`(x^4-1)^2+(x^2+1)^2`

`=x^8-2x^4+1+x^4+2x^2+1`

`=x^8-x^4+2x^2+2`

\(\left(x^4-1\right)^2+\left(x^2+1\right)^2=\left(x^2-1\right)^2.\left(x^2+1\right)^2+\left(x^2+1\right)^2\)

\(=\left(x^2+1\right)^2\left[\left(x^2-1\right)^2+1\right]=\left(x^2+1\right)^2\left(x^4-2x^2+2\right)\)

a) Ta có : A = 5 + 5² + 5³ + ... + 5¹⁰⁰ 

=> 5A = 5² + 5³ + 5⁴ + ... + 5¹⁰¹

=> 5A - A = (5² + 5³ + 5⁴ + ... + 5¹⁰¹) - (5 + 5² + 5³ + ... + 5¹⁰⁰ )

=> 4A = 5¹⁰¹ - 5

=> A = \(\frac{5^{501}-5}{4}\)

b) Ta có B = 1 + 4² + 4⁴ + ... + 4³⁰⁰

=> 4².B = 4² + 4⁴ + 4⁶ + ... + 4³⁰² = 16B

Khi đó 16B - B = (4² + 4⁴ + 4⁶ + ... + 4³⁰²) - (1 + 4² + 4⁴ + ... + 4³⁰⁰)

=> 15B = 4³⁰² - 1

=> B = \(\frac{4^{302}-1}{15}\)

c) Ta có C = 1 + 3² + 3⁴ + ... + 3²⁰²⁰

=> 3²C = 3² + 3⁴ + 3⁶ + ... + 3²⁰²² = 9C

Khi đó 9C - C = (3² + 3⁴ + 3⁶ + ... + 3²⁰²²) - (1 + 3² + 3⁴ + ... + 3²⁰²⁰)

=> 8C = 3²⁰²² - 1

=> C = \(\frac{3^{2022}-1}{8}\)

\(C=\dfrac{-5}{7}+\dfrac{-2}{7}+\dfrac{3}{4}+\dfrac{1}{4}+\dfrac{-1}{5}=-1+1-\dfrac{1}{5}=\dfrac{-1}{5}\)

KHO THE

\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)

\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)

\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)

`@` `\text {Ans}`

`\downarrow`

`a)`

\(\left(\dfrac{7}{8}-\dfrac{3}{4}\right)\cdot1\dfrac{1}{3}-\dfrac{2}{3}\cdot0,5\)

`=`\(\dfrac{1}{8}\cdot\dfrac{4}{3}-\dfrac{1}{3}\)

`=`\(\dfrac{1}{6}-\dfrac{1}{3}=-\dfrac{1}{6}\)

`b)`

\(\left(2+\dfrac{5}{6}\right)\div1\dfrac{1}{5}+\left(-\dfrac{7}{12}\right)\)

`=`\(\dfrac{17}{6}\div1\dfrac{1}{5}-\dfrac{7}{12}\)

`=`\(\dfrac{85}{36}-\dfrac{7}{12}=\dfrac{16}{9}\)

`c)`

\(75\%-1\dfrac{1}{2}+0,5\div\dfrac{5}{12}\)

`=`\(-\dfrac{3}{4}+\dfrac{6}{5}=\dfrac{9}{20}\)

a) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{3}.0,5\)

\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)

\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)

\(=\dfrac{1}{6}-\dfrac{1}{3}\)

\(=\dfrac{-1}{6}\)

b) \(\left(2+\dfrac{5}{6}\right):1\dfrac{1}{5}+\dfrac{-7}{12}\)

\(=\left(\dfrac{12}{6}+\dfrac{5}{6}\right):\dfrac{6}{5}+\dfrac{-7}{12}\)

\(=\dfrac{17}{6}.\dfrac{5}{6}+\dfrac{-7}{12}\)

\(=\dfrac{85}{36}+\dfrac{-7}{12}\)

\(=\dfrac{16}{9}\)

c) \(75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}\)

\(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{12}{5}\)

\(=\dfrac{3}{4}-\dfrac{6}{4}+\dfrac{6}{5}\)

\(=\dfrac{-3}{4}+\dfrac{6}{5}\)

\(=\dfrac{9}{20}\)

1+2-3-4+5+6-7-8+9+10-11-12+........+298-299-300+301+302 = 

1+2+(5-3)+(6-4)+(9-7)+(10-8)+…….+(301-299)+(302-300)=

Từ 302 đến 3 có số cặp là [(302-3):1+1]:2=150 cặp. Mà mỗi cặp có giá trị là 2

Vậy 1+2-3-4+5+6-7-8+9+10-11-12+........+298-299-300+301+302 = 

1+2+2×150=3+300=303

303 ok

hình như lạc đề:>

bạn cay à

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

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)

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

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

\(\Rightarrow x=\dfrac{-2}{3}\)

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

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

\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)

\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{-11}{20}\)

c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)

\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)

\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)

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

\(\Rightarrow x=\dfrac{-4}{5}\)

d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)

\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)

\(\Rightarrow x=\dfrac{2}{9}\)

e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)

\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)

\(\Rightarrow x=\dfrac{-7}{2}\)

a: \(\dfrac{x}{6}=\dfrac{8}{3}\)

=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)

b: \(\dfrac{5}{x}=\dfrac{4}{9}\)

=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)

c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)

=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)

=>x=-1-3=-4

d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)

=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)

=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)

=>\(x=-\dfrac{69}{8}\)

f: ĐKXĐ: x<>1

\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)

=>\(\left(x-1\right)^2=3\cdot27=81\)

=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)