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

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

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

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

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

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

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

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

\(dấu"="\Leftrightarrow x=y=z=1\)

\(\dfrac{AB}{AC}=\dfrac{5}{6}\Leftrightarrow AB=\dfrac{5}{6}AC\)

\(BC^2=AB^2+AC^2=\left(\dfrac{5}{6}AC\right)^2+AC^2=\dfrac{61}{36}AC^2\)

\(\Rightarrow AC^2=8784\Rightarrow AC=12\sqrt{61}\Rightarrow AB=10\sqrt{61}\)

\(BD\) là phân giác của giác của \(\widehat{ABC}\) suy ra 

\(\dfrac{AD}{AB}=\dfrac{CD}{BC}=\dfrac{AD+CD}{AB+BC}=\dfrac{AC}{AB+BC}=\dfrac{12\sqrt{61}}{10\sqrt{61}+122}=\dfrac{\sqrt{61}-5}{6}\)

\(\Rightarrow AD=\dfrac{\sqrt{61}-5}{6}.10\sqrt{61}=\dfrac{5}{3}\left(61-5\sqrt{61}\right)\)

Dựng hình vuông \(BCEF\). Lấy \(M\) thuộc \(BF\) sao cho \(PM=PQ\) khi đó suy ra \(MF=QA\). 

\(\Delta BCM=\Delta DCQ\left(c.g.c\right)\) suy ra \(\widehat{BCM}=\widehat{DCQ}\) 

suy ra \(\widehat{QCM}=\widehat{QCB}+\widehat{BCM}=\widehat{QCM}+\widehat{DCQ}=\widehat{DCB}=90^o\)

\(\Delta CPQ=\Delta CPM\left(c.c.c\right)\) suy ra \(\widehat{PCQ}=\widehat{PCM}\)

suy ra \(\widehat{PCQ}=\dfrac{1}{2}\widehat{QCM}=45^o\)