dk:\(x\ge0\)

\(A=x^2+\sqrt{x}-1\ge-1\)

dấu ''='' xảy ra khi và chỉ khi x=0

Bạn oi bài 2 hàng A thú 2 phải là \(\frac{\sqrt{a}-2}{\sqrt{a}+1}\) mình nhầm

a,\(\left(\sqrt{6}-\sqrt{10}\right)\sqrt{4+\sqrt{15}}=\sqrt{6}.\sqrt{4-\sqrt{15}}-\sqrt{10}.\sqrt{4+\sqrt{15}}\)

=\(\sqrt{24+6\sqrt{15}}-\sqrt{40+10\sqrt{15}}=\sqrt{\left(\sqrt{15}+3\right)^2}-\sqrt{\left(\sqrt{15}+5\right)^2}\)

=\(\sqrt{15}+3-\sqrt{15}-5=-2\)

b,\(\left(\sqrt{3}+\sqrt{30}\right)\sqrt{10-\sqrt{41-4\sqrt{10}}}\)

=\(\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{10-\sqrt{40-2\sqrt{40}+1}}\)

=\(\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{10-\sqrt{\left(\sqrt{40}-1\right)^2}}\)

=\(\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{10-\sqrt{40}+1}\)

=\(\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{11-2\sqrt{10}}=\sqrt{3}\left(1+\sqrt{10}\right)\sqrt{\left(\sqrt{10}-1\right)^2}\)

=\(\sqrt{3}\left(1+\sqrt{10}\right)\left(\sqrt{10}-1\right)=9\sqrt{3}\)

2,\(A=\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)-a-2}{\sqrt{a}+1}\right):\left(\frac{\sqrt{a}\left(1-\sqrt{a}\right)-\sqrt{a}+4}{1-a}\right)\)

\(A=\left(\frac{a+\sqrt{a}-a-2}{\sqrt{a}+1}\right):\left(\frac{\sqrt{a}-a-\sqrt{a}+4}{1-a}\right)=\left(\frac{\sqrt{a}+2}{\sqrt{a}+1}\right).\left(\frac{1-a}{4-a}\right)\)

\(A=\frac{\sqrt{a}-2}{\sqrt{a}+1}.\frac{a-1}{a-4}=\frac{\sqrt{a}-1}{\sqrt{a}+2}\)

b, ̣để \(A=\frac{1}{2}\Rightarrow\frac{\sqrt{a}-1}{\sqrt{a}+2}=\frac{1}{2}\Leftrightarrow2\sqrt{a}-2=\sqrt{a}+2\Leftrightarrow\sqrt{a}=4\Leftrightarrow a=16\left(t.m\right)\)

\(n_{cuso_4}=1.0.01=0.01\cdot\left(mol\right)\)

\(Fe+CuSO_4\rightarrow FeSO_4+Cu\)

\(n_{Fe}=n_{Cu}=n_{FeSO_4}=n_{CuSO_4}=0.01\left(mol\right)\)

a,chất rắn A gồm Fe và Cu\(\Rightarrow\) ta có pt \(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(\Rightarrow m_{cu}=0,01.64=0,64\left(g\right)\)

b, dung dịch B chúa FeSO4\(\Rightarrow\) ta có pt \(FeSO_4+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+Na_2SO_4\)

\(2n_{FeSO_4}=n_{NaOH}=0.02\left(mol\right)\Rightarrow V_{NaOH}\frac{0.02}{1}=0.02\left(l\right)=20\left(ml\right)\)

\(pt:zn+2HCl\rightarrow ZnCl_2+H_2\)

a,theo đề bài ta có : \(n_{zn}=\frac{13}{65}=0.2\left(mol\right),n_{hcl}=1.0,5=0,5\left(mol\right)\)

ta thấy hcl dư vì: \(\frac{n_{hcl}}{2}>\frac{n_{zn}}{1}\)

theo phương trình \(n_{zncl_2}=n_{zn}=0,2\left(mol\right)\Rightarrow m_{zncl_2}=0,2.127=25,4\left(mol\right)\)

b,\(n_{h_2}=n_{zn}=0,2\left(mol\right)\Rightarrow V_{hcl}=0,2.22,4=4,48\left(l\right)\)

c,theo phương trình \(n_{hcl_{pu}}=2n_{zn}=0,4\left(mol\right)\Rightarrow n_{hcl_{du}}=n_{hcl_{bandau}}-n_{pu}=0,5-0,4=0.1\left(mol\right)\)

\(pt:NaOH+HCl\rightarrow NaCl+H_2\)

\(\Rightarrow n_{NaOH}=n_{hcl}=0.1\Rightarrow V_{dd}=\frac{0.1}{0.5}=0.2\left(l\right)\)

Ct tính số nu \(A_1+T_1+G_1+X_1=A_2+T_2+G_2+T_2=\frac{N}{2}\)

CT tính chiều dài \(L_{ADN}=\frac{N}{2}.3,4\)

CT tính chu kì vòng xoắn \(C=\frac{N}{20}\)