b) B = \(\frac{3}{1.4}\)\(+\)\(\frac{5}{4.9}\)\(+\)\(\frac{7}{9.16}\)\(+\)\(\frac{9}{16.25}\)\(+\)\(\frac{11}{25.36}\)

\(\Rightarrow\)B\(=\) 1 \(-\)\(\frac{1}{4}\)\(+\)\(\frac{1}{4}\)\(-\)\(\frac{1}{9}\)\(+\)\(\frac{1}{9}\)\(-\)\(\frac{1}{16}\)\(+\)\(\frac{1}{16}\)\(-\)\(\frac{1}{25}\)\(+\)\(\frac{1}{25}\)\(-\)\(\frac{1}{36}\)

\(\Rightarrow\)B \(=\)1 \(-\)\(\frac{1}{36}\)

\(\Rightarrow\)B \(=\) \(\frac{35}{36}\)

Vậy B \(=\) \(\frac{35}{36}\)

k dùm mik nha

thanks bạn nhìu

chúc bạn hok tốt!!

\(1.\)

\(x\left(x+1\right)\left(x+6\right)-x^3=5x\)

\(\Leftrightarrow x\left(x^2+7x+6\right)-x^3-5x=0\)

\(\Leftrightarrow x^3+7x^2+6x-x^3-5x=0\)

\(\Leftrightarrow7x^2+x=0\)

\(\Leftrightarrow x\left(7x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\7x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{7}\end{cases}}\)

\(2.\)

\(\left(4x^2-6x\right)=\left(10x-15\right)\)

\(\Leftrightarrow2x\left(2x-3\right)-5\left(2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\2x-5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{5}{2}\end{cases}}\)

Trả lời:

1, \(x\left(x+1\right)\left(x+6\right)-x^3=5x\)

\(\Leftrightarrow\left(x^2+x\right)\left(x+6\right)-x^3=5x\)

\(\Leftrightarrow x^3+6x^2+x^2+6x-x^3=5x\)

\(\Leftrightarrow7x^2+6x=5x\)

\(\Leftrightarrow7x^2+6x-5x=0\)

\(\Leftrightarrow7x^2+x=0\)

\(\Leftrightarrow x\left(7x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\7x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{7}\end{cases}}}\)

Vậy x = 0; x = -1/7 là nghiệm của pt.

2, \(4x^2-6x=10x-15\)

\(\Leftrightarrow\left(4x^2-6x\right)-\left(10x-15\right)=0\)

\(\Leftrightarrow2x\left(2x-3\right)-5\left(2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\2x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{5}{2}\end{cases}}}\)

Vậy x = 3/2; x = 5/2 là nghiệm của pt.

mình ko biết

\(\left[3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2\right]:5\left(x-y\right)^2\)

\(=\frac{3}{5}\left(x-y\right)^3-\frac{2}{5}\left(x-y\right)^2+\frac{3}{5}\)

Trả lời:

Bài 2: 

a, 64x³ + 48x²y + 36xy² + 27y³ 

= (4x)³ + 3.(4x)².3y + 3.4x.(3y)² + (3y)³ 

= ( 4x + 3y )³ 

b, 1 - 15y + 75y² - 125y³ 

= 1³ - 3.1.5y + 3.1.(5y)² - (5y)³ 

= ( 1 - 5y )³ 

c, 8x³ + 4x²y + 2/3 xy² + 1/27 y³ 

= (2x)³ + 3.(2x)².1/3 y + 3.2x.(1/3 y)² + (1/3 y)³ 

= ( 2x + 1/3 y )³ 

d, 1/64 - 3/8 y + 3y² - 8y³ 

= (1/4)³ - 3.(1/4)².2y + 3.1/4.(2y)² - (2y)³ 

= ( 1/4 - 2y )³ 

e, ( 2x + 3y )( 4x² - 6xy + 9y² ) 

= ( 2x + 3y )[ (2x)² - 2x.3y + (3y)² ]

= (2x)³ + (3y)³ 

= 8x³ + 27y³ 

g, ( 3x + 4 )( 9x² - 12x + 16 ) 

= ( 3x + 4 )[ (3x)² - 3x.4 + 4² ]

= (3x)³ + 4³ 

= 27x³ + 64

h, ( 3 - x )( 9 + 3x + x² ) 

= ( 3 - x )( 3² + 3.x + x² )

= 3³ - x³ 

= 27 - x³ 

i, ( x² - 1/2 )( x⁴ + 1/2x² + 1/4 )

= ( x² - 1/2 )[ (x²)² + x².1/2 + (1/2)² ]

= (x²)³ - (1/2)³ 

= x⁶ - 1/8 

 Gọi F là trung điểm AD => FE là đường trung bình hình thang ABCD.

=>FE // AB // CD 

=> FE\(\perp\)AD hay FE là đường cao tam giác AED.

Mà FE cũng là đường trung tuyến tam giác AED.

=> AED là tam giác cân tại E (đpcm)

a)

AB song song CD trong hình thang cân

\(\Leftrightarrow\widehat{A}+\widehat{D}=180^o\)

\(\Leftrightarrow\widehat{D}=180^o-120^o\)

\(\Leftrightarrow\widehat{D}=60^o\)

b)

Góc D/B = \(\frac{1}{2}\)

\(\Leftrightarrow\widehat{B}=120^o\)

\(\widehat{C}=180^o-\widehat{D}=60^o\)

giúp mik please :3

( 2x - 1 )³ - 2x ( 4x² - 3x ) = 5

<=> 8x³ - 12x² + 6x - 1 - 8x³ + 6x² - 5 = 0

<=> - 6x² + 6x - 6 = 0

<=> x² - x + 1 = 0

<=> ( x - 1/2 )² = - 3/4 ( kh đúng )

=> pt vô nghiệm

\(\left(2x-1\right)^3-2x\left(4x^2-3x\right)=5\)

\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+6x^2=5\)

\(\Leftrightarrow-6x^2+6x-6=0\Leftrightarrow x^2-x+1=0\)

Ta có : \(x^2-x+1=x^2-x+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)

Vậy phương trình vô nghiệm