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

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

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

=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`

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

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

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

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

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

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

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

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

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

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

bài này đ/s là 42 đúng

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

chi tiết câu c

c) \(\left(2\sqrt{x}+1\right)^2=\left(2\sqrt{x}\right)^2+2\cdot2\sqrt{x}\cdot1+1^2=4x+4\sqrt{x}+1\)

a) \(\left(\dfrac{1}{2}m+3\right)^3\)

\(=\left(\dfrac{1}{2}m\right)^3+3\cdot\left(\dfrac{1}{2}m\right)^2\cdot3+3\cdot\dfrac{1}{2}m\cdot3^2+3^3\)

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

c) Ta có: \(\left(2\sqrt{x}+1\right)^2\)

\(=\left(2\sqrt{x}\right)^2+2\cdot2\sqrt{x}\cdot1+1^2\)

\(=4x+4\sqrt{x}+1\)