\(\dfrac{1}{6}x+\dfrac{1}{12}x+\dfrac{1}{20}x+...+\dfrac{1}{2450}x=1\)

\(x\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2450}\right)\)=1

\(x\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{49\times50}\right)\)=1

\(x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\)

\(x\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=1\)

\(x\times\)\(\dfrac{12}{25}=1\)

\(\Rightarrow x=1\div\dfrac{12}{25}\)

\(x=1\times\dfrac{25}{12}=\dfrac{25}{12}\)

vậy \(x=\dfrac{25}{12}\)

vậy \(x=2\)\(x=2\)\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3}-\dfrac{1}{9}\)\(\left(2\dfrac{2}{9}-x\right)\)=\(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{72}\)

\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{72}\)\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{8\times9}\)\(\Rightarrow\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

nè bn!!!!

máy hết chỗ đánh làm ơn chuyển 3 dòng đầu này xuống cuối

\(\left(\dfrac{20}{9}-x\right)=\dfrac{1}{3}-\dfrac{1}{9}\)

\(\left(\dfrac{20}{9}-x\right)=\dfrac{2}{9}\)

\(x=\dfrac{20}{9}-\dfrac{2}{9}\)

\(x=2\)

vậy x=2

giúp mk nha !!!THANKS!!!

=1/x*(1/2+1/6+1/12+1/20+1/30+1/42).

Ta có:

1/2+1/6+1/12+1/20+1/30+1/42.

=1/1*2+1/2*3+1/3*4+1/4*5+1/5*6+1/6*7.

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7.

=1-1/7.

=6/7.

=>1/x*6/7=36.

=>1/x=36:6/7=42.

=>x=1/42.

Vậy x=1/42.

a, Biến đổi vế trái :

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

b,\(\left(3x-2\right)\left(4x-5\right)-\left(2x-1\right)\left(6x+2\right)=0\)

\(\Leftrightarrow12x^2-15x-8x+10-\left(12x^2+4x-6x-2\right)=0\)

\(\Leftrightarrow12x^2-23x+10-12x^2+2x+2=0\)

\(\Leftrightarrow12-21x=0\)

\(\Leftrightarrow-21x=-12\)

\(\Leftrightarrow21x=12\)

\(\Leftrightarrow x=\frac{4}{7}\)

c,

a, bạn thêm

Vậy VT=VT(đpcm)

nhé

a) 6x(5x + 3) + 3x(1 – 10x) = 7  

⇒ 30x²+18x+3x-30x²=7

⇒21x=7

⇒x=\(\dfrac{7}{21}\)

⇒x= \(\dfrac{1}{3}\)

b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44

⇒15x-63x²-15+63x + 63x²-35x+36x-20=44

⇒79x-35=44

⇒79x=44+35

⇒79x=79

⇒x=1

d) 5x(12x + 7) – 3x(20x – 5) = - 100

⇒60x²+35x-60x²+15=-100

⇒35x+15=-100

⇒35x=-100-15

⇒35x=-115

⇒x=\(\dfrac{-115}{35}\)

⇒x=\(\dfrac{-23}{7}\)

(24x-1)(24x-2)(24x-3)(24x-4)=24

đặt 24x-4=a=>a(a+1)(a+2)(a+3)=24

<=>(a^2+3a)(a^2+3a+2)=24

đặt a^2+3a+1=b => (b-1)(b+1)=24

<=> b^2-5^2=0

<=> b= 5 hoặc b=-5 thay vô tìm a rồi tìm x

chúc bạn hk tốt

a)\(60x^2+35x-60x^2+15x=100\)

  35x+15x=100

  50x=100   =>x=2

b)\(10x^2-35x+16x-10x^2=5\)

  -35x+16x=5

   -19x=5    =>x=-5/19

a)

  35x+15x=100

  50x=100   

=>x=2

b)

  -35x+16x=5

   -19x=5   

=>x=-5/19

Lời giải:

PT $\Leftrightarrow 8x^3-16x^2+6x+2=0$

$\Leftrightarrow (8x^3-8x^2)-(8x^2-8x)-(2x-2)=0$

$\Leftrightarrow 8x^2(x-1)-8x(x-1)-2(x-1)=0$

$\Leftrightarrow (x-1)(8x^2-8x-2)=0$

$\Leftrightarrow 2(x-1)(4x^2-4x-1)=0$

$\Leftrightarrow x-1=0$ hoặc $4x^2-4x-1=0$

Nếu $x-1=0\Leftrightarrow x=1$ 

Nếu $4x^2-4x-1=0$

$\Leftrightarrow (2x-1)^2-2=0$

$\Leftrightarrow (2x-1-\sqrt{2})(2x-1+\sqrt{2})=0$

$\Leftrightarrow x=\frac{1\pm \sqrt{2}}{2}$

a) Ta có: \(25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3\ge3\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{2}{5}\)

b) Ta có: \(9x^2-6x+2\)

\(=9x^2-6x+1+1\)

\(=\left(3x-1\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{3}\)

c) Ta có: \(-x^2+2x-2\)

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

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

\(=-\left(x-1\right)^2-1\le-1\forall x\)

Dấu '=' xảy ra khi x-1=0

hay x=1

d) Ta có: \(x^2+12x+39\)

\(=x^2+12x+36+3\)

\(=\left(x+6\right)^2+3\ge3\forall x\)

Dấu '=' xảy ra khi x=-6

e) Ta có: \(-x^2-12x\)

\(=-\left(x^2+12x+36-36\right)\)

\(=-\left(x+6\right)^2+36\le36\forall x\)

Dấu '=' xảy ra khi x=-6

f) Ta có: \(4x-x^2+1\)

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

\(=-\left(x^2-4x+4-5\right)\)

\(=-\left(x-2\right)^2+5\le5\forall x\)

Dấu '=' xảy ra khi x=2

a) Ta có: \(25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3\ge3\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{2}{5}\)

b) Ta có: \(9x^2-6x+2\)

\(=9x^2-6x+1+1\)

\(=\left(3x-1\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{3}\)

c) Ta có: \(-x^2+2x-2\)

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

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

\(=-\left(x-1\right)^2-1\le-1\forall x\)

Dấu '=' xảy ra khi x=1

( Mình trình bày mẫu câu a các câu khác mình làm tắt lại nhưng tương tự trình bày câu a nha )

a, Ta có : \(25x^2-20x+7=\left(5x\right)^2-2.5x.2+2^2+3\)

\(=\left(5x-2\right)^2+3\)

Thấy : \(\left(5x-2\right)^2\ge0\forall x\in R\)

\(\Rightarrow\left(5x-2\right)^2+3\ge3\forall x\in R\)

Vậy \(Min=3\Leftrightarrow5x-2=0\Leftrightarrow x=\dfrac{2}{5}\)

b, \(=9x^2-2.3x+1+1=\left(3x-1\right)^2+1\ge1\)

Vậy Min = 1 <=> x = 1/3

c, \(=-x^2+2x-1-1=-\left(x^2-2x+1\right)-1=-\left(x-1\right)^2-1\le-1\)

Vậy Max = -1 <=> x = 1

d, \(=x^2+2.x.6+36+3=\left(x+6\right)^2+3\ge3\)

Vậy Min = 3 <=> x = - 6

e, \(=-x^2-2.x.6-36+36=-\left(x+6\right)^2+36\le36\)

Vậy Max = 36 <=> x = -6 .

f, \(=-x^2+4x-4+5=-\left(x^2-4x+4\right)+5=-\left(x-2\right)^2+5\le5\)

Vậy Max = 5 <=> x = 2