Câu thơ đó nói về hai bà Trưng Trắc và Trưng Nhị( Hai Bà Trưng )

TH1 : \(\left|x-2\right|=x-2\)

Ta có :

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

\(\Leftrightarrow\left(x^2-4\right)\left(x^2-1\right)=4\)

\(\Leftrightarrow x^4-5x^2=0\)

\(\Leftrightarrow x^2\left(x^2-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)

TH2 : \(\left|x-2\right|=2-x\)

Ta có :

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

\(\Leftrightarrow\left(4-x^2\right)\left(x^2-1\right)=4\)

\(\Leftrightarrow5x^2-x^4=8\)

\(\Leftrightarrow x^4-5x^2+8=0\) (Vô nghiệm)

Vậy...

ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

Ta có :

\(P=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)

\(=\dfrac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+3\sqrt{x}+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

Xem lại đề bài giúp mình với ^^

Ta có pt cân bằng nhiệt :

\(C_1m_1\left(t_1-t\right)+C_2.m_2\left(t_2-t\right)=0\)

\(\Leftrightarrow130.0,1.\left(110-70\right)+4200.m_2\left(50-70\right)=0\)

\(\Leftrightarrow m_2=0,0062\left(kg\right)\)

Vậy..

a/ Ta có : \(P.h=F.s\)

\(\Leftrightarrow F=\dfrac{P.h}{s}=\dfrac{10.10.3}{15}=20\left(N\right)\)

b/ \(A=P.h=10.10.3=300\left(J\right)\)

c/ \(P=\dfrac{A}{t}=\dfrac{300}{10}=30\left(W\right)\)

a/ \(A=ax-ay-bx+by\)

\(=a\left(x-y\right)-b\left(x-y\right)\)

\(=\left(x-y\right)\left(a-b\right)\)

\(=-50.2=-100\)

b/ \(B=ax+ay-bx-by\)

\(=a\left(x+y\right)-b\left(x+y\right)\)

\(=\left(a-b\right)\left(x+y\right)\)

\(=\left(-1\right).\left(-100\right)=100\)

\(W=W_đ=\dfrac{1}{2}mv^2=\dfrac{1}{2}m.10^2=50m\left(J\right)\)

\(W=Wt+Wđ=50m\left(J\right)\)

Mà \(W_t=W_đ\)

\(\Leftrightarrow W_t=W_đ=25m=mgz=10m.z\)

\(\Leftrightarrow z=2,5\left(m\right)\)

2003 học tốt                                                      

\(n_{Cl_2}=0,4\left(mol\right)\) \(\Leftrightarrow m_{Cl_2}=28,4\left(g\right)\)

PTHH : \(h^2_{Fe,Mg}+Cl_2\rightarrow Muối\)

Bảo toàn khối lượng :

\(m_{h^2}+m_{Cl_2}=m'\)

\(\Leftrightarrow m'=13,6+28,4=42\left(g\right)\)