2.

Áp dụng định lý hàm cosin:

\(b=\sqrt{a^2+c^2-2ac.cosB}=\sqrt{8^2+3^2-2.8.3.cos60^0}=7\)

\(S_{ABC}=\dfrac{1}{2}ac.sinB=\dfrac{1}{2}.8.3.sin60^0=6\sqrt{3}\)

4.

\(\Delta=\left(m+2\right)^2-16>0\Leftrightarrow m^2+4m-12>0\Rightarrow\left[{}\begin{matrix}m>2\\m< -6\end{matrix}\right.\) (1)

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-m-2\\x_1x_2=4\end{matrix}\right.\)

\(x_1+x_2+x_1x_2>1\)

\(\Leftrightarrow-m-2+4>1\)

\(\Rightarrow m< 1\) (2)

Kết hợp (1); (2) ta được \(m< -6\)

315 kh

>

3 tạ15 kg=315kg

4,3 > 4,27 

Mọi người giải giúp em 2 câu này với ạ

        3 tạ15 kg=..........315...........kg

Điền dấu >,<,=vào chỗ chấm sao cho thích hợp

          4,3.......>........4,27 

\(a,=7\cdot6+15:5=42+3=45\\ b,=6+3\sqrt{5}-6=3\sqrt{5}\)

\(I=\dfrac{U}{R}=\dfrac{220}{50}=4,4\left(A\right)\)

\(\Rightarrow A=Pt=UIt=220.4,4.\left(\dfrac{15}{60}\right)=242\left(Wh\right)=0,242\left(kWh\right)\)

Bài 4:

\(n_{Ba\left(OH\right)_2}=1.0,1=0,1\left(mol\right)\\ n_{HCl}=1.0,1=0,1\left(mol\right)\\ Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\\ Vì:\dfrac{0,1}{2}< \dfrac{0,1}{1}\Rightarrow Ba\left(OH\right)_2dư\\ n_{BaCl_2}=n_{Ba\left(OH\right)_2\left(p.ứ\right)}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ n_{Ba\left(OH\right)_2\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\\ \left[OH^-\left(dư\right)\right]=\left[Ba\left(OH\right)_2\left(dư\right)\right]=\dfrac{0,05}{0,1+0,1}=0,25\left(M\right)\\ \left[Cl^-\right]=2.\left[BaCl_2\right]=2.\dfrac{0,05}{0,1+0,2}=0,5\left(M\right)\\ \left[Ba^{2+}\right]=\dfrac{0,25+0,5}{2}=0,375\left(M\right)\)

MỜ VS NHỎ QUÁ BN ƠI

Bài 3:

\(n_{NaOH}=1.0,15=0,15\left(mol\right)\\ n_{HNO_3}=1.0,25=0,25\left(mol\right)\\ NaOH+HNO_3\rightarrow NaNO_3+H_2O\\ Vì:\dfrac{0,25}{1}>\dfrac{0,15}{1}\Rightarrow HNO_3dư\\ n_{HNO_3\left(dư\right)}=0,25-0,15=0,1\left(mol\right)\\ n_{NaNO_3}=n_{NaOH}=0,15\left(mol\right)\\ \left[OH^-\right]=\left[Na^+\right]=\left[NaOH\right]=\dfrac{0,15}{0,25+0,15}=0,375\left(M\right)\\ \left[NO^-_3\left(dư\right)\right]=\left[H^+\left(dư\right)\right]\left[HNO_3\left(dư\right)\right]=\dfrac{0,1}{0,15+0,25}=0,25\left(M\right)\)