nhỏ quá bn !

\(4x-2y-4x^2+4xy-y^2\)

\(=\left(4x-2y\right)-\left(4x^2-4xy+y^2\right)\)

\(=2\left(2x-y\right)-\left(2x-y\right)^2\)

\(=\left(2x-y\right)\left(2-2x+y\right)\)

9: Ta có: \(4x-2y-4x^2+4xy-y^2\)

\(=2\left(2x-y\right)-\left(2x-y\right)^2\)

\(=\left(2x-y\right)\left(2-2x+y\right)\)

KT ???

12.

\(y=\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)\le\sqrt[]{2}\)

\(\Rightarrow M=\sqrt{2}\)

13.

Pt có nghiệm khi:

\(5^2+m^2\ge\left(m+1\right)^2\)

\(\Leftrightarrow2m\le24\)

\(\Rightarrow m\le12\)

14.

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

\(\Leftrightarrow x=k2\pi\)

15.

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

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

Đáp án A

16.

\(\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=\dfrac{1}{2}\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

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

\(\left[{}\begin{matrix}2\pi\le\dfrac{\pi}{3}+k2\pi\le2018\pi\\2\pi\le\pi+k2\pi\le2018\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}1\le k\le1008\\1\le k\le1008\end{matrix}\right.\)

Có \(1008+1008=2016\) nghiệm

a: \(=24\left(25+37+38\right)=2400\)

\(3.25.8+4.37.6+2.38.12\\ =\left(3.8\right).25.\left(4.6\right).37.\left(2.12\right).38\\ =24.25.24.37.24.38\\ =24.\left(25+37+38\right)\\ =24.100\\ =2400\)

ko bt đúng ko !

bạn có tham khảo thôi nhé !

a: \(=24\left(25+37+38\right)=2400\)

a) 

136.

$48+16.272+68.20.2$

$=136.48+16.2.136+68.2.20$

$=136.48+32.136+136.20$

$=136.(48+32+20)$

$=136.100$

$=13600$

23.

\(2sin^2x+5sinx-3=0\Rightarrow\left[{}\begin{matrix}sinx=\dfrac{1}{2}\\sinx=-3\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{5}+k2\pi\end{matrix}\right.\)

Nghiệm dương bé nhất là \(x=\dfrac{\pi}{6}\)

24.

\(1-cos^2x-3cosx-4=0\)

\(\Leftrightarrow cos^2x+3cosx+3=0\)

Pt bậc 2 nói trên vô nghiệm nên pt đã cho vô nghiệm

25.

\(\Leftrightarrow\left(tanx+1\right)^2=0\)

\(\Leftrightarrow tanx=-1\)

\(\Rightarrow x=-\dfrac{\pi}{4}+k\pi\)

26.

\(\Leftrightarrow\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=1\)

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

\(\Leftrightarrow x+\dfrac{\pi}{3}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+k2\pi\)

28.

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

\(\Leftrightarrow2sin^2x+sinx-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{3}{2}\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)

Câu 3:

<tóm tắt bạn tự làm>

MCD:R1ntR2ntR3

Điện trở R3 là

ta có:\(R_{tđ}=R_1+R_2+R_3\Rightarrow R_3=R_{tđ}-R_1-R_2=55-15-30=10\left(\Omega\right)\)