`P=((3+x)/(3-x)-(3-x)/(3+x)+(4x^2)/(x^2-9)):((2x+1)/(x+3)-1)`

`=((4x^2-(3-x)^2-(3+x)^2)/(x^2-9)):((2x+1-x-3)/(x+3))`

`=((4x^2-x^2+6x-9-x^2-6x-9)/(x^2-9)):((x-2)/(x+3))`

`=((2x^2-18)/(x^2-9))*(x+3)/(x-2)`

`=((2(x^2-9))/(x^2-9))*(x+3)/(x-2)`

`=(2x+6)/(x-2)`

ĐKXĐ: \(x\ne\pm3;x\ne-\dfrac{1}{2};x\ne2\)

\(P=\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{4x^2}{\left(3-x\right)\left(3+x\right)}\right):\dfrac{2x+1-x-3}{x+3}\)

\(=\dfrac{\left(3+x\right)^2-\left(3-x\right)^2-4x^2}{\left(3+x\right)\left(3-x\right)}:\dfrac{x-2}{x+3}\)

\(=\dfrac{\left(3+x-3+x\right)\left(3+x+3-x\right)-4x^2}{\left(x+3\right)\left(3-x\right)}.\dfrac{x+3}{x-2}\)

\(=\dfrac{12x-4x^2}{3-x}\cdot\dfrac{1}{x-2}\)

\(=\dfrac{4x\left(3-x\right)}{3-x}\cdot\dfrac{1}{x-2}\) \(=\dfrac{4x}{x-2}\)

Chắc mình làm tắt quá để mình làm lại bước biến đổi.

`P=((3+x)/(3-x)-(3-x)/(3+x)+(4x^2)/(x^2-9)):((2x+1)/(x+3)-1)`

`=((x-3)/(x+3)+(4x^2)/(x^2-9)-(x+3)/(x-3)):((2x+1-x-3)/(x+3))`

`=((x-3)^2/(x^2-9)+(4x^2)/(x^2-9)-(x+3)^2/(x^2-9)):((x-2)/(x+3))`

`=(((x-3)^2+4x^2-(x+3)^2)/(x^2-9))*(x+3)/(x-2)`

`=(x^2-6x+9+4x^2-x^2-6x-9)/(x^2-9)*(x+3)/(x-2)`

`=(4x^2-12x)/(x^2-9)*(x+3)/(x-2)`

`=(4x(x-3))/((x-3)(x+3))*(x+3)/(x-2)`

`=(4x)/(x+3)*(x+3)/(x-2)`

`=(4x)/(x-2)`

\(=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}-\dfrac{132}{132}-\dfrac{84}{132}\right)\)

\(=\dfrac{115}{-161}=-\dfrac{115}{161}\)

a, \(\dfrac{5}{7}+\left(\dfrac{3}{5}+\dfrac{-5}{7}\right)\)
\(=\dfrac{5}{7}+\dfrac{-5}{7}+\dfrac{3}{5} =0+\dfrac{3}{5}=\dfrac{3}{5}\)

b, \(=\dfrac{-3}{4}-\dfrac{15}{14}:\dfrac{-5}{7}+\left(-1\right)^2=\dfrac{-3}{4}-\dfrac{-3}{2}+1=\dfrac{-3}{4}-\dfrac{-6}{4}+1=\dfrac{3}{4}+1=\dfrac{7}{4}\)

 c, \(\dfrac{-5}{9}+\left(\dfrac{-2}{3}\right)^2.\left(20\%-1.2\right)=\dfrac{-5}{9}+\dfrac{4}{9}x\left(\dfrac{1}{5}-\dfrac{6}{5}\right)=\dfrac{-5}{9}+\dfrac{4}{9}x\left(-1\right)=\dfrac{-5}{9}+\dfrac{-4}{9}=-1\)

Bài 1:

a) \(\dfrac{5}{7}+\left(\dfrac{3}{5}+\dfrac{-5}{7}\right)\)\(=\left(\dfrac{5}{7}+\dfrac{-5}{7}\right)+\dfrac{3}{5}\)\(=0+\dfrac{3}{5}=\dfrac{3}{5}\)

b) \(\dfrac{-3}{4}-\dfrac{15}{14}:\dfrac{-5}{7}+\left(-1\right)^2\)\(=\dfrac{-3}{4}-\dfrac{15}{14}:\dfrac{-5}{7}+1\)\(=\dfrac{-3}{4}-\dfrac{-3}{2}+1\)

                                            \(=\dfrac{3}{4}+1\)\(=\dfrac{7}{4}\)

\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)

x = nhân ạ

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\)

\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\times\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\times\left(\dfrac{4}{4}-\dfrac{1}{4}\right)\times\left(\dfrac{5}{5}-\dfrac{1}{5}\right)\)

\(=\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\)

\(=\dfrac{1\times2\times3\times4}{2\times3\times4\times5}\)

\(=\dfrac{1}{5}\)

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\geq \frac{9}{a+b+b+c+c+a}=\frac{9}{2(a+b+c)}\)

\(\Rightarrow \text{VT}\geq \frac{9(a^2+b^2+c^2)}{2(a+b+c)}\) \((1)\)

Theo hệ quả của BĐT Am-Gm: \(a^2+b^2+c^2\geq ab+bc+ac\)

\(\Rightarrow 3(a^2+b^2+c^2)\geq (a+b+c)^2\) \((2)\)

Từ \((1),(2)\Rightarrow \text{VT}\geq \frac{3}{2}(a+b+c)\) (đpcm)

Dấu bằng xảy ra khi \(a=b=c\)

\(\dfrac{\dfrac{4}{5}:\left(\dfrac{4}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{27}{25}-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(\dfrac{59}{9}-\dfrac{13}{4}\right)\cdot\dfrac{36}{17}}+\left(\dfrac{6}{5}\cdot\dfrac{1}{2}\right):\dfrac{4}{5}\)

\(=\dfrac{4}{5}:\dfrac{3}{5}+\dfrac{7}{4}:7+\dfrac{3}{5}:\dfrac{4}{5}\)

\(=\dfrac{4}{3}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\dfrac{7}{3}\)

1) Ta có 

\(C=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2022}\right)\)

\(C=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2021}{2022}\)

\(C=\dfrac{1}{2022}\)

2) \(A=\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\)

\(\Rightarrow3A=1-\dfrac{2}{3}+\dfrac{3}{3^2}-\dfrac{4}{3^3}+...+\dfrac{99}{3^{98}}-\dfrac{100}{3^{99}}\)

\(\Rightarrow4A=A+3A\) \(=1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)

\(\Rightarrow12A=3.4A=3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\)

\(\Rightarrow16A=12A+4A=\left(3-1+\dfrac{1}{3}-\dfrac{1}{3^2}+...-\dfrac{1}{3^{98}}-\dfrac{100}{3^{99}}\right)+\left(1-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...-\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\right)\)

\(=3-\dfrac{101}{3^{99}}-\dfrac{100}{3^{100}}\) \(< 3\). Từ đó suy ra \(A< \dfrac{3}{16}\)

\(\left(-\dfrac{2}{5}\right)^2\cdot\left|\dfrac{1}{3}-\dfrac{3}{5}\right|-\dfrac{2}{5}\cdot\sqrt{\dfrac{1}{25}}+\dfrac{4}{3}\)

\(=\dfrac{4}{25}\cdot\dfrac{4}{15}-\dfrac{2}{5}\cdot\dfrac{1}{5}+\dfrac{4}{3}\)

\(=\dfrac{16}{375}-\dfrac{2}{25}+\dfrac{4}{3}\)

\(=\dfrac{16}{375}-\dfrac{30}{375}+\dfrac{500}{375}\)

\(=\dfrac{486}{375}=\dfrac{162}{125}\)

`(2/3 x +1/2) (-2x+3)=0`

\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x+\dfrac{1}{2}=0\\-2x+3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=-\dfrac{1}{2}\\-2x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}.\dfrac{3}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\)

\(\left(\dfrac{2}{3}x+\dfrac{1}{2}\right)\cdot\left(-2x+3\right)=0\\ =>\left[{}\begin{matrix}\dfrac{2}{3}x+\dfrac{1}{2}=0\\-2x+3=0\end{matrix}\right.\\ =>\left[{}\begin{matrix}\dfrac{2}{3}x=-\dfrac{1}{2}\\-2x=-3\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\)