Làm thế này nhé 
(2x-6)y+x-3=18

(2x-6)y+x-18-3=0

(2x-6)y+x-21=0

sau đó bạn rút gọn nhé, ta làm như sau

2(x-3)=0

=> 2x= 2 x 3

=> x=3 nhé

giúp mình llàm câu b với nhé đã k cho bạn rồi nha

b, y = \(\frac{23}{x+7}\)

câu nào

cả 2

\(x^2+2x+1=x^2+2\cdot1x+1^2=\left(x+1\right)^2\)

\(4x^2+12x+9=\left(2x\right)^2+2\cdot3\cdot2x+3^2=\left(2x+3\right)^2\)

\(\dfrac{4}{9}a^2-\dfrac{4}{3}a+1=\left(\dfrac{2}{3}a\right)^2-2\cdot\dfrac{2}{3}\cdot1a+1^2=\left(\dfrac{2}{3}a-1\right)^2\)

\(a^2+5a+\dfrac{25}{4}=a^2+2\cdot2,5a+2,5^2=\left(2,5+a\right)^2\)

1: \(\Leftrightarrow x-2-7x+7=-1\)

=>-6x+5=-1

hay x=1(loại)

3: \(\Leftrightarrow\left(x+2\right)\left(x-1\right)-\left(x+1\right)\left(x+3\right)=4\)

\(\Leftrightarrow x^2+x-2-x^2-4x-3=4\)

=>-3x=9

hay x=-3(loại)

4: \(\Leftrightarrow x^2+2x+1-x^2+2x-1=3x\cdot\dfrac{x+1-x+1}{x+1}\)

\(\Leftrightarrow4x=\dfrac{6x}{x+1}\)

\(\Leftrightarrow4x^2+4x-6x=0\)

\(\Leftrightarrow4x^2-2x=0\)

=>2x(2x-1)=0

hay \(x\in\left\{0;\dfrac{1}{2}\right\}\)

Bài 1:

a) (2x+5)(x-6)=2x²+5x-12x-30=2x²-7x-30

b) (2x-1)(x²-4x+3)=2x³-8x²+6x-x²+4x-3=2x³-9x²+10x-3

c) x²-2x-(x-7)(x+2)=x²-2x-x²+7x-2x+14=3x+14

d) 3x-(x+2)(x+4)=3x-x²-2x-4x-8=-x²-3x-8

Bài 2:

a) 2(x+1)=x-1

⇒2x+2=x-1

⇒2x+2-x+1=0

⇒x+3=0

⇒x=-3

b) x(x+2)-x²=1

⇒x²+2x-x²=1

⇒2x=1

⇒x=0,5

c) 3x(x-2)=(3x-1)(x-1)-5

⇒3x²-6x=3x²-x-3x+1-5

⇒3x²-6x-3x²+x+3x-1+5=0

⇒-2x+4=0

⇒-2x=-4

⇒x=2

d) 6(x-1)(x-2)-6x(x+3)=2x

⇒6(x²-x-2x+2)-6x²-18x-2x=0

⇒6x²-6x-12x+12-6x²-18x-2x=0

⇒-38x+12=0

⇒-38x=-12

⇒x=\(\dfrac{6}{19}\)

Câu nào bạn :

đâu

\(70:\frac{2}{7}=\frac{70}{1}:\frac{2}{7}=\frac{70}{1}\times\frac{7}{2}=\frac{490}{2}=\frac{235}{1}=235\)

con rồng 

đừng hỏi tại sao me trả lời như vậy

con... MA???

Okie, xinh nên giúp :3 Đùa thui

a/ 5 nguồn mắc nối tiếp \(\left\{{}\begin{matrix}\xi_b=5.\xi=5.4=20\left(V\right)\\r_b=5r=5.0,2=1\left(\Omega\right)\end{matrix}\right.\)

b/ \(R_D=\dfrac{U^2_{dm}}{P_{dm}}=\dfrac{36}{6}=6\left(\Omega\right);I_{dm}=\dfrac{P_{dm}}{U_{dm}}=\dfrac{6}{6}=1\left(A\right)\)

Đèn sáng bình thường \(\Rightarrow I_2=I_D=I_{dm}=1\left(A\right)\)

\(\left(R_1ntR_B\right)//\left(R_2ntR_D\right)\Rightarrow R_{td}=\dfrac{\left(R_1+R_B\right)\left(R_2+R_D\right)}{R_1+R_B+R_2+R_D}=\dfrac{\left(2+4\right)\left(6+6\right)}{2+4+6+6}=4\left(\Omega\right)\)

c/ \(I=\dfrac{\xi_b}{r_b+R_{td}}=\dfrac{20}{1+4}=4\left(A\right)\)

\(I=I_1+I_2\Rightarrow I_1=I-I_2=4-1=3\left(A\right)\Rightarrow P_1=I_1^2.R_1=3^2.2=18\left(W\right)\)

\(m_{Cu}=\dfrac{A_{Cu}.I_B.t}{F.n}=\dfrac{64.3.\left(32.60+10\right)}{96500.2}=...\left(g\right)\)

ui cảm ơn cậu nhiều nhaaaa