\(\frac13+\frac16+\frac{1}{10}+\cdots+\frac{2}{x\left(x+1\right)}=\) \(\frac{2021}{2023}\)

\(\frac12.\left(\frac13+\frac16+\frac{1}{10}+\cdots+\frac{2}{x\left(x+1\right)}\right)\) = \(\frac{2021}{2.2023}\)

\(\frac16+\frac{1}{12}+\frac{1}{20}\) + ...+ \(\frac{1}{x\left(x+1\right)}\) = \(\frac{2021}{2.2023}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\) + ... + \(\frac{1}{x\left(x+1\right)}\) = \(\frac{2021}{2.2023}\)

\(\frac12\) - \(\frac13\) + \(\frac13\) - \(\frac14\) + ... + \(\frac{1}{x}\) - \(\frac{1}{x+1}\) = \(\frac{2021}{2.2023}\)

\(\frac12-\frac{1}{x+1}\) = \(\frac{2021}{2023.2}\)

\(\frac{x+1-2}{2.\left(x+1\right)}\) = \(\frac{2021}{2.2023}\)

\(\frac{x+\left(1-1\right)}{2.\left(x+1\right)}\) = \(\frac{2021}{2.2023}\)

\(\frac{x-\left(2-1\right)}{2.\left(x+1\right)}\) = \(\frac{2021}{2.2023}\)

\(\frac{x-1}{x+1}\) = \(\frac{2021}{2023}\)

2023.(\(x-1\)) = 2021.(\(x+1\))

2023\(x\) - 2023 = 2021\(x\) + 2021

2023\(x-2021x\) = 2023 + 2021

2\(x\) = 4044

\(x\) = 4044 : 2

\(x\) = 2022

Vậy \(x=2022\)

1. Biến đổi biểu thức vế trái:
   Nhận xét rằng các phân số có dạng:
   13=22×3=2(12−13)one-third equals the fraction with numerator 2 and denominator 2 cross 3 end-fraction equals 2 open paren one-half minus one-third close paren13=22×3=212−13
   16=23×4=2(13−14)one-sixth equals the fraction with numerator 2 and denominator 3 cross 4 end-fraction equals 2 open paren one-third minus one-fourth close paren16=23×4=213−14
   ………
   2x(x+1)=2(1x−1x+1)the fraction with numerator 2 and denominator x open paren x plus 1 close paren end-fraction equals 2 open paren 1 over x end-fraction minus the fraction with numerator 1 and denominator x plus 1 end-fraction close paren2𝑥(𝑥+1)=21𝑥−1𝑥+1
2. Viết lại phương trình:
   2(12−13+13−14+…+1x−1x+1)=202120232 open paren one-half minus one-third plus one-third minus one-fourth plus … plus 1 over x end-fraction minus the fraction with numerator 1 and denominator x plus 1 end-fraction close paren equals 2021 over 2023 end-fraction212−13+13−14+…+1𝑥−1𝑥+1=20212023
   2(12−1x+1)=202120232 open paren one-half minus the fraction with numerator 1 and denominator x plus 1 end-fraction close paren equals 2021 over 2023 end-fraction212−1𝑥+1=20212023
   1−2x+1=202120231 minus the fraction with numerator 2 and denominator x plus 1 end-fraction equals 2021 over 2023 end-fraction1−2𝑥+1=20212023
3. Giải phương trình tìm x:
   2x+1=1−20212023the fraction with numerator 2 and denominator x plus 1 end-fraction equals 1 minus 2021 over 2023 end-fraction2𝑥+1=1−20212023
   2x+1=22023the fraction with numerator 2 and denominator x plus 1 end-fraction equals 2 over 2023 end-fraction2𝑥+1=22023
   x+1=2023x plus 1 equals 2023𝑥+1=2023
   x=2022x equals 2022𝑥=2022 

\(57\times48+57\times53-57\)

\(=57\times\left(48+53-1\right)\)

\(=57\times100\)

\(=5700\)

=57×(48+53+1)

=57×102

= 5814

A = \(x^2\) + 2\(x\) + 3

A = \(x^2\) + 2.\(x.1\) + 1 + 2

A = (\(x+1\))\(^2\) + 2

(\(x+1\))\(^2\) ≥ 0 ∀ \(x\) ∈ R

⇒A = \(\left(x+1\right)\)\(^2\) + 2 ≥ 2 > 0 \(\forall x\) ∈ R(đpcm)

Xét \(f(x)=x^2+2x+3\). Ta có:

\(\Delta=2^2-4\cdot1\cdot3\)

\(\Delta=4-12\)

\(\Delta=-8<0\)
Vì \(a=1>0\) và \(\Delta<0\) \(\rArr f(x)>0\) \(\forall\) \(x\in\R\)

Vậy \(x^2+2x+3\) luôn dương với mọi \(x\in\R\) (đpcm)

A = \(x^2\) + 2\(x\) + 3

A = \(x^2\) + 2.\(x.1\) + 1 + 2

A = (\(x+1\))\(^2\) + 2

(\(x+1\))\(^2\) ≥ 0 ∀ \(x\) ∈ R

⇒A = \(\left(x+1\right)\)\(^2\) + 2 ≥ 2 > 0 \(\forall x\) ∈ R(đpcm)

Số táo Hùng còn lại là:

20 - 10 =10 (quả táo)

Đáp số :10 quả táo

bài giải

hùng có số quả táo là :

20-10=10(quả)

đáp số: 10 quả táo

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86% = \(\frac{86}{100}\) = 8,6

367456

15,3 - 21,5 - 3.1,5

= -6,2 - 4,5

= - 10,7

\(15,3-21,5-3\cdot1,5\)

\(=15,3-21,5-4,5\)

\(=15,3+\left(-21,5-4,5\right)\)

\(=15,3+\left(-26\right)\)

\(=-10,7\)

= 0,39

Bài 5:

a: ĐKXĐ: x≠-2

Ta có: \(1+\frac{1}{x+2}=\frac{12}{x^3+8}\)

=>\(1+\frac{1}{x+2}=\frac{12}{\left(x+2\right)\left(x^2-2x+4\right)}\)

=>\(\frac{x^3+8}{\left(x+2\right)\left(x^2-2x+4\right)}+\frac{x^2-2x+4}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{12}{\left(x+2\right)\left(x^2-2x+4\right)}\)

=>\(x^3+8+x^2-2x+4=12\)

=>\(x^3+x^2-2x=0\)

=>\(x\left(x^2+x-2\right)=0\)

=>x(x+2)(x-1)=0

=>\(\left[\begin{array}{l}x=0\\ x+2=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(nhận\right)\\ x=-2\left(loại\right)\\ x=1\left(nhận\right)\end{array}\right.\)

b: ĐKXĐ: x<>2/7

Ta có: \(\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)

=>\(\left(2x+3\right)\cdot\frac{3x+8+2-7x}{2-7x}=\left(x-5\right)\cdot\frac{3x+8+2-7x}{2-7x}\)

=>\(\left(2x+3\right)\cdot\frac{-4x+10}{2-7x}=\left(x-5\right)\cdot\frac{-4x+10}{2-7x}\)

=>\(\left(2x+3\right)\left(-4x+10\right)-\left(x-5\right)\left(-4x+10\right)=0\)

=>(-4x+10)(2x+3-x+5)=0

=>-2(2x-5)(x+8)=0

=>(2x-5)(x+8)=0

=>\(\left[\begin{array}{l}2x-5=0\\ x+8=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac52\left(nhận\right)\\ x=-8\left(nhận\right)\end{array}\right.\)

Bài 4:

a: ĐKXĐ: x∉{2;-1}

Ta có: \(\frac{x+2}{x+1}+\frac{3}{x-2}=\frac{3}{x^2-x-2}+1\)

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

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

=>(x-2)(x+2)+3(x+1)=3+(x-2)(x+1)

=>\(x^2-4+3x+3=3+x^2-x-2\)

=>3x-1=-x+1

=>4x=2

=>\(x=\frac12\) (nhận)

b: ĐKXĐ: x∉{5;-6}

Ta có: \(\frac{x+6}{x-5}+\frac{x-5}{x+6}=\frac{2x^2+23x+61}{x^2+x-30}\)

=>\(\frac{x+6}{x-5}+\frac{x-5}{x+6}=\frac{2x^2+23x+61}{\left(x+6\right)\left(x-5\right)}\)

=>\(\frac{\left(x+6\right)^2+\left(x-5\right)^2}{\left(x+6\right)\left(x-5\right)}=\frac{2x^2+23x+61}{\left(x+6\right)\left(x-5\right)}\)

=>\(\left(x+6\right)^2+\left(x-5\right)^2=2x^2+23x+61\)

=>\(x^2+12x+36+x^2-10x+25=2x^2+23x+61\)

=>2x+61=23x+61

=>-21x=0

=>x=0(nhận)

Bài 3:

a: ĐKXĐ: x∉{5;-6}

Ta có: \(\frac{x+6}{x-5}+\frac{x-5}{x+6}=\frac{2x^2+23x+61}{x^2+x-30}\)

=>\(\frac{x+6}{x-5}+\frac{x-5}{x+6}=\frac{2x^2+23x+61}{\left(x+6\right)\left(x-5\right)}\)

=>\(\frac{\left(x+6\right)^2+\left(x-5\right)^2}{\left(x+6\right)\left(x-5\right)}=\frac{2x^2+23x+61}{\left(x+6\right)\left(x-5\right)}\)

=>\(\left(x+6\right)^2+\left(x-5\right)^2=2x^2+23x+61\)

=>\(x^2+12x+36+x^2-10x+25=2x^2+23x+61\)

=>2x+61=23x+61

=>-21x=0

=>x=0(nhận)

b: ĐKXĐ: x∉{3;-3}

Ta có: \(\frac{x^2-x}{x+3}-\frac{x_{}^2}{x-3}=\frac{7x^2-3x}{9-x^2}\)

=>\(\frac{\left(x^2-x\right)\left(x-3\right)-x^2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{-7x^2+3x}{\left(x-3\right)\left(x+3\right)}\)

=>\(\left(x^2-x\right)\left(x-3\right)-x^2\left(x+3\right)=-7x^2+3x\)

=>\(x^3-3x^2-x^2+3x-x^3-3x^2+7x^2-3x=0\)

=>0x=0(luôn đúng)

Vậy: x∉{3;-3}

Bài 2:

a: ĐKXĐ: x∉{-1;2}

ta có: \(\frac{x+2}{x+1}+\frac{3}{x-2}=\frac{3}{x^2-x-2}+1\)

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

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

=>\(x^2-4+3x+3=x^2-x+1\)

=>3x-1=-x+1

=>4x=2

=>\(x=\frac12\) (nhận)

b: ĐKXĐ: x∉{0;2}

ta có: \(\frac{5-x}{4x^2-8x}+\frac78=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\)

=>\(\frac{5-x}{4x\left(x-2\right)}+\frac78=\frac{x-1}{2x\left(x-2\right)}+\f...