1: \(=\left(y-1\right)^2\)

2: \(=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)

3: =(1-2x)(1+2x)

\(=\left(2-3x\right)\left(4+6x+9x^2\right)\)

5: \(=\left(x+3\right)^3\)

6: \(=\left(2x-y\right)^3\)

\(1.1-2y+y^2=\left(1-y\right)^2\\ 2.\left(x+1\right)^2-25=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\\ 3.1-4x^2=\left(1-2x\right)\left(1+2x\right)\\ 4.8-27x^3=\left(2-3x\right)\left(4+6x+9x^2\right)\\ 5.27+27x+9x^2+x^3=\left(3+x\right)^3\\ 6.8x^3-12x^2y+6xy^2-y^3=\left(2x-y\right)^3\\ 7.x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\\ 8.3x^2-6x+9x^2=12x^2-6x=6x\left(2x-1\right)\\ 9.10x\left(x-y\right)-6y\left(y-x\right)=10x\left(x-y\right)+6y\left(x-y\right)=\left(10x+6y\right)\left(x-y\right)=2\left(5x+3y\right)\left(x-y\right)\)

\(10.3x^2+5y-3xy-5x=\left(3x^2-3xy\right)-\left(5x-5y\right)=3x\left(x-y\right)-5\left(x-y\right)=\left(3x-5\right)\left(x-y\right)\)

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2abc=abc x 9

2000+abc=abc x 9

2000=abc x 9-abc

2000=abc x (9-1)

2000=abc x 8

abc=2000:8

abc=250

bằng không

abc*9=2000+abc

abc*9-abc=2000

abc*(9-1)=2000

abc*8=2000

abc=2000/8

abc=250

Vậy abc=250

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

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

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

sẽ không cứu ai cả

cuu me 

cứu mẹ là đúng nhất .

\(y=tan\left(\sqrt{x^2+4}\right)\Rightarrow y'=\dfrac{1}{cos^2\left(\sqrt{x^2+4}\right)}.\left(\sqrt{x^2+4}\right)'\)

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

Suy ra : \(y'=\dfrac{x}{cos^2\left(\sqrt{x^2+4}\right).\sqrt{x^2+4}}\)

2^x=25-3^9/3^7

2^x=16

2^x=2⁴

x=4

thank

\(\sqrt[n]{y}=4x+1\)

\(y^{\dfrac{1}{n}}=4x+1\)

đạo cấp 1

\(\dfrac{1}{n}y^{\left(\dfrac{1}{n}-1\right)}=\dfrac{1}{n}\sqrt[n]{y^{\left(1-n\right)}}=4\)

thay y=(4x+1)^n vào

\(\dfrac{1}{n}\sqrt[n]{\left(4x+1\right)^{n\left(1-n\right)}}=\dfrac{1}{n}\left(4x+1\right)^{\left(1-n\right)}\)

từ đó: \(y'=\dfrac{4}{\dfrac{1}{n}\left(4x+1\right)^{\left(1-n\right)}}=4.n\left(4x+1\right)^{n-1}\)

Có đúng không: cấp n có thể phải làm lấy vài cái--> quy luật nào đó