\(x^7+x^6+x^4+x^3+x^2+1\)

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

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

=x²-6x+9-4y²

=x²-6x+3²-4y²

=(x-3)²-4y²

=(x-3)²-(2y)²

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

\(x^2-6x+9-4y^2\)

\(=\left(x-3\right)^2-4y^2\)

\(=\left(x-3-2y\right)\left(x-3+2y\right)\)

`(x-1)/3+(3x-5)/2+(2x)/9+(-5x)/9`

`=(x-1)/3+(3x-5)/2+x/3`

`=(2x-2+9x-15+2x)/6`

`=(13x-17)/6`

cảm ơn bạn nhiều^^

\(\left|2x+7\right|-\dfrac{1}{2}=4\)

\(\Rightarrow\left|2x+7\right|=\dfrac{9}{2}\)

\(\Rightarrow\left[{}\begin{matrix}2x+7=\dfrac{9}{2}\\2x+7=-\dfrac{9}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=-\dfrac{5}{2}\\2x=-\dfrac{23}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{4}\\x=-\dfrac{23}{4}\end{matrix}\right.\)

Vậy: \(x\in\left\{-\dfrac{5}{4};-\dfrac{23}{4}\right\}\).

Hơi muộn xíu mà cảm ơn anh lần nữa nhe :33

c) \(\left(2\sqrt{x}+1\right)^2=4x+4\sqrt{x}+1\)

a: x-2 là ước của 2x+5

=>2x+5⋮x-2

=>2x-4+9⋮x-2

=>9⋮x-2

=>x-2∈{1;-1;3;-3;9;-9}

=>x∈{3;1;5;-1;11;-7}

b: 3x-5 là bội của x+1

=>3x-5⋮x+1

=>3x+3-8⋮x+1

=>-8⋮x+1

=>x+1∈{1;-1;2;-2;4;-4;8;-8}

=>x∈{0;-2;1;-3;3;-5;7;-9}

c: 5x+8⋮x+2

=>5x+10-2⋮x+2

=>-2⋮x+2

=>x+2∈{1;-1;2;-2}

=>x∈{-1;-3;0;-4}

d: 3x+1⋮2x-3

=>6x+2⋮2x-3

=>6x-9+11⋮2x-3

=>11⋮2x-3

=>2x-3∈{1;-1;11;-11}

=>2x∈{4;2;14;-8}

=>x∈{2;1;7;-4}

\(a, A=2\sqrt{x-1}-3\sqrt{x-1}-4\sqrt{x-1}=\left(2-3-4\right)\sqrt{x-1}=-5\sqrt{x-1}\)

\(b, B=\frac{2}{x+y}.\left(x+y\right)\sqrt{\frac{3}{4}}=2\sqrt{\frac{3}{4}}=2.\frac{1}{2}.\sqrt{3}=\sqrt{3}\)

a) \(\left(\dfrac{1}{2}m+3\right)^3=\dfrac{m^3}{8}+\dfrac{9m^2}{4}+\dfrac{27m}{2}+9\)

b) \(2\left(m+\dfrac{1}{2}\right)=2m+1\)

c) \(\left(2\sqrt{x}+1\right)^2=4x+4\sqrt{x}+1\)

d) \(\left(2\sqrt{2}+\dfrac{1}{4}\right)^2=8+\sqrt{2}+\dfrac{1}{16}=\dfrac{129}{16}+\sqrt{2}\)

mình cám ơn

y+y*1/3*9/2+y*7/2

y+y*3/2+y*7/2=252

y*(1+3/2+7/2)=252

y*      6         =252

y                  =252:6

y                  =42

sai ngay bước đầu tiên nhưng cách làm sai phần đầu còn lại đúng