Theo đầu bài ta có:
\(2x\left(x-\frac{1}{7}\right)=0\)
\(\Rightarrow\hept{\begin{cases}2x=0\\x-\frac{1}{7}\end{cases}=0}\)
\(\Rightarrow\hept{\begin{cases}x=0\\x=\frac{1}{7}\end{cases}}\)

\(2x.\left(x-\frac{1}{7}\right)=0\Rightarrow2x=0\)hoặc \(x-\frac{1}{7}=0\)

\(\Rightarrow x=0\)hoặc \(x=\frac{1}{7}\)

=>x+6=0 hoặc 2x-3=0

=>x=-6 hoặc x=3/2

phá ngoặc ra mà tính đi bn ơi

a. \(\left(x^2-2x+1\right)-3x\left(x-1\right)=0\)

\(\Leftrightarrow x^2-2x+1-3x^2+3x=0\)

\(\Leftrightarrow-2x^2+x+1=0\)

\(\Leftrightarrow-2x^2+2x-x+1=0\)

\(\Leftrightarrow-2x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow-\left(2x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)

Vậy \(x\in\left\{-\frac{1}{2};1\right\}\)

b. \(4\left(7x-3\right)-\left(7x^2-3x\right)=0\)

\(\Leftrightarrow4\left(7x-3\right)-x\left(7x-3\right)=0\)

\(\Leftrightarrow\left(4-x\right)\left(7x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4-x=0\\7x-3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{3}{7}\end{cases}}\)

Vậy \(x\in\left\{4;\frac{3}{7}\right\}\)

c.\(\left(5-x\right)\left(2+3x\right)=4-9x^2\)

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

\(\Leftrightarrow\left(2+3x\right)\left(5-x-2+3x\right)=0\)

\(\Leftrightarrow\left(2+3x\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2+3x=0\\2x+3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=-\frac{3}{2}\end{cases}}\)

Vậy \(x\in\left\{-\frac{2}{3};-\frac{3}{2}\right\}\)

d. \(7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Leftrightarrow7-2x-4=-x-4\)

\(\Leftrightarrow7-4+4=-x+2x\)

\(\Leftrightarrow7=x\)

Vậy x = 7

e. \(\left(x-1\right)-\left(2x-1\right)=9\)

\(\Leftrightarrow x-1-2x+1=9\)

\(\Leftrightarrow-x=9\)

\(\Leftrightarrow x=-9\)

g. \(x^3+x^2+x+1=0\)

\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x+1=0\end{cases}}\)Mà : \(x^2+1\ge1>0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy x = -1

1) \(\left(x-3\right)^2-4=0\)

\(\Leftrightarrow\left(x-3-2\right)\left(x-3+2\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

2) \(x^2-2x=24\)

\(\Leftrightarrow x^2-2x-24=0\)

\(\Leftrightarrow x^2+4x-6x-24=0\)

\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

Câu 3 số xấu rồi e

alo =0 nhé anh pop pop

Lời giải:

$2x+xy-2y=7$

$x(2+y)-2y=7$

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

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

Do $x,y$ là số nguyên nên $x-2, y+2$ cũng là số nguyên. Do đó ta có bảng sau:

\(2x+xy-2y=7\)

\(\Rightarrow x\left(2+y\right)-2y-4+4=7\)

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

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

\(\Rightarrow\left(x-2\right);\left(y+2\right)\in\left\{-1;1;-3;3\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(1;-5\right);\left(3;1\right);\left(-1;-3\right);\left(5;-1\right)\right\}\left(x;y\inℤ\right)\)

Mik sẽ k cho bạn đó mik viết nhầm

\(A=\frac{x}{\sqrt{x}-1}+\frac{\sqrt{x}-2x}{x-\sqrt{x}}\)

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

bn làm tắt à

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

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

a: Diện tích hình thang ABCD là:

3,6*9,15=32,94m²

b: AB+CD=9,15*2=18,3(m)

CD-AB=7,5

=>CD=12,9m và AB=5,4m

AD=2/3DE

nên EA/ED=1/3

Xét ΔEDC có AB//DC

nên ΔEAB đồng dạng với ΔEDC

=>S EAB/S EDC=(EA/ED)^2=1/9

=>S EAB/S ABCD=1/8

=>S EAB=1/8*32,94=4,1175m²