Hỏi đáp môn Toán

\(\left(m+1\right)\left(2sin^2x\right)-2sin2x+2cos2x=0\)

\(\Leftrightarrow\left(m+1\right)\left(1-cos2x\right)-2sin2x+2cos2x=0\)

\(\Leftrightarrow2sin2x+\left(m-1\right)cos2x=m+1\)

Theo điều kiện có nghiệm của pt lượng giác bậc nhất:

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

\(\Leftrightarrow m\le1\)

\(\Rightarrow\) có \(1-\left(-2018\right)+1=2020\) giá trị nguyên của m

C1: a^4 + b^4 + 2 ≥ 4ab
<=> a^4 - 2a^2 + 1 + b^2 - 2b^2 + 1 + 2a^2 + 2b^2 + 4ab
<=> (a^2 - 1)^2 + (b^2 -1)^2 + 2( a^2 -2ab+ b^2)
<=> (a^2 -1)^2 + (b^2 -1)^2 + 2(a-b) >= 0 (với mọi a, b)
Vậy nên a^4 + b^4 + 2 ≥ 4ab (với mọi số nguyên a, b)

C2:Xét (a + b)^2 - 4ab
= a^2 + 2ab +b^2 - 4ab = a^2 - 2ab + b^2 = (a-b)^2 >= 0
=> (a+b)^2 >= 4ab
Mà ta có:
a^4 + b^4 + 2 - (a+b)^2
= a^4 + b^4 +2 -a^2 - b^2 - 2ab
=a^4 - 2a^2 + 1 + a^2 + b^4 - 2b^2 +1 + b^2 - 2ab
= (a^2 - 1)^2 + (b^2 - 1)^2 + (a-b)^2 >= 0
=> a^4 + b^4 +2 >= (a+b)^2
=> a^4 + b^4 +2 >= 4ab

bạn thấy cánh nào dễ hơn thì chọn nha

\(a^4+b^4+2\ge4ab\)

\(\Leftrightarrow a^4-2a^2b^2+b^4+2a^2b^2-4ab+2\ge0\)

\(\Leftrightarrow\left(a^2-b^2\right)^2+2\left(ab-1\right)^2\ge0^{\left(1\right)}\)

\(^{\left(1\right)}\) đúng vậy ta có đpcm

3b.

\(\left(2cosx-1\right)\left(2sinx+cosx\right)=2sinx.cosx-sinx\)

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

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

\(\Leftrightarrow\left(2cosx-1\right).\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{4}+k\pi\end{matrix}\right.\)

3.

a.

\(\Leftrightarrow\left(cos3x-cosx\right)+\left(cos2x-1\right)=0\)

\(\Leftrightarrow-2sin2x.sinx+1-2sin^2x-1=0\)

\(\Leftrightarrow sin2x.sinx+sin^2x=0\)

\(\Leftrightarrow2sin^2x.cosx+sin^2x=0\)

\(\Leftrightarrow sin^2x\left(2cosx+1\right)=0\)

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

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

ĐKXĐ: ...

\(\Leftrightarrow\frac{3x-3\sqrt{3}y-2x-2\sqrt{3}y}{x^2-3y^2}=7-20\sqrt{3}\)

\(\Leftrightarrow x-5\sqrt{3}y=\left(7-20\sqrt{3}\right)\left(x^2-3y^2\right)\)

\(\Leftrightarrow x-5\sqrt{3}y=7x^2-21y^2-20\sqrt{3}\left(x^2-3y^2\right)\)

\(\Leftrightarrow7x^2-21y^2-x=5\sqrt{3}\left(4x^2-12y^2-y\right)\)

Do \(5\sqrt{3}\) vô tỉ còn các đại lượng khác hữu tỉ nên đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}7x^2-21y^2-x=0\\4x^2-12y^2-y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}28\left(x^2-3y^2\right)-4x=0\\28\left(x^2-3y^2\right)-7y=0\end{matrix}\right.\)

\(\Rightarrow4x-7y=0\Rightarrow y=\frac{4}{7}x\)

Thay vào pt đầu:

\(7x^2-21\left(\frac{4}{7}x\right)^2-x=0\)

\(\Leftrightarrow\frac{x^2}{7}-x=0\Rightarrow\left[{}\begin{matrix}x=y=0\left(ktm\right)\\x=7\Rightarrow y=4\end{matrix}\right.\)

\(6x^4+5x^3-38x^2+5x+6=0\)

\(6x^4-12x^3+17x^3-34x^2-4x^2+8x-3x+6=0\)

\(6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)=0\)

\(\left(x-2\right)\left(6x^3+17x^2-4x-3\right)=0\)

\(\left(x-2\right)\left[6x^3-3x^2+20x^2-10x+6x-3\right]=0\)

\(\left(x-2\right)\left[6x^2\left(x-\dfrac{1}{2}\right)+20x\left(x-\dfrac{1}{2}\right)+6\left(x-\dfrac{1}{2}\right)\right]=0\)

\(\left(x-2\right)\left(x-\dfrac{1}{2}\right)\left(6x^2+20x+6\right)=0\)

=> \(\left[{}\begin{matrix}x-2=0\\x-\dfrac{1}{2}=0\\6x^2+20x+6=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\\\left(3x+1\right)\left(x+3\right)=0\end{matrix}\right.\)

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

Do tam giác SAB đều \(\Rightarrow SI\perp AB\Rightarrow SI\perp\left(ABC\right)\)

\(SI=\frac{a\sqrt{3}}{2}\) ; \(BC=AB.tan60^0=a\sqrt{3}\)

\(V_{S.ABC}=\frac{1}{3}SI.\frac{1}{2}AB.BC=\frac{a^3}{4}\)

Qua S kẻ đường thẳng vuông góc SB cắt AB kéo dài tại P

\(\Rightarrow IH//SP\Rightarrow IH//\left(SCP\right)\Rightarrow d\left(IH;SC\right)=d\left(IH;\left(SCP\right)\right)=d\left(I;\left(SCP\right)\right)\)

\(BP=\frac{SB}{cos60^0}=2a\)

Từ I kẻ \(IQ\perp CP\) , từ \(I\) kẻ \(IK\perp SQ\Rightarrow IK\perp\left(SCP\right)\)

\(\Rightarrow IK=d\left(I;\left(SCP\right)\right)\)

\(IP=BP-IB=\frac{3a}{2}\) ; \(IQ=IP.sin\widehat{BPC}=IP.\frac{BC}{CP}=\frac{IP.BC}{\sqrt{BP^2+BC^2}}=\frac{3a\sqrt{21}}{14}\)

\(\frac{1}{IK^2}=\frac{1}{SI^2}+\frac{1}{IQ^2}\Rightarrow IK=\frac{SI.IQ}{\sqrt{SI^2+IQ^2}}=\frac{3a\sqrt{3}}{8}\)

Câu a tiếp tục ko dịch được đề :)

b.

\(\Leftrightarrow1+cos3x=sin^2\frac{x}{2}+cos^2\frac{x}{2}+2sin\frac{x}{2}.cos\frac{x}{2}\)

\(\Leftrightarrow1+cos3x=1+sinx\)

\(\Leftrightarrow cos3x=sinx\)

\(\Leftrightarrow cos3x=cos\left(\frac{\pi}{2}-x\right)\)

\(\Leftrightarrow...\)

c.

\(\Leftrightarrow cos8x+cos2x+sinx=cos8x\)

\(\Leftrightarrow cos2x+sinx=0\)

\(\Leftrightarrow cos2x=-sinx\)

\(\Leftrightarrow cos2x=cos\left(\frac{\pi}{2}+x\right)\)

\(\Leftrightarrow...\)

d.

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

\(\Leftrightarrow sin\left(3x-\frac{\pi}{6}\right)=-cosx\)

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

\(\Leftrightarrow...\)

\(A=1+x^4+y^4+x^4y^4=1+\left(x^2+y^2\right)^2-2x^2y^2+x^4y^4\)

Đặt \(xy=t\Rightarrow\left\{{}\begin{matrix}0< t\le\frac{1}{4}\left(x+y\right)^2=\frac{5}{2}\\x^2+y^2=\left(x+y\right)^2-2xy=10-2t\end{matrix}\right.\)

\(\Rightarrow A=1+\left(10-2t\right)^2-2t^2+t^4\)

\(A=t^4+2t^2-40t+101=\left(t-2\right)^2\left(t^2+4t+14\right)+45\ge45\)

\(A_{min}=45\) khi \(t=2\)

a/

\(y'=-\frac{4}{\left(x-2\right)^2}\Rightarrow\left\{{}\begin{matrix}y'\left(3\right)=-4\\y\left(3\right)=6\end{matrix}\right.\)

Pt tiếp tuyến: \(y=-4\left(x-3\right)+6\Leftrightarrow y=-4x+18\)

b.

\(y'=\frac{-5}{\left(x-1\right)^2}\)

Tiếp tuyến song song với \(y=-5x-3\) nên có hệ số góc \(k=-5\)

\(\Rightarrow\frac{-5}{\left(x-1\right)^2}=-5\Rightarrow\left(x-1\right)^2=1\Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

Có 2 tiếp tuyến thỏa mãn

\(\dfrac{\left(\sqrt{X}+\sqrt{Y}\right)\left(1+\sqrt{XY}\right)+\left(\sqrt{X}-\sqrt{Y}\right)\left(1-\sqrt{XY}\right)}{1-XY}\cdot\dfrac{1-XY}{1-XY+\sqrt{X}+\sqrt{Y}+2\sqrt{XY}}=\dfrac{\sqrt{X}+X\sqrt{Y}+\sqrt{Y}+Y\sqrt{X}+\sqrt{X}-X\sqrt{Y}-\sqrt{Y}+Y\sqrt{X}}{1-XY}\cdot\dfrac{1-XY}{XY+X+Y+1}=\dfrac{2\sqrt{X}\left(1+Y\right)}{\left(1+Y\right)\left(X+1\right)}=\dfrac{2\sqrt{X}}{X+1}\)