đầy đủ phần ngoặc hay gì đó

\(\dfrac{11}{2}\): \(\dfrac{1}{4}\) \(\times\) \(\dfrac{5}{3}\)

= \(\dfrac{11}{2}\) \(\times\) \(\dfrac{4}{1}\) \(\times\) \(\dfrac{5}{3}\)

= 22 \(\times\) \(\dfrac{5}{3}\)

= \(\dfrac{110}{3}\)

\(\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{5}{3}\)

= \(\dfrac{30}{12}-\dfrac{3}{12}+\dfrac{20}{12}\)

= \(\dfrac{7}{12}\) 

\(\dfrac{14}{5}\times\dfrac{2}{3}\)+ 5

= \(\dfrac{28}{15}\) + 5

= \(\dfrac{28}{15}\) + \(\dfrac{75}{15}\)

= \(\dfrac{103}{15}\)

a) Ta có: \(P=\dfrac{x-2}{x^2-1}-\dfrac{x+2}{x^2+2x+1}\cdot\dfrac{1-x^2}{2}\)

\(=\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\cdot\dfrac{-\left(x-1\right)\left(x+1\right)}{2}\)

\(=\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x+2\right)\left(x-1\right)}{2\left(x+1\right)}\)

\(=\dfrac{2\left(x-2\right)}{2\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)^2\cdot\left(x+2\right)}{2\left(x-1\right)\left(x+1\right)}\)

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

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

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

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

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

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

1. ĐKXĐ: \(x\ne\pm1\)

2. \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x+3}{x+1}\right)\cdot\dfrac{x+1}{2}\)

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

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

\(=\dfrac{6x-2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)

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

\(=\dfrac{x-3}{x-1}\)

3. Tại x = 5, A có giá trị là:

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

4. \(A=\dfrac{x-3}{x-1}\) \(=\dfrac{x-1-3}{x-1}=1-\dfrac{3}{x-1}\)

Để A nguyên => \(3⋮\left(x-1\right)\) hay \(\left(x-1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\left(tmđk\right)\\x=0\left(tmđk\right)\\x=4\left(tmđk\right)\\x=-2\left(tmđk\right)\end{matrix}\right.\)

Vậy: A nguyên khi \(x=\left\{2;0;4;-2\right\}\)

1: ĐKXĐ: x∉{1;-1}

2: \(A=\left(\frac{x+1}{x-1}-\frac{x+3}{x+1}\right)\cdot\frac{x+1}{2}\)

\(=\frac{\left(x+1\right)^2-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x+1}{2}\)

\(=\frac{x^2+2x+1-\left(x^2+2x-3\right)}{x-1}\cdot\frac12=\frac{4}{2\left(x-1\right)}=\frac{2}{x-1}\)

3: THay x=5 vào A, ta được:

\(A=\frac{2}{5-1}=\frac24=\frac12\)

4: Để A là số nguyên thì 2⋮x-1

=>x-1∈{1;-1;2;-2}

=>x∈{2;0;3;-1}

Kết hợp ĐKXĐ, ta được: x∈{0;2;3}

giá trị của biểu thức là: 13/3.      nhớ ủng hộ mk đó

Bằng 5/6 bạn nhé.

`@` `\text {Ans}`

`\downarrow`

`(1/2-1/3+1/4-1/5):(1/4-1/5)`

`=`\(\left(\dfrac{1}{6}+\dfrac{1}{20}\right)\div\dfrac{1}{20}\)

`=`\(\dfrac{1}{20}\div\dfrac{1}{20}+\dfrac{1}{6}\div\dfrac{1}{20}\)

`= 1+10/3`

`= 13/3`

A = (\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)): (\(\dfrac{1}{4}\) - \(\dfrac{1}{5}\))

A = ( \(\dfrac{30}{60}\) - \(\dfrac{20}{60}\) + \(\dfrac{15}{60}\) - \(\dfrac{12}{60}\)):(\(\dfrac{5}{20}\) - \(\dfrac{4}{20}\))

A = \(\dfrac{13}{60}\): \(\dfrac{1}{20}\)

A = \(\dfrac{13}{60}\times\dfrac{20}{1}\) 

A = \(\dfrac{13}{3}\)

Thanks kiuuuu

\(2\dfrac{1}{20}\)

\(\dfrac{13}{18}\)

\(\text{A}=1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{100}{2^{100}}\)

\(\frac{1}{2}.\text{A}=\frac{1}{2}+\frac{3}{2^4}+\frac{4}{2^5}+...+\frac{99}{2^{100}}+\frac{100}{2^{101}}\)

\(=\left[\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right]-\frac{100}{2^{101}}\left(\text{do}\frac{3}{2^3}=\frac{1}{2^2}+\frac{1}{2^3}\right)\)

\(=\frac{\left[1-\left(\frac{1}{2}\right)^{101}\right]}{\left(1-\frac{1}{2}\right)}-\frac{100}{2^{101}}\)

\(=\frac{\left(2^{101}-1\right)}{2^{100}}-\frac{100}{2^{101}}\)

\(\Rightarrow\text{A}=\frac{\left(2^{101}-1\right)}{2^{99}}-\frac{100}{2^{101}}\)

P/s: Sai đâu thì bn sửa nhé.

Bài này là ttoan nâng cao hả bạn

`a)` Thay `x=2` vào `B` có: `B=[-10]/[2-4]=5`

`b)` Với `x ne -1;x ne -5` có:

`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+1)(x+5)]`

`A=[x^2+x+2x+2-5x-1-x-5]/[(x+1)(x+5)]`

`A=[x^2-3x-4]/[(x+1)(x+5)]`

`A=[(x+1)(x-4)]/[(x+1)(x+5)]`

`A=[x-4]/[x+5]`

`c)` Với `x ne -5; x ne -1; x ne 4` có:

`P=A.B=[x-4]/[x+5].[-10]/[x-4]`

           `=[-10]/[x+5]`

Để `P` nguyên `<=>[-10]/[x+5] in ZZ`

    `=>x+5 in Ư_{-10}`

Mà `Ư_{-10}={+-1;+-2;+-5;+-10}`

`=>x={-4;-6;-3;-7;0;-10;5;-15}` (t/m đk)

\(\dfrac{5}{2}+\dfrac{1}{4}.\dfrac{1}{8}\)

\(=\dfrac{5}{2}+\dfrac{1}{32}=\dfrac{5.16+1}{32}=\dfrac{80+1}{32}=\dfrac{81}{32}\)

    \(\dfrac{5}{2}\)  + \(\dfrac{1}{4}\) x \(\dfrac{1}{8}\)

=  \(\dfrac{5}{2}\) + \(\dfrac{1}{32}\)

= \(\dfrac{80}{32}\) +  \(\dfrac{1}{32}\)

= \(\dfrac{81}{32}\)

31/12

= 5/2 + 1/12

= 30/12 + 1/12

= 31/12