a: Số cần tìm là 5,32:0,125=42,56

b: \(A=1+\dfrac{1}{2019}-1-\dfrac{1}{2018}+\dfrac{1}{2018}-\dfrac{1}{2019}=0\)

Đặt \(2017=a\)

\(A=\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1\right)^2-2a+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1\right)^2-2\left(a+1\right)\cdot\dfrac{a}{a+1}+\left(\dfrac{a}{a+1}\right)^2}+\dfrac{a}{a+1}\\ A=\sqrt{\left(a+1-\dfrac{a}{a+1}\right)^2}+\dfrac{a}{a+1}\\ A=\left|a+1-\dfrac{a}{a+1}\right|+\dfrac{a}{a+1}\)

Ta có \(\dfrac{a}{a+1}< 1\Leftrightarrow a+1-\dfrac{a}{a+1}>0\)

\(\Leftrightarrow A=a+1-\dfrac{a}{a+1}+\dfrac{a}{a+1}=a+1=2018\)

\(E=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{2}+\dfrac{1}{12}\)

\(E=\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{24}\right)\)

\(E=\dfrac{2}{2}+\dfrac{3}{6}+\left(\dfrac{1}{8}+\dfrac{3}{24}\right)\)

\(E=1+\dfrac{1}{2}+\left(\dfrac{1}{8}+\dfrac{1}{8}\right)\)

\(E=\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\dfrac{2}{8}\)

\(E=\dfrac{3}{2}+\dfrac{1}{4}\)

\(E=\dfrac{6}{4}+\dfrac{1}{4}\)

\(E=\dfrac{7}{4}\)

a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)

=1-2/4=1/2

b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)

c: x-y=0 nên x=y

\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)

=2019

\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\left(đk:x\ge0,x\ne1\right)\)

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

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

\(=\dfrac{\left(\sqrt{x}-1\right)^2.2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)

Để A nguyên thì: \(x+\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Mà \(x+\sqrt{x}+1=\left(x+\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(\Rightarrow x+\sqrt{x}+1\in\left\{1;2\right\}\)

+ Với \(x+\sqrt{x}+1=1\)

\(\Leftrightarrow\sqrt[]{x}\left(\sqrt{x}+1\right)=0\)

\(\Leftrightarrow x=0\left(tm\right)\left(do.\sqrt{x}+1\ge1>0\right)\)

+ Với \(x+\sqrt{x}+1=2\)

\(\Leftrightarrow\left(x+\sqrt{x}+\dfrac{1}{4}\right)=\dfrac{5}{4}\)

\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+\dfrac{1}{2}=\dfrac{\sqrt{5}}{2}\\\sqrt{x}+\dfrac{1}{2}=-\dfrac{\sqrt{5}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{5}-1}{2}\\\sqrt{x}=-\dfrac{\sqrt{5}+1}{2}\left(VLý\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{3-\sqrt{5}}{2}\left(tm\right)\)

Vậy \(S=\left\{1;\dfrac{3-\sqrt{5}}{2}\right\}\)

\(=\dfrac{2}{3}+\dfrac{1}{3}.\left(\dfrac{7}{18}\right):\dfrac{7}{12}\)

\(=\dfrac{2}{3}+\dfrac{7}{54}:\dfrac{7}{12}\)

\(=\dfrac{2}{3}+\dfrac{2}{9}\)

\(=\dfrac{8}{9}\)

Cách 1: Tính giá trị từng biểu thức trong ngoặc

A=

Cách 2: Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp

A =

= (6-5-3) -

= -2 -0 - = - (2 + ) = -2

Lời giải:

Cách 1: Tính giá trị từng biểu thức trong ngoặc

A=

Cách 2: Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp

A =

= (6-5-3) -

= -2 -0 - = - (2 + ) = -2

cách 1:

A = \(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

= \(\left(\dfrac{6}{1}-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(\dfrac{5}{1}+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(\dfrac{3}{1}-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

= \(\left(\dfrac{18}{3}-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(\dfrac{15}{3}+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(\dfrac{9}{3}-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

= \(\left(\dfrac{16}{3}+\dfrac{1}{2}\right)-\left(\dfrac{20}{3}-\dfrac{3}{2}\right)-\left(\dfrac{2}{3}+\dfrac{5}{2}\right)\)

= \(\left(\dfrac{32}{6}+\dfrac{3}{6}\right)-\left(\dfrac{40}{6}-\dfrac{9}{6}\right)-\left(\dfrac{4}{6}+\dfrac{15}{6}\right)\)

= \(\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}\)

= \(-\dfrac{15}{6}=-\dfrac{5}{2}\)

cách 2:

A = \(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

= \(\left(6-5-3\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}-\dfrac{7}{3}\right)-\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)

= \(\left(-2\right)-0+\dfrac{1}{2}\)

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