Ta có : \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{1989}{1991}\)

=> \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1989}{1991}\)

=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1989}{1991}\)

=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1989}{1991}\)

=> \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1989}{3982}\)

=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1989}{3982}\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{1989}{3982}\)

=> \(\frac{1}{x+1}=\frac{1}{1991}\)

=> x + 1 = 1991

=> x = 1990

Vậy x = 1990

\(2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{x\left(x+1\right)}\right)=\frac{3980}{1991}\) 

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{1990}{1991}\) 

\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1990}{1991}\) 

\(1-\frac{1}{x+1}=\frac{1990}{1991}\) 

\(\frac{1}{x+1}=1-\frac{1990}{1991}\) 

\(\frac{1}{x+1}=\frac{1}{1991}\) 

\(x+1=1991\) 

\(x=1990\)  

a) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.........+\frac{2}{x\left(x+1\right)}=\frac{1998}{2000}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.......+\frac{2}{x\left(x+1\right)}=\frac{1998}{2000}\)

\(\Leftrightarrow2.\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+......+\frac{1}{x\left(x+1\right)}\right]=\frac{1998}{2000}\)

\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+......+\frac{1}{x\left(x+1\right)}=\frac{999}{2000}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{x}-\frac{1}{x+1}=\frac{999}{2000}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{999}{2000}\)\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2000}\)

\(\Leftrightarrow x+1=2000\)\(\Leftrightarrow x=1999\)

Vậy \(x=1999\)

b) \(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+......+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

\(\Leftrightarrow\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.....+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{15.2}{93}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+......+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\Leftrightarrow\frac{1}{2x+3}=\frac{1}{93}\)\(\Leftrightarrow2x+3=93\)

\(\Leftrightarrow2x=90\)\(\Leftrightarrow x=45\)

Vậy \(x=45\)

Mình viết nhầm, phần a là 1/2x nhé mn!

a) \(x^2=\frac{1}{16}\Rightarrow x^2=\left(\pm\frac{1}{4}\right)^2\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{1}{4}\end{cases}}\)

b) \(x^3=\frac{27}{64}\Rightarrow x^3=\left(\frac{3}{4}\right)^3\Rightarrow x=\frac{3}{4}\)

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

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

=> vô nghiệm

d) (2x - 1)³ - 27 = 0 => (2x - 1)³ = 27 => (2x - 1)³ = 3³ => 2x - 1 = 3 => 2x = 4 => x = 2

e) Vì \(x^2\ge0\forall x\)

\(\left(2,5-y\right)^2\ge0\forall y\)

=> x² + (2,5 - y)² \(\ge\)0 với mọi x,y

Dấu " = " xảy ra khi x² = 0 => x = 0 và (2,5 - y)² = 0 => y = 2,5

Vậy x = 0,y = 2,5

f) 2^x = 8 => 2^x = 2³ => x = 3

g) (-3)^x + 1 = -26

=> (-3)^x = -26 - 1 = -27

=> (-3)^x = (-3)³

=> x = 3

h) 2^4-x = 32

=> 2^4-x = 2⁵

=> 4 - x = 5

=> x = -1

j) 2^x + 2^x+1 = 18

=> 2^x + 2^x . 2 = 18

=> 2^x (1 + 2) = 18

=> 2^x . 3 = 18

=> 2^x = 6

=> x không thỏa mãn

j) Thiếu dữ liệu

Với x<−2

⇒−x−2+2.(1−x)=5

⇒−x−2+2−2x=5

⇒−3x=5

⇒x=−5/3( Loại vì −5/3>−63=-2)

Với −2≤x<1

⇒x+2+2.(1−x)=5

⇒x+2+2−2x=5

⇒−x+4=5

⇒−x=1

⇒x=−1 < Chọn>

Với x≥1

⇒x+2+2.(x−1)=5

⇒x+2+2x−2=5

⇒3x=5

⇒x=5/3 < Chọn> 

Ta có: \(\left|2x-\frac{5}{6}\right|=\left|x+\frac{1}{3}\right|\)

\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{5}{6}=x+\frac{1}{3}\\2x-\frac{5}{6}=-x-\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{6}\\3x=\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}}\)

\(|2x-\frac{5}{6}|=|x+\frac{1}{3}|\) 

\(\orbr{\begin{cases}2x-\frac{5}{6}=x+\frac{1}{3}\\2x-\frac{5}{6}=-\left(x+\frac{1}{3}\right)\end{cases}}\)  

\(\orbr{\begin{cases}2x-x=\frac{5}{6}+\frac{1}{3}\\2x-\frac{5}{6}=-x-\frac{1}{3}\end{cases}}\) 

\(\orbr{\begin{cases}x=\frac{7}{6}\\2x+x=-\frac{1}{3}+\frac{5}{6}\end{cases}}\) 

\(\orbr{\begin{cases}x=\frac{7}{6}\\3x=\frac{1}{2}\end{cases}}\) 

\(\orbr{\begin{cases}x=\frac{7}{6}\\x=\frac{1}{6}\end{cases}}\)

\(2^{24}=(2^3)^8=8^8\)

\(3^{16}=\left(3^2\right)^8=9^8\)

vì \(8^8< 9^8\Rightarrow2^{24}< 3^{16}\)

\(2^{24}=\left(2^3\right)^8=8^8\) 

\(3^{16}=\left(3^2\right)^8=9^8\) 

\(8< 9\) 

\(\Rightarrow8^9< 9^9\) 

\(\Rightarrow2^{24}< 3^{16}\)

\(C=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

\(\Rightarrow3C=3\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)

\(\Rightarrow3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}+\frac{1}{3^{98}}\)

\(\Rightarrow3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+..+\frac{1}{3^{97}}+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)

\(\Rightarrow2C=1-\frac{1}{3^{99}}\)

MÀ \(2C=1-\frac{1}{3^{99}}< 1\Rightarrow C=\frac{1-\frac{1}{3^{99}}}{2}< \frac{1}{2}\)

Từ đó ta suy ra điều phải chứng minh

A = 2¹⁰⁰- 2⁹⁹ + 2⁹⁸ - 2⁹⁷ + ... + 2² - 2

=> 2A =  2¹⁰¹ - 2¹⁰⁰ + 2⁹⁹ - 2⁹⁸ + ... + 2³ - 2² 

Khi đó 2A  + A = (2¹⁰¹ - 2¹⁰⁰ + 2⁹⁹ - 2⁹⁸ + ... + 2³ - 2²) + (2¹⁰⁰- 2⁹⁹ + 2⁹⁸ - 2⁹⁷ + ... + 2² - 2)

=> 3A = 2¹⁰¹ - 2

=> \(A=\frac{2^{201}-2}{3}\)

b) Ta có B = 3¹⁰⁰- 3⁹⁹ + 3⁹⁸ - 3⁹⁷ + ... + 3² - 3 + 1

=> 3B = 3¹⁰¹ - 3¹⁰⁰ + 3⁹⁹ - 3⁹⁸  + ... + 3³ - 3² + 3

Khi đó 3B + B = (3¹⁰¹ - 3¹⁰⁰ + 3⁹⁹ - 3⁹⁸  + ... + 3³ - 3² + 3) + (3¹⁰⁰- 3⁹⁹ + 3⁹⁸ - 3⁹⁷ + ... + 3² - 3 + 1)

=> 4B = 3¹⁰¹ + 1

=> B = \(\frac{3^{101}+1}{4}\)

a) \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)

=> \(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)

=> \(2A+A=\left(2^{101}-2^{100}+...-2^2\right)+\left(2^{100}-2^{99}+...-2\right)\)

<=> \(3A=2^{101}-2\)

=> \(A=\frac{2^{101}-2}{3}\)

b) \(B=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)

=> \(3A=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\)

=> \(3A+A=\left(3^{101}-3^{100}+...+3\right)+\left(3^{100}-3^{99}+...+1\right)\)

<=> \(4A=3^{101}+1\)

=> \(A=\frac{3^{101}+1}{4}\)

Ta có : |x| + x = 3

=> |x| = 3 - x

Đk : \(3-x\ge0\Rightarrow x\le3\)

Khi đó |x| = 3 - x

<=> \(\orbr{\begin{cases}x=3-x\\x=-3+x\end{cases}}\Rightarrow\orbr{\begin{cases}2x=3\\0x=-3\left(\text{loại}\right)\end{cases}}\Rightarrow x=\frac{3}{2}\)(tm)

Vậy x = 3/2

Ta có:

Nếu \(x< 0\)

=> \(\left|x\right|+x=-x+x=0\) (vô lý)

Vì vậy nên \(x\ge0\)

Khi đó: \(\left|x\right|+x=3\)

\(\Leftrightarrow x+x=3\)

\(\Leftrightarrow2x=3\)

\(\Rightarrow x=\frac{3}{2}\)