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24 tháng 9 2021

bó tay luôn

 

AH
Akai Haruma
Giáo viên
24 tháng 9 2021

Bài 1:

a. 

$\frac{2}{3}\times \frac{x}{y}=\frac{8}{15}$

$\frac{x}{y}=\frac{8}{15}: \frac{2}{3}=\frac{4}{5}$

b.

$\frac{x}{y}: \frac{3}{4}=\frac{2}{5}$

$\frac{x}{y}=\frac{3}{4}\times \frac{2}{5}=\frac{3}{10}$

c.

$\frac{3}{5}: \frac{x}{y}=\frac{4}{7}$

$\frac{x}{y}=\frac{3}{5}: \frac{4}{7}=\frac{21}{20}$

AH
Akai Haruma
Giáo viên
24 tháng 9 2021

Bài 2:

Chiều dài hình chữ nhật là:

$\frac{3}{5}: \frac{3}{4}=\frac{4}{5}$ (m)

Chu vi hình chữ nhật:

$2\times (\frac{3}{4}+\frac{4}{5})=\frac{31}{10}$ (m)

5 tháng 4 2022

`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) \(4a^2-4b^2+16bc-16c^2\)

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

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

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

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

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

20 tháng 7 2023

Bài 2:

\(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)

\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}=\dfrac{a+b+a-b}{c+a+c-a}=\dfrac{a}{c}\) (T/c dãy tỷ số = nhau)

\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a}{c}\Rightarrow c\left(a+b\right)=a\left(c+a\right)\)

\(\Rightarrow ac+bc=ac+a^2\Rightarrow a^2=bc\)

a: \(\Leftrightarrow x\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;9;-9;12;-12;18;-18;36;-36\right\}\)

mà -3<x<30

nên \(x\in\left\{-2;-1;1;2;3;4;6;9;12;18\right\}\)

b: \(\Leftrightarrow x\in\left\{0;4;-4;8;-8;12;-12;...\right\}\)

mà -16<=x<20

nên \(x\in\left\{-16;-12;-8;-4;0;4;8;12;16\right\}\)

c: \(\Leftrightarrow x-1+4⋮x-1\)

\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{2;0;3;-1;5;-3\right\}\)

d: \(\Leftrightarrow2x+4-5⋮x+2\)

\(\Leftrightarrow x+2\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{-1;-3;3;-7\right\}\)

20 tháng 8 2023

c)\(\dfrac{3}{8}\times\dfrac{5}{8}+y=\dfrac{5}{4}\) 

   \(\dfrac{15}{64}+y=\dfrac{5}{4}\) 

           \(y=\dfrac{5}{4}-\dfrac{15}{64}\) 

           \(y=\dfrac{65}{64}\)

d, \(\dfrac{3}{8}+\dfrac{5}{8}\times y=\dfrac{5}{4}\) 

          \(\dfrac{5}{8}\times y=\dfrac{5}{4}-\dfrac{3}{8}\) 

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

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

                \(y=\dfrac{7}{5}\)

   

21 tháng 8 2023

 a, 3/4 x y = 3/5 + 3/10   

3/4 x y = 9/10

y = 9/10 : 3/4

y = 6/5

b, 3/5 : y = 3/4 - 2/5

3/5 : y = 7/20

y = 3/5 : 7/20 

y = 12/7

 

20 tháng 12 2020

Bai 2 : 

\(\hept{\begin{cases}x+y=7\\x-7=13\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=7\\x=20\end{cases}}}\)

Thay x vào phương trình đầu ta có : 

\(20+y=7\Leftrightarrow y=-13\)

Vậy \(\left\{x;y\right\}=\left\{20;-13\right\}\)

Thử \(20-13=7\)\(20-7=13\)( thỏa mãn ) 

24 tháng 4 2022

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

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

\(y\times\dfrac{2}{3}=\dfrac{5}{4}\)

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

\(y=\dfrac{5}{4}\times\dfrac{3}{2}\)

\(y=\dfrac{15}{8}\)