Bạn thêm nhầm tag môn học rồi á

Ta có: \(\omega=\sqrt{\dfrac{g}{l}}=\pi\left(\dfrac{rad}{s}\right)\)

Phương trình dao động: \(x=14,3cos\left(\pi t-\dfrac{\pi}{2}\right)\) (cm)

Thay t=0,25 ta được x=10,11cm

\(\Rightarrow tan\varphi=\dfrac{x}{l}\Rightarrow\varphi\approx5,8^0\)

Chọn D

a. ĐKXĐ: \(x\ge0\)

\(\sqrt{x}-\sqrt{x\left(x+2\right)}=0\Leftrightarrow\sqrt{x}=\sqrt{x\left(x+2\right)}\)

\(\Rightarrow x=x\left(x+2\right)\Leftrightarrow x=x^2+2x\)\(\Rightarrow\left[{}\begin{matrix}x=-1\left(KTM\right)\\x=0\left(TM\right)\end{matrix}\right.\)

b. ĐKXĐ: \(x\ge1\)

\(\sqrt{x^2-1}-\sqrt{x-1}=0\Leftrightarrow\sqrt{x^2-1}=\sqrt{x-1}\)

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

Khi A=B thì \(\left(2x-x^2\right)\left(2x^2-3x-2\right)=\left(2x^2+x\right)\left(3x-12m\right)\)

\(\Leftrightarrow x\left(2-x\right)\left(x-2\right)\left(2x+1\right)=x\left(2x+1\right)\left(3x-12m\right)\)

\(\Leftrightarrow x\left(2x+1\right)\left(3x-12m+x^2-4x+4\right)=0\)

\(\Leftrightarrow x\left(2x+1\right)\left(x^2-x+4-12m\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x+1=0\\x^2-x+4-12m=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\\m=\dfrac{x^2-x+4}{12}\left(\forall x\in R\right)\end{matrix}\right.\)

56=2³.7

84=2².3.7

UCLN(56,84)=2².7=28

a. \(-2x+3\ge0\Leftrightarrow x\le\dfrac{3}{2}\)

b.\(x\ne0\)

c.\(x+3>0\Leftrightarrow x>-3\)

d.\(x^2+6< 0\Leftrightarrow x^2< -6\) (Vô lí, không có giá trị x nào thỏa mãn)

\(\sqrt{sin^2x-2sinx+2}=2sinx-1\)

\(\Leftrightarrow sin^2x-2sinx+2=4sin^2x-4sinx+1\) \(\left(sinx\ge\dfrac{1}{2}\right)\)

\(\Leftrightarrow3sin^2x-2sinx-1=0\)

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

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

Ta có: \(y'=\left(m+1\right)x^2+2\left(2m-1\right)x-\left(3m+2\right)\)

Để hàm số nghịch biến trên khoảng có độ dài bằng 4 thì y'=0 phải có hai nghiệm phân biệt, (m+1)>0 và \(\left|x_1-x_2\right|=4\)

Ta có: m+1>0 \(\Rightarrow m>-1\) (1)

TH1: m=-1

\(\Rightarrow y'=-6x+1=0\Rightarrow x=\dfrac{1}{6}\) (loại)

TH2: \(m\ne-1\)

\(\Rightarrow\Delta=4\left(2m-1\right)^2+4\left(3m+2\right)\left(m+1\right)\)

\(\Leftrightarrow\Delta=4\left(4m^2-4m+1\right)+4\left(3m^2+3m+2m+2\right)\)

\(\Leftrightarrow\Delta=16m^2-16m+4+12m^2+20m+8\)

\(\Leftrightarrow\Delta=28m^2+4m+12>0\forall m\in R\)

Theo định lí Vi-et ta có: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-2\left(2m-1\right)}{\left(m+1\right)}\\x_1x_2=\dfrac{-\left(3m+2\right)}{\left(m+1\right)}\end{matrix}\right.\)

Ta có: \(\left|x_1-x_2\right|=4\Rightarrow\left(x_1-x_2\right)^2=16\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=16\)

\(\Leftrightarrow\left(\dfrac{-2\left(2m-1\right)}{m+1}\right)^2+\dfrac{4\left(3m+2\right)}{m+1}=16\)

\(\Leftrightarrow\dfrac{4\left(2m-1\right)^2}{\left(m+1\right)^2}+\dfrac{12m+8}{m+1}=16\)

\(\Leftrightarrow\dfrac{4\left(4m^2-4m+1\right)}{\left(m+1\right)^2}+\dfrac{12m+8}{m+1}=16\)

\(\Leftrightarrow\dfrac{16m^2-16m+4}{\left(m+1\right)^2}+\dfrac{12m+8}{m+1}=16\)

\(\Leftrightarrow16m^2-16m+4+\left(m+1\right)\left(12m+8\right)=16\left(m+1\right)^2\)

\(\Leftrightarrow16m^2-16m+4+12m^2+8m+12m+8=16m^2+32m+16\)

\(\Leftrightarrow12m^2-28m-4=0\)

\(\Rightarrow\left[{}\begin{matrix}m=\dfrac{7+\sqrt{61}}{6}\\m=\dfrac{7-\sqrt{61}}{6}\end{matrix}\right.\)(TM)

\(sin\left(\dfrac{3\pi}{5}+2x\right)=2sin^2\left(\dfrac{\pi}{5}-x\right)\)

\(\Leftrightarrow sin\left(\dfrac{3\pi}{5}+2x\right)=1-cos\left(\dfrac{2\pi}{5}-2x\right)\)

\(\Leftrightarrow sin\left(\dfrac{3\pi}{5}+2x\right)=1-sin\left(\dfrac{2\pi}{5}-2x+\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow sin\left(\dfrac{3\pi}{5}+2x\right)+sin\left(\dfrac{9\pi}{10}-2x\right)=1\)

\(\Leftrightarrow2sin\left(\dfrac{\dfrac{3\pi}{5}+\dfrac{9\pi}{10}}{2}\right)cos\left(2x+\dfrac{\dfrac{3\pi}{5}-\dfrac{9\pi}{10}}{2}\right)=1\)

\(\Leftrightarrow2sin\left(\dfrac{3\pi}{4}\right)cos\left(2x-\dfrac{3\pi}{20}\right)=1\)

\(\Leftrightarrow\sqrt{2}cos\left(2x-\dfrac{3\pi}{20}\right)=1\)

\(\Leftrightarrow cos\left(2x-\dfrac{3\pi}{20}\right)=\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3\pi}{20}=\dfrac{\pi}{4}+k2\pi\\2x-\dfrac{3\pi}{20}=\dfrac{-\pi}{4}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{5}+k\pi\\x=\dfrac{-\pi}{20}+k\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)