bó tay luôn

\(a,\Rightarrow y=\dfrac{2\cdot5}{1}=10\\ b,\Rightarrow\dfrac{y}{5}=\dfrac{42}{25}:\dfrac{6}{5}=\dfrac{7}{5}\Rightarrow y=7\)

\(a,\dfrac{242}{363}+\dfrac{1616}{2121}=\dfrac{2}{7}\times y\)

\(\dfrac{2}{7}\times y=\dfrac{2\times121}{3\times121}+\dfrac{16\times101}{21\times101}\)

\(\dfrac{2}{7}\times y=\dfrac{2}{3}+\dfrac{16}{21}\)

\(\dfrac{2}{7}\times y=\dfrac{14}{21}+\dfrac{16}{21}\)

\(\dfrac{2}{7}\times y=\dfrac{30}{21}\)

\(\dfrac{2}{7}\times y=\dfrac{10}{7}\)

\(y=\dfrac{10}{7}:\dfrac{2}{7}\)

\(y=\dfrac{10}{7}\times\dfrac{7}{2}\)

\(y=5\)

\(---\)

\(b,\left(y+\dfrac{1}{4}\right)+\left(y+\dfrac{1}{16}\right)+\left(y+\dfrac{1}{16}\right)=2\)

\(\left(y+y+y\right)+\left(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{16}\right)=2\)

\(3\times y+\left(\dfrac{4}{16}+\dfrac{2}{16}\right)=2\)

\(3\times y+\dfrac{6}{16}=2\)

\(3\times y+\dfrac{3}{8}=2\)

\(3\times y=2-\dfrac{3}{8}\)

\(3\times y=\dfrac{16}{8}-\dfrac{3}{8}\)

\(3\times y=\dfrac{13}{8}\)

\(y=\dfrac{13}{8}:3\)

\(y=\dfrac{13}{8}\times\dfrac{1}{3}\)

\(y=\dfrac{13}{24}\)

#\(Toru\)

` 242/363 + 1616/2121 = 2/7 xxy`

`2/7 xxy= 2/3 + 16/21`

`2/7 xxy= 14/21 +16/21`

`2/7 xxy= 30/21`

`y=10/7 : 2/7`

`y=10/7 xx 7/2`

`y=70/14`

`y=5`

__

` (y + 1/4) + (y + 1/16) + (y + 1/16) =2`

`(y+y+y)+(1/4 + 1/16+1/16)=2`

`3y + (4/16 +1/16 +1/16)=2`

`3y + 6/16=2`

`3y=2-6/16`

`3y= 32/16-6/16`

`3y= 26/16`

`y=26/16 : 3`

`y=26/48`

`y=13/24`

`Answer:`

1) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

\(=[x\left(x+3\right)][\left(x+1\right)\left(x+2\right)]+1\)

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

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

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

2) \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)

\(=[\left(4x+1\right)\left(3x+2\right)][\left(12x-1\right)\left(x+1\right)]-4\)

\(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)

\(=[\left(12x^2+11x+0,5\right)+1,5][\left(12x^2+11x+0,5\right)-1,5]-4\)

\(=\left(12x^2+11x+0,5\right)^2-\left(1,5\right)^2-4\)

\(=\left(12x^2+11x+0,5\right)^2-\left(2,5\right)^2\)

\(=\left(12x^2+11x+0,5-2,5\right)\left(12x^2+11x+0,5+2,5\right)\)

\(=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)

3) \(\left(x^2+6x+5\right)\left(x^2+10x+21\right)+15\)

\(=\left(x^2+x+5x+5\right)\left(x^2+3x+7x+21\right)+15\)

\(=\left(x+1\right)\left(x+5\right)\left(x+3\right)\left(x+7\right)+15\)

\(=[\left(x+1\right)\left(x+7\right)][\left(x+5\right)\left(x+3\right)]+15\)

\(=\left(x^2+x+7x+7\right)\left(x^2+3x+5x+15\right)+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt \(v=x^2+=8x+11\)

Đa thức có dạng sau: \(\left(v-4\right)\left(v+4\right)+15\)

\(=v^2-4^2+15\)

\(=v^2-1\)

\(=\left(v+1\right)\left(v-1\right)\)

\(=\left(x^2+8x+11+1\right)\left(x^2+8x+11-1\right)\)

\(=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)

4) \(\left(x^2-a\right)^2-6x^2+4x+2a\)

\(=\left(x^2-a\right)\left(x^2-a\right)-6x^2+4x+2a\)

\(=\left(x^2-a\right).x^2-a\left(x^2-a\right)-6x^2+4x+2a\)

\(=x^4-ax^2-a.\left(x^2-a\right)-6x^2+4x+2a\)

\(=x^4-ax^2-\left(ax^2-aa\right)-6x^2+4x+2a\)

\(=x^4-2ax^2+a^2-6x^2+2a+4x\)

6) \(a^2-b^2-c^2+2bc-2a+1\)

\(=\left(a^2-2a+1\right)-\left(b^2-2bc+c^2\right)\)

\(=\left(a-1\right)^2-\left(b-c\right)^2\)

\(=\left(a-b+c-1\right)\left(a+b-c-1\right)\)

7) \(4a^2-4b^2+16bc-16c^2\)

\(=4a^2-\left(4b^2-16bc+16c^2\right)\)

\(=\left(2a\right)^2-\left(2b-4c\right)^2\)

\(=\left(2a-2b+4c\right)\left(2a+2b-4c\right)\)

\(=2.\left(a-b-2c\right).2\left(a+b-2c\right)\)

\(=4\left(a-b-2c\right)\left(a+b-2c\right)\)

bài 1:

a) (x+1)^2-(x-1)^2-3(x+1)(x-1)

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

=2x^2-3x^2-3

=-x^2-3

Bài 1:

A = 3(x + 1)² + 5 

Ta có: (x + 1)² \(\ge\) 0 Với mọi x

\(\Rightarrow\) 3(x + 1)² \(\ge\) 0 với mọi x

\(\Rightarrow\) 3(x + 1)² + 5 \(\ge\) 5 với mọi x

Hay A \(\ge\) 5

Dấu "=" xảy ra khi và chỉ khi x + 1 = 5 hay x = -1

Vậy...

B = 2|x + y| + 3x² - 10

Ta có: 2|x + y| \(\ge\) 0 với mọi x, y

3x² \(\ge\) 0 với mọi x

\(\Rightarrow\) 2|x + y| + 3x² - 10 \(\ge\) -10 với mọi x,y

Dấu "=" xảy ra khi và chỉ khi x + y = 0; x = 0

\(\Rightarrow\) x = y = 0

Vậy ...

C = 12(x - y)² + x² - 6

Ta có: 12(x - y)² \(\ge\) 0 với mọi x; y

x² \(\ge\) 0 với mọi x

\(\Rightarrow\) 12(x - y)² + x² - 6 \(\ge\) -6 với mọi x, y

Dấu "=" xảy ra khi và chỉ khi x = y = 0

Phần D ko rõ đầu bài nha vì D luôn có một giá trị duy nhất

Bài 2:

Phần A ko rõ đầu bài!

B = 3 - (x + 1)² - 3(x + 2y)²

Ta có: -(x + 1)² \(\le\) 0 với mọi x

-3(x + 2y)² \(\le\) 0 với mọi x, y

\(\Rightarrow\) 3 - (x + 1)² - 3(x + 2y)² \(\le\) 3 với mọi x, y

Dấu "=" xảy ra khi và chỉ khi x = 2y; x + 1 = 0

\(\Rightarrow\) x = -1; y = \(\dfrac{-1}{2}\)

Vậy ...

C = -12 - 3|x + 1| - 2(y - 1)²

Ta có: -3|x + 1| \(\le\) 0 với mọi x

-2(y - 1)² \(\le\) 0 với mọi y

\(\Rightarrow\)  -12 - 3|x + 1| - 2(y - 1)² \(\le\) -12 với mọi x, y

Dấu "=" xảy ra khi và chỉ khi x + 1 = 0; y - 1 = 0

\(\Rightarrow\) x = -1; y = 1

Vậy ...

Phần D đề ko rõ là \(\dfrac{5}{2x^2}-3\) hay \(\dfrac{5}{2}\)x² - 3 nữa

F = \(\dfrac{-5}{3}\) - 2x²

Ta có: -2x² \(\le\) 0 với mọi x

\(\Rightarrow\) \(\dfrac{-5}{3}-2x^2\) \(\le\) \(\dfrac{-5}{3}\) với mọi x

Dấu "=" xảy ra khi và chỉ khi x = 0

Vậy ...

Chúc bn học tốt!

Bài 1:

Tổng của 2 số là

  \(36\times2=72\) 

Số lớn là

  \(72-17=55\) 

Bài 2:

a) \(4567+y\div34=10987\) 

                 \(y\div34=10987-4567\) 

                 \(y\div34=6420\) 

                         \(y=6420\times34\) 

                         \(y=218280\) 

b) \(\dfrac{4}{3}+\dfrac{1}{2}\div y=2\) 

             \(\dfrac{1}{2}\div y=2-\dfrac{4}{3}\) 

              \(\dfrac{1}{2}\div y=\dfrac{2}{3}\) 

                      \(y=\dfrac{1}{2}\div\dfrac{2}{3}\) 

                      \(y=\dfrac{3}{4}\) 

Bài 3:

a) \(\dfrac{2}{5}\times\dfrac{2}{5}+\dfrac{9}{8}\div3=\dfrac{4}{25}+\dfrac{9}{8}\times\dfrac{1}{3}=\dfrac{4}{25}+\dfrac{3}{8}=\dfrac{107}{200}\) 

b) \(2-\left(\dfrac{1}{7}\times4+\dfrac{5}{21}\right)=2-\left(\dfrac{4}{7}+\dfrac{5}{21}\right)=2-\dfrac{17}{21}=\dfrac{25}{21}\)

Bài 1 : Gọi a là số lớn, b là số bé, theo đề bài ta có :

(a+b):2=36⇒a+b=72

mà b=17

Nên a=72-17=55

Bài 2 :

a) 4567+y:34=10987

⇒ y:34=10987-4567

⇒ y:34=6420

⇒ y=6420x34

⇒ y=218280

b) \(\dfrac{4}{3}+\dfrac{1}{2}:y=2\)

\(\Rightarrow\dfrac{1}{2}:y=2-\dfrac{4}{3}\)

\(\Rightarrow\dfrac{1}{2}:y=\dfrac{2}{3}\)

\(\Rightarrow y=\dfrac{1}{2}:\dfrac{2}{3}\)

\(\Rightarrow y=\dfrac{1}{2}x\dfrac{3}{2}\)

\(\Rightarrow y=\dfrac{3}{4}\)

Bài 3 :

\(\dfrac{2}{5}x\dfrac{2}{5}+\dfrac{9}{8}:3=\dfrac{4}{25}+\dfrac{9}{8}x\dfrac{1}{3}=\dfrac{4}{25}+\dfrac{3}{8}\)

= \(\dfrac{4x8}{25x8}+\dfrac{25x3}{25x8}=\dfrac{32}{200}+\dfrac{75}{200}=\dfrac{107}{200}\)

\(2-\left(\dfrac{1}{7}x4+\dfrac{5}{21}\right)=2-\left(\dfrac{4}{7}+\dfrac{5}{21}\right)=2-\left(\dfrac{12}{21}+\dfrac{5}{21}\right)=2-\dfrac{17}{21}=\dfrac{42}{21}-\dfrac{17}{21}=\dfrac{25}{21}\)

\(a,2\dfrac{2}{5}:y\times1\dfrac{3}{4}=\dfrac{7}{8}\\ \dfrac{12}{5}:y\times\dfrac{7}{4}=\dfrac{7}{8}\\ \dfrac{12}{5}:y=\dfrac{7}{8}:\dfrac{7}{4}\\ \dfrac{12}{5}:y=\dfrac{1}{2}\\ y=\dfrac{12}{5}:\dfrac{1}{2}=\dfrac{24}{5}\\ b,3\dfrac{2}{5}:y:1\dfrac{1}{4}=2\dfrac{3}{5}\\ \dfrac{17}{5}:y:\dfrac{5}{4}=\dfrac{13}{5}\\ y:\dfrac{5}{4}=\dfrac{17}{5}:\dfrac{13}{5}\\ y:\dfrac{5}{4}=\dfrac{17}{13}\\ y=\dfrac{17}{13}\times\dfrac{5}{4}=\dfrac{85}{52}\)

\(c,\dfrac{12}{5}-2\dfrac{2}{5}\times y=1\dfrac{1}{4}\\ \dfrac{12}{5}-\dfrac{12}{5}\times y=\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{12}{5}-\dfrac{5}{4}\\ \dfrac{12}{5}\times y=\dfrac{23}{20}\\ y=\dfrac{23}{20}:\dfrac{12}{5}\\ y=\dfrac{23}{48}\)

a, 2\(\dfrac{2}{5}\): y \(\times\)1\(\dfrac{3}{4}\) = \(\dfrac{7}{8}\)

     \(\dfrac{12}{5}\) : y   \(\times\dfrac{7}{4}\)    = \(\dfrac{7}{8}\)

   \(\dfrac{12}{5}\)  : y         = \(\dfrac{7}{8}\) : \(\dfrac{7}{4}\)

          \(\dfrac{12}{5}\) : y        = \(\dfrac{1}{2}\)       

                  y         = \(\dfrac{12}{5}\) : \(\dfrac{1}{2}\)

                  y         =    \(\dfrac{24}{5}\)

b, 3\(\dfrac{2}{5}\): y : 1\(\dfrac{1}{4}\) = 2\(\dfrac{3}{5}\) 

     \(\dfrac{17}{5}\): y: \(\dfrac{5}{4}\)    = \(\dfrac{13}{5}\)

       \(\dfrac{17}{5}\):y         = \(\dfrac{13}{5}\times\dfrac{5}{4}\)

       \(\dfrac{17}{5}\) : y       = \(\dfrac{13}{4}\)

               y        =   \(\dfrac{17}{5}\) : \(\dfrac{13}{4}\)

               y        =  \(\dfrac{68}{65}\)

c, \(\dfrac{12}{5}\) - 2\(\dfrac{2}{5}\)\(\times y\) = 1\(\dfrac{1}{4}\)

     \(\dfrac{12}{5}\) - \(\dfrac{12}{5}\)\(\times\)y = \(\dfrac{5}{4}\)

            \(\dfrac{12}{5}\times y\) =  \(\dfrac{12}{5}\) - \(\dfrac{5}{4}\)

           \(\dfrac{12}{5}\) \(\times\) y =    \(\dfrac{23}{20}\)

                     \(y\)  = \(\dfrac{23}{20}\): \(\dfrac{12}{5}\)

                     y   = \(\dfrac{23}{48}\)