\(x+x+x\times1+x\times6=120\\ x\times\left(1+1+1+6\right)=120\\ x\times9=120\\ x=\dfrac{120}{9}\\ x=\dfrac{40}{3}\)

\(x+x+x\text{×}1+x\text{×}6=120\)

\(x\text{×}\left(1+1+1\right)+x\text{×}6=120\)

\(x\text{×}3+x\text{×}6=120\)

\(x\text{×}\left(3+6\right)=120\)

\(x\text{×}9=120\)

\(x=\dfrac{120}{9}\)

x+x+x×1+x×6=120x×(1+1+1+6)=120x×9=120x=1209x=403

\(\dfrac{x}{3}=\dfrac{y}{7}\Rightarrow x=\dfrac{3}{7}y\\ xy=175\Rightarrow\dfrac{3}{7}y\cdot y=175\\ \Rightarrow y^2=175:\dfrac{3}{7}=\dfrac{1225}{3}\Rightarrow y=\dfrac{35\sqrt{3}}{3}\\ \Rightarrow x=\dfrac{35\sqrt{3}}{7}\)

Hình như bạn ghi sai đề vì lớp 7 chưa học căn 

a: Thay x=-3 vào B, ta được:

\(B=\dfrac{2\cdot\left(-3\right)^2}{3\cdot\left(-3\right)+6}=\dfrac{2\cdot9}{-9+6}=\dfrac{18}{-3}=-6\)

b: \(A=\dfrac{2x^2+20+3x-6-7x-14}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x^2-4x}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x}{x+2}\)

a) \(a^2\cdot a^3\cdot a^7\cdot b^2\cdot b\)

\(=\left(a^2\cdot a^3\cdot a^7\right)\cdot\left(b^2\cdot b\right)\)

\(=a^{12}\cdot b^3\)

b) \(b^6\cdot b\cdot c^7\cdot c^8\)

\(=\left(b^6\cdot b\right)\cdot\left(c^7\cdot c^8\right)\)

\(=b^7\cdot c^{15}\)

c) \(a^8\cdot a^9\cdot a\cdot c\cdot c^{20}\)

\(=\left(a^8\cdot a^9\cdot a\right)\cdot\left(c\cdot c^{20}\right)\)

\(=a^{18}\cdot c^{21}\)

d) \(a^2\cdot a^3\cdot b^4\cdot c\cdot c^3\)

\(=\left(a^2\cdot a^3\right)\cdot b^4\cdot\left(c\cdot c^3\right)\)

\(=a^5\cdot b^4\cdot c^4\)

a) Kiểm tra lại nhé

b) \(b^6.b^7.c^8\)

\(=b^{6+7}.c^8=b^{13}.c^8\)

c) \(a^8.a^9.a.c.c^{20}\)

\(=a^{8+9+1}.c^{1+20}\)

\(=a^{18}.c^{21}\)

d) \(a^2.a^3.b^4.c.c^3\)

\(=a^{2+3}.b^4.c^{1+3}\)

\(=a^5.b^4.c^4\)

\(#WendyDang\)

Câu b mình cũng không hiểu nha ( phần b mình làm bị sai á ).

2+ 6/ căn x -1

Ta có: \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2003\right)=4+1007\cdot2003\)

\(\Leftrightarrow2004x+\dfrac{2003\cdot2004}{2}=4+1007\cdot2003\)

\(\Leftrightarrow2004x=10019\)

hay \(x=\dfrac{10019}{2004}\)

\(x^2+\left(x+3\right)\left(x-9\right)=-27\\ \Rightarrow x^2+x^2+3x-9x-27=-27\\ \Rightarrow2x^2-6x=0\\ \Rightarrow2x\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

\(x^2+\left(x+3\right)\left(x-9\right)=-27\)

\(\Rightarrow2x^2-6x=0\)

\(\Rightarrow2x\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

\(\Rightarrow x^2+x^2-6x-27=-27\\ \Rightarrow2x^2-6x=0\\ \Rightarrow2x\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)