a) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}\)

\(=\dfrac{5}{7}\times\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\)

\(=\dfrac{5}{7}\times1\)

\(=\dfrac{5}{7}\)

b) \(\dfrac{1}{10}+\dfrac{5}{9}+\dfrac{4}{9}+\dfrac{9}{10}-1\)

\(=\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\left(\dfrac{1}{10}+\dfrac{9}{10}-1\right)\)

\(=1+0\)

\(=1\)

c) \(\dfrac{5}{7}\times\dfrac{5}{9}+\dfrac{4}{9}\times\dfrac{5}{7}+\dfrac{2}{7}\)

\(=\dfrac{5}{7}\times\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{2}{7}\)

\(=\dfrac{5}{7}+\dfrac{2}{7}\)

\(=1\)

d) \(\dfrac{2}{7}+\dfrac{2}{8}+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{4}{7}\)

\(=\left(\dfrac{2}{8}+\dfrac{1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{4}{7}\right)\)

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

\(=\dfrac{1}{2}+1\)

\(=\dfrac{3}{2}\)

e) \(\dfrac{4}{5}+\dfrac{3}{10}+\dfrac{2}{10}+0,7\)

\(=\dfrac{4}{5}+\dfrac{5}{10}+\dfrac{7}{10}\)

\(=\dfrac{4}{5}+\dfrac{12}{10}\)

\(=\dfrac{4}{5}+\dfrac{6}{5}\)

\(=\dfrac{10}{5}\)

\(=2\)

g) \(362\times728+326\times272\)

\(=326\times\left(728+272\right)\)

\(=326\times1000\)

\(=326000\) 

a) 

x -  17 = -5 

x        = ( - 5) + 17 

x        = 12

b ) 

2x + 1/4 = 3/2

2x          = 3/2 - 1/4

2x          = 6/4 - 1/4

2x          = 5/4

 x           = 5/4 : 2

x            = 5/4 x 1/2

x           = 5/8

tk và  kb với mk nha! mơn ạ!  

a)\(x-17=-5\Rightarrow x=-5+17\Rightarrow x=12\)

b)\(2x+\frac{1}{4}=\frac{3}{2}\Rightarrow2x=\frac{3}{2}-\frac{1}{4}=\frac{5}{4}\)\(\Rightarrow x=\frac{5}{4}:2\Rightarrow x=\frac{5}{8}\)

a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)

b, \(-2x+2=2\Leftrightarrow x=0\)

c, \(-2x-6=-8\Leftrightarrow x=1\)

ơ

Rõ ràng có câu hỏi mà

bn đăng lại ik

câu hỏi sửa được đấy

\(BPT\Leftrightarrow1+\frac{1}{x+2}

Bài \(3\)

\(A=\left(x-5\right)\left(2x+3\right)-2x\left(x-3\right)+x+7\)

\(=2x^2+3x-10x-15-\left(2x^2-6x\right)+x+7\)

\(=2x^2+3x-10x-15-2x^2+6x+x+7\)

\(=\left(2x^2-2x^2\right)+\left(3x-10x+6x+x\right)+\left(-15+7\right)\)

\(=-8\)

Vậy biểu thức không phụ thuộc vào biến

\(B=4\left(y-6\right)-y^2\left(2+3y\right)+y\left(5y-4\right)+3y^2\)

Đề như này à?

Bài \(4\)

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

\(b,4x^2-4x+1=\left(2x\right)^2-2.2x.1+1^2=\left(2x+1\right)^2\)

\(c,\) ?

\(d,\left(x-y\right)^2-\left(2x-y\right)^2\\ =\left[\left(x-y\right)-\left(2x-y\right)\right]\left[\left(x-y\right)+\left(2x-y\right)\right]\\ =\left(x-y-2x+y\right)\left(x-y+2x-y\right)\\ =\left(-x\right)\left(3x-2y\right)\)

\(e,8x^3-y^3=\left(2x\right)^3-y^3\\ =\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(i,3x+6y+\left(x+2y\right)\\ =3\left(x+2y\right)+\left(x+2y\right)\\ =4\left(x+2y\right)\)

\(j,ax-ay-x+y=\left(ãx-ay\right)-\left(x-y\right)\\ =a\left(x-y\right)-\left(x-y\right)=\left(x-y\right)\left(a-1\right)\)

`k,` `y` hay `y^2` ạ? vì nó mới phân tích được nhân tử.

Tớ xin làm câu k nhé!

\(k)2x^2-y+6x^2y-3y^2\\=(2x^2-y)+(6x^2y-3y^2)\\=(2x^2-y)+3y(2x^2-y)\\=(2x^2-y)(1+3y)\)

#\(Toru\)

\(c)\\1)(2x+y)^2-x^2\\=(2x+y-x)(2x+y+x)\\=(x+y)(3x+y)\\2)?\)

Dấu _ là sao cậu?

#\(Toru\)

`a,x(x-1)-(x+2)^2=1`

`<=>x^2-x-x^2-4x-4=1`

`<=>-5x=5`

`<=>x=-1`

`b,(x+5)(x-3)-(x-2)^2=-1`

`<=>x^2+2x-15-x^2+4x-4+1=0`

`<=>6x-18=0`

`<=>x-3=0`

`<=>x=3`

`c,x(2x-4)-(x-2)(2x+3)=0`

`<=>2x(x-2)-(x-2)(2x+3)=0`

`<=>(x-2)(2x-2x-3)=0`

`<=>-3(x-2)=0`

`<=>x-2=0`

`<=>x=2`

`d,x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12`

`<=>3x^2+2x+x^2+2x+1-4x^2+25=-12`

`<=>4x+26=-12`

`<=>4x=-38`

`<=>x=-19/2`

c: Ta có: \(\left(2x-3\right)^2-\left(2x-3\right)\left(x-10\right)=7\)

\(\Leftrightarrow4x^2-12x+9-2x^2+20x+3x-30=7\)

\(\Leftrightarrow11x=28\)

hay \(x=\dfrac{28}{11}\)

d: Ta có: \(\left(3x-4\right)^2-9\left(x-3\right)\left(x+3\right)=8\)

\(\Leftrightarrow9x^2-24x+16-9x^2+81=8\)

\(\Leftrightarrow-24x=-89\)

hay \(x=\dfrac{89}{24}\)

f: Ta có: \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Leftrightarrow x^2+8x+16-x^2+1=16\)

\(\Leftrightarrow8x=-1\)

hay \(x=-\dfrac{1}{8}\)