Gọi vận tốc ca nô là x ( x > 0 ) 

Theo bài ra ta có pt \(\dfrac{72}{x+3}+\dfrac{54}{x-3}=6\Rightarrow x=21\left(tm\right)\)

1.

a, \(sin2x-\sqrt{3}cos2x=-1\)

\(\Leftrightarrow\dfrac{1}{2}sin2x-\dfrac{\sqrt{3}}{2}cos2x=-\dfrac{1}{2}\)

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)=-\dfrac{1}{2}\)

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{3}\right)=sin\left(-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=-\dfrac{\pi}{6}+k2\pi\\2x-\dfrac{\pi}{3}=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{3\pi}{4}+k\pi\end{matrix}\right.\)

Do tổng các hệ số thứ 1,2,3 là 46 nên ta có:\(C_n^0+C_n^1+C_n^2=46\)

\(\Leftrightarrow1+\dfrac{n!}{1!\left(n-1\right)!}+\dfrac{n!}{2!\left(n-2\right)!}=46\)

\(\Leftrightarrow1+n+\dfrac{\left(n-1\right)n}{2}=46\)

\(\Leftrightarrow n^2+n-90=0\)

\(\Leftrightarrow\left[{}\begin{matrix}n=9\\n=-10\left(loai\right)\end{matrix}\right.\)

Khai triển biểu thức: \(\left(x+\dfrac{1}{x}\right)^9\)

Hạng tử thứ k+1 trong biểu thức trên

\(\left(x+\dfrac{1}{x}\right)^9=C_9^{k+1}+\left(x^2\right)^{10-k}.\left(\dfrac{1}{x}\right)^{k+1}\)

đến đây mình chịu rùi hjhj b nào làm được giúp b kia với

1.

b, \(cos^2\left(x+\dfrac{\pi}{3}\right)+cos^2\left(x+\dfrac{2\pi}{3}\right)=\dfrac{1}{2}\left(sinx+1\right)\)

\(\Leftrightarrow cos^2\left(x+\dfrac{\pi}{3}\right)+cos^2\left(x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\left(sinx+1\right)\)

\(\Leftrightarrow2cos^2\left(x+\dfrac{\pi}{3}\right)-1+2cos^2\left(x-\dfrac{\pi}{3}\right)-1=sinx-1\)

\(\Leftrightarrow cos\left(2x+\dfrac{2\pi}{3}\right)+cos\left(2x-\dfrac{2\pi}{3}\right)=sinx-1\)

\(\Leftrightarrow2cos2x.cos\dfrac{2\pi}{3}=sinx-1\)

\(\Leftrightarrow-cos2x=sinx-1\)

\(\Leftrightarrow2sin^2x-sinx=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

Đến đây dễ rồi, tự làm tiếp.

c: Ta có: \(\dfrac{1}{x^2+x+1}-\dfrac{1}{x-x^2}+\dfrac{2x}{1-x^3}\)

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

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

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

Đây bạn nhé!

a) Do \(\left(3x-\dfrac{1}{2}\right)^2\ge0\forall x\)

 \(\Rightarrow A=\left(3x-\dfrac{1}{2}\right)^2-4\ge-4\)

\(minA=-4\Leftrightarrow x=\dfrac{1}{6}\)

b) Do \(\left(2x+1\right)^4\ge0\forall x,\left(y-\dfrac{1}{2}\right)^6\ge0\forall y\)

\(\Rightarrow B=\left(2x+1\right)^4+3\left(y-\dfrac{1}{2}\right)^6\ge0\)

\(minB=0\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\)

a: \(A=\left(3x-\dfrac{1}{2}\right)^2-4\ge-4\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{6}\)

b: \(B=\left(2x+1\right)^4+3\left(y-\dfrac{1}{2}\right)^6\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\left(x,y\right)=\left(-\dfrac{1}{2};\dfrac{1}{2}\right)\)

ko có ảnh bn ơi

Đề bài là gì zạ? 

Program HOC24;

var t: longint;

n,i: integer;

begin

write('Nhap n= '); readln(n);

t:=1;

for i:=1 to n do t:=t*i;

write(n,'! =',t);

readln

end.

a) Cl₂ + 2NaOH --> NaClO + NaCl + H₂O

Chất oxh: Cl₂, chất khử: Cl₂

b) \(n_{Cl_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right);n_{NaOH}=0,5.4=2\left(mol\right)\)

PTHH: Cl₂ + 2NaOH --> NaClO + NaCl + H₂O

_____0,8---->1,6--------->0,8---->0,8

=> \(\left\{{}\begin{matrix}C_{M\left(NaCl\right)}=\dfrac{0,8}{0,5}=1,6M\\C_{M\left(NaClO\right)}=\dfrac{0,8}{0,5}=1,6M\\C_{M\left(NaOH\right)}=\dfrac{2-1,6}{0,5}=0,8M\end{matrix}\right.\)