a. \(\sqrt{x^2+1}=4\)

\(\Leftrightarrow x^2+1=16\)

\(\Leftrightarrow x^2=15\)

\(\Leftrightarrow x=\pm\sqrt{15}\)

b. \(\sqrt{x^2+5x+20}=4\)

\(\Leftrightarrow x^2+5x+20=16\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-4\end{matrix}\right.\)

c. \(x-5\sqrt{x}+4=0\)

Đặt \(t=\sqrt{x}\ge0\)

\(\Rightarrow t^2-5t+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=4\\t=1\end{matrix}\right.\) (TM)

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

d. \(\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)=8\)

\(\Leftrightarrow x+3\sqrt{x}+\sqrt{x}+3=8\)

\(\Leftrightarrow x+4\sqrt{x}-5=0\)

Tương tự câu c ta được: \(\left[{}\begin{matrix}x=1\left(TM\right)\\x=-1\left(KTM\right)\end{matrix}\right.\)

a) \(\sqrt{x^2+1}=4\)

\(\Leftrightarrow x^2+1=16\)

\(\Leftrightarrow x^2=15\)

\(\Leftrightarrow x=\pm\sqrt{15}\)

b) \(\sqrt{x^2+5x+20}=4\)

\(\Leftrightarrow x^2+5x+20=16\)

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

\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-4\end{matrix}\right.\)

c) \(x-5\sqrt{x}+4=0\)(ĐK: \(x\ge0\))

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=16\end{matrix}\right.\)(thỏa mãn) 

d) \(\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)=8\)(ĐK: \(x\ge0\))

\(\Leftrightarrow x+4\sqrt{x}-5=0\)

\(\Leftrightarrow\left(\sqrt{x}+5\right)\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow x=1\) (thỏa mãn) 

Câu 11: 

Gọi độ dài cạnh hình vuông là \(x\).

\(cot\widehat{MBQ}=\dfrac{BQ}{MQ}\Rightarrow cot60^o=\dfrac{1-x}{2x}\Rightarrow2x=\sqrt{3}\left(1-x\right)\)

\(\Leftrightarrow x=2\sqrt{3}-3\)

\(S_{MNPQ}=\left(2\sqrt{3}-3\right)^2=21-12\sqrt{3}\)

Câu 12: 

Gọi vận tốc của ô tô đi từ A đến B là \(x\left(km/h\right),x>10\).

Ta có: 

\(\dfrac{40}{x}+\dfrac{8}{x-10}=1\)

\(\Rightarrow40\left(x-10\right)+8x=x\left(x-10\right)\)

\(\Leftrightarrow x^2-58x+400=0\)

\(\Leftrightarrow\left(x-50\right)\left(x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=50\left(tm\right)\\x=8\left(l\right)\end{matrix}\right.\)

Vậy vận tốc của ô tô đi từ A đến B là \(50km/h\).

Đk: \(x\ge2+\sqrt{3}\)

Ta có: \(x+1+\sqrt{x^2-4x+1}=3\sqrt{x}\)

<=> \(x-4+\sqrt{x^2-4x+1}-1-3\left(\sqrt{x}-2\right)=0\)

<=> \(x-4+\dfrac{x\left(x-4\right)}{\sqrt{x^2-4x+1}+1}-\dfrac{3\left(x-4\right)}{\sqrt{x}+2}=0\)

<=> \(\left(x-4\right).\left(1+\dfrac{x}{\sqrt{x^2-4x+1}+1}-\dfrac{3}{\sqrt{x}+2}\right)=0\)

<=> \(x=4\)

Vì \(x\ge2+\sqrt{3}\) -> \(\dfrac{x}{\sqrt{x^2-4x+1}}>0\); \(-\dfrac{3}{\sqrt{x}+2}>-1\)

=> \(1+\dfrac{x}{\sqrt{x^2-4x+1}}-\dfrac{3}{\sqrt{x}+2}>0\)

\(n_{H_2SO_4}=0,2.1,5=0,3\left(mol\right)\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Có \(n_{H_2SO_4}>n_{H_2}\)

=> Trong A chứa H₂SO₄

Vậy kim loại R không tan trong H₂SO₄ loãng

\(n_{SO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH: 2R + 2nH₂SO₄ --> R₂(SO₄)_n + nSO₂ + 2nH₂O

                                            \(\dfrac{0,05}{n}\)<---0,05

=> \(M_{R_2\left(SO_4\right)_n}=\dfrac{8}{\dfrac{0,05}{n}}=160n\left(g/mol\right)\)

=> \(M_R=32n\left(g/mol\right)\)

Xét n = 2 thỏa mãn => M_R = 64 (g/mol) => R là Cu

\(n_{Cu}=n_{CuSO_4}=\dfrac{8}{160}=0,05\left(mol\right)\)

Gọi số mol Al, Fe là a,b (mol)

=> 27a + 56b = 8,7 - 0,05.64 = 5,5 (1)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

              a----------------------------->1,5a

             Fe + H₂SO₄ --> FeSO₄ + H₂

              b--------------------------->b

=> 1,5a + b = 0,2 (2)

(1)(2) => a = 0,1 (mol); b = 0,05 (mol)

\(\left\{{}\begin{matrix}m_{Al}=0,1.27=2,7\left(g\right)\\m_{Fe}=0,05.56=2,8\left(g\right)\\m_{Cu}=0,05.64=3,2\left(g\right)\end{matrix}\right.\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(n_{H_2SO_4\left(bđ\right)}=0,2.1,5=0,3\left(mol\right)\)

\(\xrightarrow[]{\text{BTNT H}}n_{H_2SO_4\left(pư\right)}=n_{H_2}=0,2\left(mol\right)< 0,3\)

=> H₂SO₄ dư, mà vẫn có chất rắn B không tan là kim loại R

=> R là kim loại yếu

Đặt R có hoá trị n

\(n_{SO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ n_{R_2\left(SO_4\right)_n}=\dfrac{8}{2M_R+96n}\left(mol\right)\)

PTHH:

\(2R+2nH_2SO_{4\left(đặc,nóng\right)}\xrightarrow[]{t^o}R_2\left(SO_4\right)_n+nSO_2+2nH_2O\)

                                            \(\dfrac{0,05}{n}\)<-----0,05

\(\rightarrow\dfrac{8}{2M_R+96n}=\dfrac{0,05}{n}\\ \Leftrightarrow M_R=32n\left(g\text{/}mol\right)\)

Xét n = 2 thoả mãn

=> M_R = 32.2 = 64 (g/mol)

=> R là Cu

\(n_{Cu}=\dfrac{0,05.2}{2}=0,05\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\)

\(\rightarrow27a+56b+0,05.64=8,7\left(1\right)\)

PTHH:

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

a---->1,5a

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2O\)

b----->b

\(\rightarrow1,5a+b=0,2\left(2\right)\)

\(\left(1\right),\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,05\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{8,7}.100\%=31,03\%\\\%m_{Fe}=\dfrac{0,05.56}{8,7}.100\%=32,18\%\\\%m_{Cu}=100\%-31,03\%-32,18\%=36,79\%\end{matrix}\right.\)

Ta có:

a + b + c = 0

<=> a² + b² + c² + 2(ab + bc + ac) = 0

<=> ab + bc + ac = -7

<=> a²b² + b²c² + a²c²  + 2abc(a + b + c) = 49

<=> a²b² + b²c² + a²c² = 49 (vì a + b + c = 0)

<=> 2(a²b² + b²c² + a²c²) = 98

<=> (a² + b² + c²)² = 98 + a⁴ + b⁴ + c⁴

<=> a⁴ + b⁴ + c⁴ = 14² - 98 = 98

swim

swim

\(x-2\sqrt{x-1}=16\)

Điều kiện: \(x\ge1\)

\(\Leftrightarrow\left(x-1\right)-2\sqrt{x-1}+1=16\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2=16\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2=\left(\pm4\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=4\\\sqrt{x-1}=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=5\\\sqrt{x-1}=-3\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\left(\sqrt{x-1}\right)^2=5^2\)

\(\Leftrightarrow x=26\)

tìm x

\(\Sigma\dfrac{1}{\sqrt[3]{x\left(y+z\right)}}\ge\dfrac{9}{\sqrt[3]{x\left(y+z\right)}+\sqrt[3]{y\left(x+z\right)}+\sqrt[3]{\left(z\left(x+y\right)\right)}}=\dfrac{9\sqrt[3]{4}}{\sqrt[3]{4}.\Sigma\sqrt[3]{x\left(y+z\right)}}\)

\(có:\sqrt[3]{4}\sqrt[3]{x\left(y+z\right)}=\sqrt[3]{2.2x.\left(y+z\right)}\le\dfrac{2+2x+y+z}{3}\)

\(tương\) \(tự\Rightarrow\Sigma\dfrac{1}{\sqrt[3]{x\left(y+z\right)}}\ge\dfrac{9.\sqrt[3]{4}}{\dfrac{2+2x+y+z}{3}+\dfrac{2+2y+x+z}{3}+\dfrac{2+2z+x+y}{3}}=\dfrac{27\sqrt[3]{4}}{6+4\left(x+y+z\right)}\ge\dfrac{27.\sqrt[3]{4}}{6+4\sqrt{3\left(x^2+y^2+z^2\right)}}=\dfrac{27\sqrt[3]{4}}{6+4\sqrt{3.3}}=\dfrac{3}{\sqrt[3]{2}}\)

\(dấu"="\Leftrightarrow x=y=z=1\)