\(\frac{7x+2}{5x+7}=\frac{7x-1}{5x+1}=\frac{7x+2-7x+1}{5x+7-5x-1}=\frac{3}{6}=\frac{1}{2}\)

\(2\left(7x+2\right)=5x+7\)

\(14x-5x=7-4\)

\(9x=3\)

\(\Rightarrow x=3\div9\)

\(\Rightarrow x=\frac{1}{3}\)

nhân chéo ta được:

\(\left(7x+2\right)\left(5x+1\right)=\left(5x+7\right)\left(7x-1\right)\)

\(35x^2+7x+10x+2=35x^2-5x+49x-7\) ( nhân phân phối)

\(35x^2-35x^2+7x+10x-5x+49x=-2-7\)

\(-27x=-9\)

\(x=\frac{-9}{-27}\)

\(x=\frac{1}{3}\)

=>x=1/3

7x+2/5x+7=7x-1/5x+1=>37/5x+7=34/5x+1=>37/5x-34/5x=1-7=>3/5x=-6=>x=-6:3/5=-10 vay x=-10 nho ****

\(\frac{7x+2}{5x+7}=\frac{7x-1}{5x+1}=\frac{7x+2-\left(7x-1\right)}{5x+7-\left(5x+1\right)}=\frac{7x+2-7x+1}{5x+7-5x-1}=\frac{3}{6}=\frac{1}{2}\)

tu \(\frac{7x+2}{5x+7}=\frac{1}{2}=>\left(7x+2\right).2=5x+7=>14x+4=5x+7=>14x-5x=7-4=>9x=3=>x=\frac{3}{9}=\frac{1}{3}\)

vay x=1/3

tick nha

mình thi toán volympic lớp 4 , mình hok lớp 4 

k nha 

Mình cũng thế nè

Mình cũng thế nè

a) =(x-y)*(x+y)-(5*(x+y))

=(x+y)*(x-y-5)

Mấy bài còn lại cũng tương tự nha bạn = cách đặt nhân tử chung 

bai nao khong hieu thi pan nhan tin vào nick minh minh se giai đùm ban

a) (x² - y²) - 5(x + y)

= (x - y)(x + y) - 5 (x + y)

= (x + y) (x - y -5)

b) 5x³ - 5x²y - 10x² + 10 xy

= 5[(x³ - x²y) - (2x² - 2 xy)]

=5[x²(x - y) - 2x(x - y)]

=5x(x-y)(x - 2)

c) 2x² - 5x = x(2x - 5)

d) x³ - 3x² +1 - 3x 

= (x³ + 1) - (3x² + 3x)

= (x + 1)(x² - x + 1) - 3x(x + 1)

= (x + 1) [x² - x + 1 - 3x]

= (x + 1)[x² - 4x + 1]

= (x + 1)[x² - 2.x.2 + 2² - 2² + 1]

= (x + 1)[(x - 2)² - 3]

= \(\left(x+1\right)\left(x-2+\sqrt{3}\right)\left(x-2-\sqrt{3}\right)\)

e) 3x² - 6xy + 3y² - 12z²

= 3[ x² - 2xy + y² - 4z²]

= 3[ (x - y)² - (2z)²]

= 3(x - y + 2z)(x - y - 2z)

f) 3x² - 7x - 10

= 3x² - 7x - 7 - 3

= (3x² -3) - (7x + 7)

= 3(x² - 1) - 7(x + 1)

= 3 (x + 1)(x - 1) - 7(x + 1)

= (x + 1)[3(x - 1) - 7]

= (x +1)(3x - 8)

g) x⁴ + 1 - 2x² = (x²)² - 2.x² + 1 = (x² - 1)²

= (x + 1)²(x - 1)²

h) 3x² - 3y² - 12x + 12y

= 3(x² - y²) - 12(x - y)

= 3(x - y)(x + y) - 12(x -y)

= (x - y) [3(x + y) - 12]

= (x - y). 3. (x+y - 4)

j) x² - 3x + 2 = x² - x - 2x +2

= x(x - 1) - 2(x -1)

=(x - 1)(x - 2)

a) =(x-y)*(x+y)-(5*(x+y))

=(x+y)*(x-y-5)

Mấy bài còn lại cũng tương tự nha bạn = cách đặt nhân tử chung 

a) \(\frac{x^2+5x+6}{x^2+7x+12}\)=\(\frac{x^2+2x+3x+6}{x^2+3x+4x+12}\)=\(\frac{x\left(x+2\right)+3\left(x+2\right)}{x\left(x+3\right)+4\left(x+3\right)}\)=\(\frac{\left(x+3\right)\left(x+2\right)}{\left(x+4\right)\left(x+3\right)}\)

b) \(\frac{7x^2+14x+7}{3x^2+3x}\)=\(\frac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\)=\(\frac{7\left(x+1\right)^2}{3x\left(x+1\right)}\)=\(\frac{7\left(x+1\right)\left(x+1\right)}{3x\left(x+1\right)}\)=\(\frac{7\left(x+1\right)}{3x}\)