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.

(145x99+145)-(143x101-143)

=145x(99+1)-143x(101-1)

=145x100-143x100

=14500-14300

=200

=145x(99+1)-143x(101-1)

=145x100-143x100

=14500-14300

=200

KHO THE

\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)

\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)

\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)

BTVN Nâng cao thì tự làm hoặc là  ghi không biết vào đó

Chứ t chịu

Bài 4: 

a) Ta có: \(\dfrac{x}{13}=\dfrac{35}{91}\)

\(\Leftrightarrow\dfrac{x}{13}=\dfrac{5}{13}\)

hay x=5

Vậy: x=5

b) Ta có: \(\dfrac{9+x}{13-x}=\dfrac{5}{6}\)

\(\Leftrightarrow6\left(x+9\right)=5\left(13-x\right)\)

\(\Leftrightarrow6x+54=65-5x\)

\(\Leftrightarrow6x+5x=65-54\)

\(\Leftrightarrow11x=11\)

hay x=1

Vậy: S={1}

c) Ta có: \(x+x:5\cdot7.5+x:2\cdot9=315\)

\(\Leftrightarrow x+\dfrac{1}{5}x\cdot\dfrac{15}{2}+\dfrac{1}{2}x\cdot9=315\)

\(\Leftrightarrow7x=315\)

hay x=45

Vậy: x=45

câu 18

=6,46 m²

câu 19:

=150 l

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

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