x^2+2x+3=0\)

\(\Rightarrow S=\varnothing\)

giúp em tl những câu tl trên vs

c: Ta có: \(\dfrac{1}{3}-\dfrac{7}{8}x=\dfrac{1}{4}\)

\(\Leftrightarrow x\cdot\dfrac{7}{8}=\dfrac{1}{12}\)

\(\Leftrightarrow x=\dfrac{1}{12}\cdot\dfrac{8}{7}=\dfrac{2}{21}\)

d: Ta có: \(\dfrac{3}{2}x+\dfrac{1}{7}=\dfrac{7}{8}\cdot\dfrac{64}{49}\)

\(\Leftrightarrow x\cdot\dfrac{3}{2}=1\)

hay \(x=\dfrac{2}{3}\)