vẽ lại mạch ik bạn 

đoạn cuối có điểm B mk không hiểu 

\(P=\sqrt{\left(a+b\right)^2+b^2}+\sqrt{\left(b+c\right)^2+c^2}+\sqrt{\left(a+c\right)^2+a^2}\)

\(\Leftrightarrow\sqrt{5}P=\sqrt{5\left[\left(a+b\right)^2+b^2\right]}+\sqrt{5\left[\left(b+c\right)^2+c^2\right]}+\sqrt{5\left[\left(a+c\right)^2+a^2\right]}\)

Có : \(\sqrt{5\left[\left(a+b\right)^2+b^2\right]}=\sqrt{\left[\left(a+b\right)^2+b^2\right]\left(2^2+1^2\right)}\ge2\left(a+b\right)+b\)

(BĐT Bunyakovski) 

Tương tự được \(\sqrt{5}P\ge2\left(a+b\right)+b+2\left(b+c\right)+c+2\left(a+c\right)+a\)

\(\Leftrightarrow\sqrt{5}P\ge5\left(a+b+c\right)\Leftrightarrow P\ge\sqrt{5}\)

"=" khi \(\left\{{}\begin{matrix}\dfrac{a+b}{2}=b\\\dfrac{b+c}{2}=c\\\dfrac{a+c}{2}=a\\a+b+c=1\end{matrix}\right.\Leftrightarrow a=b=c=\dfrac{1}{3}\)

Thả hợp kim vào nước => nước + nhiệt lượng kế thu ; hợp kim tỏa nhiệt

Lại có Q_n = 10 Q_nlk

<=> \(Q_{nlk}=\dfrac{m_n.c_n.\Delta t}{10}\)

=> Phương trình cân bằng 

Q_tỏa = Q_thu

<=> \(m_n.c_n.\Delta t+\dfrac{m_n.c_n.\Delta t}{10}=m_t.c_t.\Delta t+m_{Al}.c_{Al}.\Delta t\)

<=> \(1.4200.\left(30-25\right)+\dfrac{1.4200.\left(30-25\right)}{10}=m_t230.\left(80-30\right)+m_{Al}.880.\left(80-30\right)\)<=>  23100 = m_t.11500 + m_Al.44000

<=> 115m_t + 440m_Al = 231 (1) 

Lại có m_t + m_Al = 0,9 (2) 

Từ (1) (2) => m_t = 0,51 (kg) ; m_Al = 0,39 (kg) 

Ta có : \(\sin\alpha=\cos\beta\); với \(\left(\alpha+\beta=90\right)\)

\(\Rightarrow\sin^2\alpha=\cos^2\beta\)

Khi đó \(P=\dfrac{sin^210+sin^220+...+sin^280}{cos^210+cos^220+...+cos^280}=\dfrac{cos^280+cos^270+...+cos^210}{cos^210+cos^220+...+cos^280}=1\)

b) ĐKXĐ : \(-\dfrac{1}{3}\le x\le6\)

Ta có \(\sqrt{6-x}-\sqrt{3x+1}=3x^2-14x-8\)

<=> \(\left(\sqrt{6-x}-1\right)+\left(4-\sqrt{3x+1}\right)=3x^2-14x-5\)

<=> \(\dfrac{5-x}{\sqrt{6-x}+1}+\dfrac{15-3x}{4+\sqrt{3x+1}}=\left(x-5\right)\left(3x+1\right)\)

\(\Leftrightarrow\left(x-5\right)\left(3x+1+\dfrac{1}{\sqrt{6-x}+1}+\dfrac{3}{4+\sqrt{3x+1}}\right)=0\)

\(\Leftrightarrow x=5\left(\text{vì với }x\ge-\dfrac{1}{3}\text{thì }3x+1+\dfrac{1}{\sqrt{6-x}+1}+\dfrac{3}{4+\sqrt{3x+1}}>0\right)\)

Tập nghiệm phương trình \(S=\left\{5\right\}\)

c) ĐKXĐ : \(x\ge1\)

\(x^2-1=2\sqrt{2x+1}\Leftrightarrow x^2+2x+1-\left(2x+1\right)-2\sqrt{2x+1}-1=0\)

\(\Leftrightarrow\left(x+1\right)^2-\left(\sqrt{2x+1}+1\right)^2=0\)

\(\Leftrightarrow\left(x-\sqrt{2x+1}\right)\left(x+\sqrt{2x+1}+2\right)=0\)

\(\Leftrightarrow x=\sqrt{2x+1}\left(\text{vì }x+\sqrt{2x+1}+2>0\text{ với }x\ge1\right)\)

Khi \(x=\sqrt{2x+1}\Leftrightarrow\left\{{}\begin{matrix}x^2-2x-1=0\\x\ge1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=2\\x\ge1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm\sqrt{2}+1\\x\ge1\end{matrix}\right.\Leftrightarrow x=\sqrt{2}+1\)

3. ĐKXĐ : \(x\ge-2\)

Đặt \(\sqrt{x+5}=a;\sqrt{x+2}=b\left(a;b\ge0\right)\)

Khi đó phương trình tương đương

\(\left(a-b\right)\left(ab+1\right)=a^2-b^2\)

\(\Leftrightarrow\left(a-b\right)\left(a-1\right)\left(b-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=1\\b=1\end{matrix}\right.\)

Khi a = 1 => x = -4 (loại)

Khi b = 1 => x = -1 (tm)

Khi a = b => \(\sqrt{x+5}=\sqrt{x+2}\Leftrightarrow\left\{{}\begin{matrix}x+5=x+2\\x\ge-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}∄x\\x\ge-2\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Tập nghiệm phương trình \(S=\left\{-1\right\}\)

\(P=\sqrt{3+\sqrt{2}}.\sqrt{3+\sqrt{6+\sqrt{2}}}.\sqrt{3+\sqrt{6+\sqrt{6+\sqrt{2}}}}.\sqrt{3-\sqrt{6+\sqrt{6+\sqrt{2}}}}\)

\(=\sqrt{3+\sqrt{2}.}\sqrt{3+\sqrt{6+\sqrt{2}}}.\sqrt{\left(3+\sqrt{6+\sqrt{6+\sqrt{2}}}\right).\left(3-\sqrt{6+\sqrt{6+\sqrt{2}}}\right)}\)

\(=\sqrt{3+\sqrt{2}}.\sqrt{3+\sqrt{6+\sqrt{2}}}.\sqrt{9-\left(6+\sqrt{6+\sqrt{2}}\right)}\)

\(=\sqrt{3+\sqrt{2}}.\sqrt{3+\sqrt{6+\sqrt{2}}}.\sqrt{3-\sqrt{6+\sqrt{2}}}\)

\(=\sqrt{3+\sqrt{2}}.\sqrt{\left(3+\sqrt{6+\sqrt{2}}\right)\left(3-\sqrt{6+\sqrt{2}}\right)}=\sqrt{3+\sqrt{2}}.\sqrt{3-\sqrt{2}}\)

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

Có x + y + z = 6

=> (x  + y + z)² = 36

<=> x² + y² + z² + 2xy + 2yz + 2zx = 36

<=> x² + y² + z² + 2xy + 2yz + 2zx = 3(xy + yz + zx) 

<=> x² + y² + z² - xy - yz - zx = 0

<=> 2(x² + y² + z² - xy - yz - zx) = 0

<=> (x - y)² + (y - z)² + (z - x)² = 0

<=> x = y = z = 2 (ĐPCM)

Ta có : \(\sqrt{x^2+x+\dfrac{5}{4}}=\sqrt{\left(x+\dfrac{1}{2}\right)^2+1}\ge1\)(1) 

\(\sqrt{x^2+x+\dfrac{17}{4}}=\sqrt{\left(x+\dfrac{1}{2}\right)^2+4}\ge2\) (2) 

Từ (1) ; (2) => VT \(\ge3=VP\)

Dấu "=" xảy ra <=> \(x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy tập nghiệm phương trình \(S=\left\{-\dfrac{1}{2}\right\}\)

 Ba can plays the piano better than his friends can.
 A. plays           B. better            C. than              D. can

Chọn A vì can + Vinf