a: \(\left(18\dfrac{1}{3}:\sqrt{225}+8\dfrac{2}{3}\cdot\sqrt{\dfrac{49}{4}}\right):\left[\left(12\dfrac{1}{3}+8\dfrac{6}{7}\right)-\dfrac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\right]:\dfrac{1704}{445}\)

\(=\left(\dfrac{55}{3}:15+\dfrac{26}{3}\cdot\dfrac{7}{4}\right):\left[\left(12+\dfrac{1}{3}+8+\dfrac{6}{7}\right)-\dfrac{7}{18}\right]\cdot\dfrac{445}{1704}\)

\(=\left(\dfrac{55}{45}+\dfrac{91}{6}\right):\left[20+\dfrac{101}{126}\right]\cdot\dfrac{445}{1704}\)

\(=\dfrac{295}{18}:\dfrac{2621}{126}\cdot\dfrac{445}{1704}\)

\(=\dfrac{295}{18}\cdot\dfrac{126}{2621}\cdot\dfrac{445}{1704}\simeq0,21\)

b: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

c: \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{n+1}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{n}{n+1}\)

\(=\dfrac{1}{n+1}\)

d: \(-66\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)+124\cdot\left(-37\right)+63\cdot\left(-124\right)\)

\(=-66\cdot\dfrac{33-22+6}{66}+124\left(-37-63\right)\)

\(=-17-12400=-12417\)

e: \(\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)

\(=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)

\(=\dfrac{7}{4}\cdot33\cdot\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)

\(=33\cdot\dfrac{7}{4}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)

\(=33\cdot\dfrac{7}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)

\(=33\cdot\dfrac{7}{4}\cdot\dfrac{4}{21}=\dfrac{33\cdot1}{3}=11\)

Câu 1: 

=>15(2x+1)-8(3x-1)=100

=>30x+15-24x+8=100

=>6x+23=100

hay x=77/6

Câu 2:

=>2(5x-3)+12-3(7x-1)=x+2

=>10x-6+12-21x+3-x-2=0

=>-12x=-7

hay x=7/12

Câu 3: 

\(\Leftrightarrow2\left(x^2-1\right)+3\left(x+1\right)=2\left(x^2-4x+4\right)\)

\(\Leftrightarrow2x^2-2+3x+3-2x^2+8x-8=0\)

=>11x-7=0

hay x=-7/11

Câu 4:

(x - 4)^3/6 + 1 = x(x + 1)/2 - (x - 5)(x + 5)/3

<=> (x - 4)^3 + 6/6 = x^2 + x/2 - x^2 - 25/3

<=> (x - 4)^3 + 6/6 = 3x^2 + 3x - 2x^2 + 50/6

<=> (x - 4)^3 + 6 = 3x^2 + 3x - 2x^2 + 50

<=> x^3 - 12x^2 + 48x - 58 = x^2 + 3x + 50

<=> x^3 -13x^2 + 45x - 108 = 0

Đến đây bạn bấm máy nhẩm nghiệm là ra nhé

Câu 5:

3(x + 2)^3/5 - (x - 1)^2/10 = (x - 3)(x + 3)/2

<=> 6(x + 2)^3 - (x - 1)^2/10 = 5(x^2 - 9)/10

<=> 6(x + 2)^3 - (x - 1)^2 = 5(x^2 - 9)

<=> 6x^3 + 36x^2 + 72x + 48 - x^2 + 2x - 1 - 5x^2 + 45 = 0

<=> 6x^3 + 30x^2 + 74x + 92 = 0

Đến đây bạn bấm máy nhẩm nghiệm như câu 4 nhé

Bài 3:

\(a,A=\dfrac{x^2+xy-xy+y^2}{\left(x-y\right)\left(x+y\right)}:\dfrac{x^2+2xy+y^2-2xy}{\left(x-y\right)\left(x+y\right)}\\ A=\dfrac{x^2+y^2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x-y\right)\left(x+y\right)}{x^2+y^2}=1\\ b,=\left[\dfrac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right]\left[\dfrac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]^2\\ =\left(a+2\sqrt{a}+1\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\\ =\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}=1\)

\(i.\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)

\(\Leftrightarrow\dfrac{4x^2+4x+1}{5}-\dfrac{x^2-2x+1}{3}=\dfrac{7x^2-14x-5}{15}\)

\(\Leftrightarrow\dfrac{12x^2+12x+3}{15}-\dfrac{5x^2-10x+5}{15}=\dfrac{7x^2-14x-5}{15}\)

\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)

\(\Leftrightarrow36x=-3\)

\(\Leftrightarrow x=-\dfrac{1}{12}\)

\(k.x+\dfrac{2x+\dfrac{x-1}{5}}{3}=1-\dfrac{3x-\dfrac{1-2x}{3}}{5}\)

\(\Leftrightarrow\dfrac{15x}{15}+\dfrac{10x+x-1}{15}=\dfrac{15}{15}-\dfrac{9x-1+2x}{15}\)

\(\Leftrightarrow15x+9x-1=14-7x\)

\(\Leftrightarrow31x=15\)

\(\Leftrightarrow x=\dfrac{15}{31}\)

câu h tương tự bài trước câu g mik làm nhé

A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)

Xét: n⁴ + 4 = (n²+2)² - 4n² = (n²-2n+2)(n²+2n+2) = [(n-1)²+1][(x+1)²+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)

B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)

Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)

= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)

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

= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)

= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)

Vậy ...

\(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{19}\right)\left(1-\dfrac{1}{20}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{18}{19}.\dfrac{19}{20}=\dfrac{1}{20}>\dfrac{1}{21}\)