3x = 2y ;  7y = 5z

=>x/2=y/3;y/5=z/7

=>x/10=y/15;y/15=z/21

=>x/10=y/15=z/21

áp dụng tính chất của dãy tỉ số bằng nhau ta có:

x/10=y/15=z/21=x-y+z/10-15+21=32/16=2

suy ra x/10=2 => x=20

y/15=2 =>y=30

z/21=2 => z=42

SSSSSSSSSSSSSSSSSSSSSSSS

cách trả lời của công chúa giá bằng s

ai đề

\(1,\frac{x+1}{x-2}=\frac{3}{4}\)

\(\Rightarrow3x-6=4x+4\)

\(\Rightarrow3x-4x=4+6\)

\(\Rightarrow-x=10\Leftrightarrow x=-10\)

\(2,\frac{x-1}{3}=\frac{x+3}{5}\)

\(\Rightarrow5x-5=3x+9\)

\(\Rightarrow5x-3x=9+5\)

\(\Rightarrow2x=14\Leftrightarrow x=7\)

\(3,\frac{2x+3}{24}=\frac{3x-1}{32}\)

\(\Rightarrow64x+96=72x-24\)

\(\Rightarrow72x-64x=24+96\)

\(\Rightarrow8x=120\)

\(\Rightarrow x=15\)

x/2 = 3/4

Phân tích đa thức thành nhân tử(tách hạng tử)
1)x^2+2x-3=x^2-x+3x-3=x(x-1)+3(x-1)=(x-1)(x+3)
2)x^2-5x+6=x^2-2x-3x+6=x(x-2)-3(x-2)=(x-2)(x-3)
3)x^2+7x+12=(x+3)(x+4)
4)x^2-x-12=(x-4)(x+3)
5)3x^2+3x-36=3[(x-3)(x+4)]
6)5x^2-5x-10=5[(x-2)(x+1) ]       
7)3x^2-7x-6=(x-3)(3x+2)
8)4x^2+4x-3=4x^2+6x-2x-3=(2x-1)(2x+3)
9)8x^2-2x-3=8x^2+4x-6x-3=(4x-3)(2x+1)
 

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

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

3: \(x^2+7x^2+12x=4x\left(2x+3\right)\)

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

5: \(3x^2+3x-36=3\left(x^2+x-12\right)=3\left(x+4\right)\left(x-3\right)\)

6: \(5x^2-5x-10=5\left(x^2-x-2\right)=5\left(x-2\right)\left(x+1\right)\)

a) ( -2x - 28 )⁵ = -32 = ( -2 )⁵

-2x - 28 = -2

-2x = ( -2 ) + 28 = 26

x = 26 : ( -2 ) = -13

b) ( 7x - 11 )³ = 1000 = 10³

7x - 11 = 10

7x = 10 + 11 = 21

x = 21 : 7 = 3

c) ( 3x - 15 )² - 6 = 75

( 3x - 15 )² = 75 + 6 = 81 = 9²

3x - 15 = 9

3x = 9 + 15 = 24

x = 24 : 3 = 8

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

<=> \(9x^2-9x+2=9x^2+6x+1\)

<=>  \(15x=1\) <=> \(x=\frac{1}{15}\)

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

<=> \(4x^2+3x-1=4x^2-12x+9\)

<=> \(15x^2=10\) <=> \(x=\frac{2}{3}\)

c) \(\left(5x+1\right)^2=\left(7x-3\right)\left(7x+2\right)\) <=> \(25x^2+10x+1=49x^2-7x-6\)

<=> \(24x^2-17x-7=0\) <=> \(24x^2-24x+7x-7=0\)

<=> \(\left(24x+7\right)\left(x-1\right)=0\) <=> \(\orbr{\begin{cases}x=-\frac{7}{24}\\x=1\end{cases}}\)

d) (4 - 3x)(4 + 3x) = (9x - 3)(1 - x)

<=> 16 - 9x² = 12x - 9x2 - 3

<=> 12x = 19

<=> x = 19/12

e) x(x + 1)(x + 2)(x + 3) = 24

<=> (x² + 3x)(x² + 3x + 2) = 24

<=> (x² + 3x)²  + 2(x² + 3x) - 24 = 0

<=> (x² + 3x)² + 6(x² + 3x) - 4(x² + 3x) - 24 = 0

<=> (x² + 3x + 6)(x² + 3x - 4) = 0

<=> \(\orbr{\begin{cases}x^2+3x+6=0\\x^2+3x-4=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x+\frac{3}{2}\right)^2+\frac{15}{4}=0\left(vn\right)\\\left(x+4\right)\left(x-1\right)=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-4\\x=1\end{cases}}\)

g) (7x - 2)² = (7x - 3)(7x + 2)

<=> 49x² - 28x + 4 = 49x² - 7x - 6

<=> 21x = 10 <=> x = 10/21

a) x = 4                

b) x = 7     

c) x = 2                

d) x = 5

e) x = 2                

f) x= 1. 

Mua tài khoản VIP để học tập không giới hạn trên OLM!

a) x = 4

b) x = 7

c) x = 2

d) x = 5

e) x = 2

 f) x= 1

\(\Rightarrow7x^3+7x^2-4x^2-4x+x+1=0\\ \Rightarrow\left(x+1\right)\left(7x^2-4x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\7x^2-4x+1=0\left(1\right)\end{matrix}\right.\\ \left(1\right)\Rightarrow7\left(x^2-2\cdot\dfrac{2}{7}x+\dfrac{4}{49}\right)+\dfrac{3}{7}=0\\ \Rightarrow7\left(x-\dfrac{2}{7}\right)^2+\dfrac{3}{7}=0\left(\text{vô lí}\right)\)

Vậy x=-1

\(x^2\left(7x-3\right)=0\)

I don't now 

sorry 

...................

nha

a)   \(\left(3x-1\right)^2+2\left(3x-1\right)\left(2x+1\right)+\left(2x+1\right)^2=0\)

\(\Leftrightarrow\)\(\left[\left(3x-1\right)+\left(2x-1\right)\right]^2=0\)

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

\(\Leftrightarrow\)\(5x-2=0\)

\(\Leftrightarrow\)\(x=\frac{2}{5}\)

Vậy...

b)  \(\left(7x+2\right)^2+\left(7x-2\right)^2-2\left(7x+2\right)\left(7x-2\right)=0\)

\(\Leftrightarrow\)\(\left[\left(7x+2\right)-\left(7x-2\right)\right]^2=0\)

\(\Leftrightarrow\)\(4^2=0\)  vô lí

Vậy pt vô nghiệm

-7X-3X