Mình làm đc k nhỉ ?

GIẢI CÁI GÌ????????????????????????????????????????

ĐIÊN À????????????????????????????????????????????

dien tich tang them la:62853-60639=2214

chieu cao hinh binh hanh la:2214*2:18=246

do dai day manh vuon la:60639:246=246,5

các bạn làm sai rồi phải là 1973 mới đúng 

Ta có : \(\overline{abcdeg}=10000.\overline{ab}+100.\overline{cd}+\overline{eg}\)

\(=\left(9999+1\right).\overline{ab}+\left(99+1\right).\overline{cd}+\overline{eg}\)

\(=9999.\overline{ab}+\overline{ab}+99.\overline{cd}+\overline{cd}+\overline{eg}\)

\(=11.909.\overline{ab}+ab+11.9.\overline{cd}+\overline{cd}+\overline{eg}\)

Vì \(11.909.\overline{ab}⋮11;11.9.\overline{cd}⋮11;\overline{ab}+\overline{cd}+\overline{eg}⋮11\) nên \(\overline{abcdeg}⋮11\)

1+4+9+16+25+36+49+64+81+100 = 385

Với n = 1 , ta có : VT = VP = 1 (Luôn đúng)

Với n = 2 , ta có : VT = VP = 4 (mệnh đề đúng)

Giả sử mệnh đề đúng với n = k \(\ge\) 1 (k \(\in\) N*) , tức là :

1 + 3 + 5 + ... + 2k - 1 = k²

Ta phải CM mệnh đề đúng vs n = k + 1

Ta có : VT = 1 + 3 + 5 + (2k - 1) + (2(k + 1) - 1)

= (1 + 3 + 5 + 2k - 1) + (k + 1)

= k² + 2(k + 1) - 1

= k² + 2k + 2 - 1

= k² + 2k + 1

= (k + 1)²

= n² = VP (đpcm)

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

\(\Leftrightarrow4sinx+2cosx+2=sinx-2cosx+3\)

\(\Leftrightarrow3sinx+4cosx=1\)

\(\Leftrightarrow\dfrac{3}{5}sinx+\dfrac{4}{5}cosx=\dfrac{1}{5}\)

Đặt \(\left\{{}\begin{matrix}\dfrac{3}{5}=sin\varphi\\\dfrac{4}{5}=cos\varphi\end{matrix}\right.\)

\(pt\Leftrightarrow sin\varphi\cdot sinx+cos\varphi\cdot cos=\dfrac{1}{5}\)

\(\Leftrightarrow cos\cdot\left(\varphi-x\right)=\dfrac{1}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}\varphi-x=arc\cdot cos\dfrac{1}{5}+k2\pi\\\varphi-x=-arc\cdot cos\dfrac{1}{5}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\varphi+arc\cdot cos\dfrac{1}{5}+k2\pi\\x=\varphi-arc\cdot cos\dfrac{1}{5}+k2\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)

b, \(tanx-3cotx=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)

\(\Leftrightarrow\dfrac{sinx}{cosx}-\dfrac{3cos}{sinx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)

\(\Leftrightarrow\dfrac{sin^2x-3cos^2x}{sinx-cosx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)

\(\Leftrightarrow sin^2x-3cos^2x=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)

\(\Leftrightarrow\left(sinx-\sqrt{3}\cdot cosx\right)\cdot\left(sinx+\sqrt{3}\cdot cosx\right)=4\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)

\(\Leftrightarrow\left(sinx+\sqrt{3}\cdot cosx\right)\cdot\left[\left(sinx-\sqrt{3}\cdot cosx\right)-4sinx\cdot cosx\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+\sqrt{3}\cdot cosx=0\text{ (1) }\\sinx-\sqrt{3}\cdot cosx-4sinx\cdot cosx=0\text{ (2) }\end{matrix}\right.\)

(1) : \(sinx+\sqrt{3}\cdot cosx=0\)

\(\Leftrightarrow\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=0\)

\(\Leftrightarrow cos\dfrac{\pi}{3}\cdot sinx+sin\dfrac{\pi}{3}\cdot cosx=0\)

\(\Leftrightarrow sin\cdot\left(x+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow x+\dfrac{\pi}{3}=k\pi\Leftrightarrow x=\dfrac{-\pi}{3}+k\pi\left(k\in Z\right)\)

(2) : \(sinx-\sqrt{3}cosx-4sinx\cdot cosx=0\)

\(\Leftrightarrow sinx-\sqrt{3}cos=2sin2x\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cos2=sin2x\)

\(\Leftrightarrow cos\dfrac{\pi}{3}-sinx-sin\dfrac{\pi}{3}\cdot cosx=sin2x\)

\(\Leftrightarrow sin\cdot\left(x-\dfrac{\pi}{3}\right)=sin2x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=2x+k2\pi\\x-\dfrac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\left(k\in Z\right)\end{matrix}\right.\)

a, \(sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2cos^2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2\cdot\left[1+cos2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\right]=0\)

\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-1-cos\left(\dfrac{\pi}{2}-x\right)=0\)

\(\Leftrightarrow sin\dfrac{s}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x-sinx=0\)

\(\Leftrightarrow sinx\cdot\left(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\text{ (1) }\\sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\text{ (2) }\end{matrix}\right.\)

(1) : \(sinx=0\Leftrightarrow x=k\pi\left(k\in Z\right)\)

(2) : \(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-cos\dfrac{x}{2}\cdot2sin\dfrac{x}{2}\cdot cos\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot cos^2\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot\left(1-sin^2\dfrac{x}{2}\right)-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}+2sin^3\dfrac{x}{2}-1=0\)

\(\Leftrightarrow2sin^3\dfrac{x}{2}-sin\dfrac{x}{2}-1=0\)

\(\Leftrightarrow sin\dfrac{x}{2}=1\Leftrightarrow\dfrac{x}{2}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\pi+k4\pi\left(k\in Z\right)\)