a, ⇔ x⁴ - 2x³ + 4x³ - 8x² + 4x² - 8x + 3x - 6 = 0

⇔ (x - 2)(x³ + 4x² + 4x + 3) = 0

⇔ (x - 2)(x³ + 3x² + x² + 3x + x + 3) = 0

⇔ (x - 2)(x + 3)(x² + x + 1) = 0 mà x² + x + 1 > 0 ∀ x

⇔ \(\left\{{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Vậy tập nghiệm phương trình S = {2; -3}

b) \(\dfrac{7}{2}-\left(\dfrac{x}{5}-\dfrac{1}{4}\right)=\dfrac{9}{2}\)

<=> \(\dfrac{7}{2}-\dfrac{x}{5}+\dfrac{1}{4}=\dfrac{9}{2}\)

<=> \(\dfrac{15}{4}-\dfrac{x}{5}-\dfrac{9}{2}=0\)

<=> \(\dfrac{x}{5}=\dfrac{5}{4}\)

<=> x = 6,25

Vậy,...

c) ( x + 2)( x + 3)( x - 5)( x - 6) = 180

<=> ( x + 2)( x - 5)( x + 3)( x - 6) = 180

<=> ( x² - 3x - 10 )( x² - 3x - 18 ) = 180

Đặt : x² - 3x - 14 = a , ta có :

( a + 4)( a - 4) = 180

<=> a² - 16 - 180 = 0

<=> a² - 196 = 0

<=> ( a - 14)( a + 14 ) = 0

<=> a = 14 hoặc a = -14

* Với , a = 14 , ta có :

x² - 3x - 14 = 14

<=> x² - 3x - 28 = 0

<=> x² - 7x + 4x - 28 = 0

<=> x( x - 7) + 4( x - 7) = 0

<=> ( x + 4)( x - 7) = 0

<=> x = -4 hoặc : x = 7

* Với : a = -14 , ta có :

x² - 3x - 14 = -14

<=> x( x - 3) = 0

<=> x = 0 hoặc : x = 3

Vậy,...

b: \(\Leftrightarrow x-10\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{53\cdot55}\right)=\dfrac{3}{11}\)

\(\Leftrightarrow x-10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)

\(\Leftrightarrow x-10\cdot\dfrac{4}{55}=\dfrac{3}{11}\)

=>x=3/11+20/55=3/11+4/11=7/11

c: \(\Leftrightarrow\left(\dfrac{x-1}{99}-1\right)+\left(\dfrac{x-2}{98}-1\right)+\left(\dfrac{x-5}{95}-1\right)=\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{95}\)

\(\Leftrightarrow x-100=1\)

hay x=101

\(\Rightarrow\left(\frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110}\right).x=10.\left(\frac{10}{1.10}+\frac{10}{2.12}+...+\frac{10}{100.110}\right)\)

\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{10}-\frac{1}{110}\right).x=10.\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)

\(\Rightarrow\left(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right).x=10.\left(\left(1+\frac{1}{2}+..+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)-\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{110}\right)\right)\)

\(\Rightarrow\left(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right).x=10.\left(\left(1+\frac{1}{2}+..+\frac{1}{10}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{110}\right)\right)\)=> x = 10

Nhân 100 vào 2 vế ta được :

(100/1.101 + 100/2.102 + 100/3.103 +....+100/10.110) . x = (10/1.11 + 10/2.12 + 10/100.110 )10

=>(1+1/2+1/3+...+1/10-1/101-...-1/110)x=(1+1/2+1/3+...+1/10+1/11+...+1/100-1/11-...-1/100-1/101-...-1/110)10

=>(1+1/2+1/3+...+1/10-1/101-...-1/110)x=(1+1/2+1/3+...+1/10-1/101-...-1/110)10

=>x=10

Hay thì like nha ! hj hj

bài này bạn lấy trong sách nào vậy

Câu 3: 

a: \(A=-\left|x-10\right|+2018< =2018\)

Dấu '=' xảy ra khi x=10

\(B=-\left(x+2\right)^2+1999< =1999\)

Dấu '=' xảy ra khi x=-2

b: \(A=\left(2x-8\right)^2+3>=3\)

Dấu '=' xảy ra khi x=4

\(B=\left|x^2-25\right|-2017>=-2017\)

Dấu '=' xảy ra khi x=5 hoặc x=-5