Đặt \(\int f\left(x\right)dx=F\left(x\right)\Rightarrow\int\limits^{17}_1f\left(x\right)dx=F\left(17\right)-F\left(1\right)\)

Từ giả thiết:

\(2x.f\left(x^2+1\right)+\dfrac{f\left(\sqrt{x}\right)}{2\sqrt{x}}=2lnx\)

Lấy nguyên hàm 2 vế:

\(F\left(x^2+1\right)+F\left(\sqrt{x}\right)=2xlnx-2x+C\)

Thay \(x=4\):

\(F\left(17\right)+F\left(2\right)=16ln2-8+C\) (1)

Thay \(x=1\):

\(F\left(2\right)+F\left(1\right)=-2+C\) (2)

Trừ vế cho vế (1) cho (2):

\(F\left(17\right)-F\left(1\right)=16ln2-6\)

Vậy \(\int\limits^{17}_1f\left(x\right)dx=16ln2-6\)

Em cảm ơn thầy nhiều ạ 💕💕

\(\left(x^2-2x+3\right)\left(\frac{1}{2x}-5\right)\)

\(=\frac{x^2}{2x}-5x^2-\frac{2x}{2x}+10x+\frac{3}{2x}-15\)

\(=\frac{x^2}{2x}-5x^2-16+10x+\frac{3}{2x}\)

\(=-5x^2+\frac{x^2}{2x}+\frac{20x^2}{2x}+\frac{3}{2x}-16\)

\(=-5x^2+\frac{x^2+20x+3}{2x}-16\)

học tốt

(x^2-2x+3)(1/2x-5)=1/2x^3-5x^2-x^2+10x+3/2x-15=1/2x^3-6x^2+11,5x-15

a: \(\frac{2x}{3}:\frac{5}{6x^2}=\frac{2x}{3}\cdot\frac{6x^2}{5}=\frac{12x^3}{15}=\frac{4x^3}{5}\)

b: \(16x^2y^2:\left(-\frac{18x^2y^5}{5}\right)\)

\(=16x^2y^2\cdot\frac{-5}{18x^2y^5}=\frac{-80x^2y^2}{18x^2y^5}=\frac{-40}{9y^3}\)

c: \(\frac{25x^3y^5}{3}:15xy^2=\frac{25x^3y^5}{3\cdot15xy^2}=\frac{25x^3y^5}{45xy^2}=\frac59x^2y^3\)

d: \(\frac{x^2-y^2}{6x^2y}:\frac{x+y}{3xy}=\frac{\left(x-y\right)\left(x+y\right)}{6x^2y}\cdot\frac{3xy}{x+y}=\frac{x-y}{2x}\)

e: \(\frac{a^2+ab}{b-a}:\frac{a+b}{2a^2-2b^2}\)

\(=\frac{a\left(a+b\right)}{b-a}\cdot\frac{2\left(a^2-b^2\right)}{a+b}=\frac{a\cdot2\cdot\left(a-b\right)\left(a+b\right)}{b-a}=-2a\left(a+b\right)\)

f: \(\frac{x+y}{y-x}:\frac{x^2+xy}{3x^2-3y^2}\)

\(=\frac{-\left(x+y\right)}{x-y}\cdot\frac{3\left(x^2-y^2\right)}{x\left(x+y\right)}=\frac{-3\left(x-y\right)\left(x+y\right)}{x\left(x-y\right)}=\frac{-3\cdot\left(x+y\right)}{x}\)

g: \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}\cdot\frac{3x}{2\left(1-2x\right)}=\frac{3\left(1+2x\right)}{2\left(x+4\right)}\)

h: \(\frac{5x-15}{4x+4}:\frac{x^2-9}{x^2+2x+1}\)

\(=\frac{5\left(x-3\right)}{4\left(x+1\right)}\cdot\frac{\left(x+1\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{5\left(x+1\right)}{4\left(x+3\right)}\)

i: \(\frac{6x+48}{7x-7}:\frac{x^2-64}{x^2-2x+1}\)

\(=\frac{6\left(x+8\right)}{7\left(x-1\right)}\cdot\frac{\left(x-1\right)^2}{\left(x-8\right)\left(x+8\right)}=\frac{6\left(x-1\right)}{7\left(x-8\right)}\)

k: \(\frac{4x-24}{5x+5}:\frac{x^2-36}{x^2+2x+1}\)

\(=\frac{4\left(x-6\right)}{5\left(x+1\right)}\cdot\frac{\left(x+1\right)^2}{\left(x-6\right)\left(x+6\right)}=\frac{4\left(x+1\right)}{5\left(x+6\right)}\)

l: \(\frac{3x+21}{5x+5}:\frac{x^2-49}{x^2+2x+1}\)

\(=\frac{3\left(x+7\right)}{5\left(x+1\right)}\cdot\frac{\left(x+1\right)^2}{\left(x-7\right)\left(x+7\right)}=\frac{3\left(x+1\right)}{5\left(x-7\right)}\)

m: \(\frac{3-3x}{\left(1+x\right)^2}:\frac{6x^2-6}{x+1}\)

\(=\frac{-3\left(x-1\right)}{\left(x+1\right)^2}\cdot\frac{x+1}{6\left(x-1\right)\left(x+1\right)}=\frac{-3}{6\left(x+1\right)^2}=\frac{-1}{2\left(x+1\right)^2}\)

\(a,=x^2+x+4x+4=\left(x+1\right)\left(x+4\right)\\ b,=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\\ c,=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\\ d,=3\left(x^2-2x+5x-10\right)=3\left(x-2\right)\left(x+5\right)\\ e,=-3x^2+6x-x+2=\left(x-2\right)\left(1-3x\right)\\ f,=x^2-x-6x+6=\left(x-1\right)\left(x-6\right)\\ h,=4\left(x^2-3x-6x+18\right)=4\left(x-3\right)\left(x-6\right)\\ i,=3\left(3x^2-3x-8x+5\right)=3\left(x-1\right)\left(3x-8\right)\\ k,=-\left(2x^2+x+4x+2\right)=-\left(2x+1\right)\left(x+2\right)\\ l,=x^2-2xy-5xy+10y^2=\left(x-2y\right)\left(x-5y\right)\\ m,=x^2-xy-2xy+2y^2=\left(x-y\right)\left(x-2y\right)\\ n,=x^2+xy-3xy-3y^2=\left(x+y\right)\left(x-3y\right)\)

Bào quan riboxom trong chất tế bào có chức năng gì? 

1, are having

2,does your train leave

3,Does-begin

4,am going

does-start

5, does-open

Opens, finishes

6, are-finishing

Am not going, am going

7, Are-doing

Am going, arrives

8,are-going

Are learning, leaves

9 does-finish

Last

10 am meeting

Anh/chị ơi, em ko hiểu cho lắm ạ

Anh chị viết rõ giùm em đc ko ạ. Em cảm ơn nhiều ạ

Câu 10: B

dạ cảm ơn anh hoặc chị ạ.Anh hoặc chị có thể giải thích vì sao ra vậy ko ạ?

a,\(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)

PTHH: 2Na + 2H₂O → 2NaOH + H₂

Mol:     0,2                        0,2       0,1

\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b,m_NaOH=0,2.40=8 (g)

\(C\%_{ddNaOH}=\dfrac{8.100\%}{4,6+200-0,1.2}=3,91\%\)

Dạ em cảm rất nhiều ạ