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15 tháng 11 2021

a) \(=5x\left(y+2x\right)\)

b) \(=x\left(x-6y\right)+3\left(x-6y\right)=\left(x-6y\right)\left(x+3\right)\)

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

d) \(=\left(x^2-5x\right)-\left(4x-20\right)=x\left(x-5\right)-4\left(x-5\right)=\left(x-4\right)\left(x-5\right)\)

15 tháng 11 2021

a) 5x( y+ 2)

b)(x2+3x)-(6xy+ 18y)

=x(x+3)-6y(x+3)

=(x+3)(x-6y)

c)(x2-y2)+ (6x+9)

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

d) chịu  nha

 

29 tháng 7 2021

a, \(f\left(x\right)=2x^4-x^3+4x^2-x\)

\(\Rightarrow f'\left(x\right)=\left(2x^4-x^3+4x^2-x\right)'\)

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

\(=2.4x^3-3x^2+4.2x-1\)

\(=8x^3-3x^2+8x-1\)

b, \(f\left(x\right)=2sinx\)

\(\Rightarrow f'\left(x\right)=\left(2sinx\right)'=2cosx\)

c, \(f\left(x\right)=\dfrac{3x^2+2x-5}{x}\)

\(\Rightarrow f'\left(x\right)=\left(\dfrac{3x^2+2x-5}{x}\right)'\)

\(=\left(3x+2-\dfrac{5}{x}\right)'\)

\(=\left(3x\right)'+\left(2\right)'-\left(\dfrac{5}{x}\right)'\)

\(=3+0+\dfrac{5}{x^2}=\dfrac{5}{x^2}+3\)

2 tháng 1 2022

1, The homework is too difficult for him to do

2, The homework isn't easy enough for him to do

3, The homework is so difficult that he can't do it

4, It is such difficult homework that he can't do it

7 tháng 1 2022

1. My sister says she went to school by bus this morning.

    My sister said she had gone to school by bus that morning.

2. Nga says she have done her homework.

    Nga said she had done her homework.

7 tháng 1 2022

1, My sister says that she goes to school by bus this morning

My sister said that she had gone to school by bus this morning

2, Nga says that she have done her homework

Nga said that she had done her homework

2 tháng 1 2022

is a big supermarket next to our school

not met him for 3 months

10-day Tet holiday

30 tháng 8 2022

Bài 1: \(n_{CuO}=\dfrac{3,2}{80}=0,04\left(mol\right)\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

0,04  →  0,04

\(\Rightarrow m_{H_2SO_4}=0,04\cdot98=3,92\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{3,92}{80}\cdot100\%=4,9\%\)

Bài 2: \(n_{HCl}=0,2\cdot2=0,4\left(mol\right)\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2\uparrow\)

\(\dfrac{1}{15}\)     ←    0,4

\(\Rightarrow m_{Fe_2O_3}=\dfrac{1}{15}\cdot160=\dfrac{32}{3}\left(g\right)\)

24 tháng 12 2021

1.........................................

=>the computer hasn't been fixed by them yet

2.......................................

=>Last night I was helped by a stranger.

 

17 tháng 1 2022

\(PT\Leftrightarrow\left(3x+5\right)\dfrac{x+1}{2}+\left(3x+5\right)\dfrac{2x-2}{3}-\left(3x+5\right)=0\)

\(\Leftrightarrow\left(3x+5\right)\left(\dfrac{x+1}{2}+\dfrac{2x-2}{3}-1\right)=0\)

\(\Leftrightarrow\left(3x+5\right)\dfrac{3\left(x+1\right)+2\left(2x-2\right)-6}{6}=0\)

\(\Leftrightarrow\left(3x+5\right)\dfrac{7x-7}{6}=0\)

\(\Leftrightarrow\left(3x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{3}\\x=1\end{matrix}\right.\)

KL: Phương trình có tập nghiệm \(S=\left\{\dfrac{-5}{3};1\right\}\)

17 tháng 1 2022

Bn oie còn 2 câu dưới,bn giúp mik luôn đc k?

 

23 tháng 10 2023

3: \(\left(\dfrac{1}{\sqrt{2}-1}-\dfrac{1}{\sqrt{2}+1}\right):\sqrt{3-2\sqrt{2}}\)

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

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

5: 

\(\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)

\(=\dfrac{\left(\sqrt{5}+\sqrt{3}\right)^2+\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)

\(=\dfrac{8+2\sqrt{15}+8-2\sqrt{15}}{2}=\dfrac{16}{2}=8\)

6:

\(\dfrac{\sqrt{5}-1}{\sqrt{5}+1}-\dfrac{\sqrt{5}+1}{\sqrt{5}-1}\)

\(=\dfrac{\left(\sqrt{5}-1\right)^2-\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)

\(=\dfrac{6-2\sqrt{5}-6-2\sqrt{5}}{4}=\dfrac{-4\sqrt{5}}{4}=-\sqrt{5}\)

4:

\(\dfrac{6}{\sqrt{2}-\sqrt{3}+3}\)

\(=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{\left(\sqrt{2}-\sqrt{3}-3\right)\left(\sqrt{2}-\sqrt{3}+3\right)}\)

\(=\dfrac{6\left(\sqrt{2}-\sqrt{3}-3\right)}{5-2\sqrt{6}-9}=\dfrac{-6\left(\sqrt{2}-\sqrt{3}-3\right)}{4+2\sqrt{6}}\)

\(=\dfrac{-3\left(\sqrt{2}-\sqrt{3}-3\right)}{\sqrt{6}+2}=\dfrac{-3\left(\sqrt{2}-\sqrt{3}-3\right)\left(\sqrt{6}-2\right)}{2}\)

\(=\dfrac{-3\left(2\sqrt{3}-2\sqrt{2}-3\sqrt{2}+2\sqrt{3}-6\sqrt{3}+6\right)}{2}\)

\(=\dfrac{-3\left(-2\sqrt{3}-5\sqrt{2}+6\right)}{2}\)