Khá dễ!

Ta có: \(\left(a+b\right)\left(a^3+b^3\right)\le2\left(a^4+b^4\right)\)

<=> \(a^4+a^3b+ab^3+b^4\le a^4+b^4+a^4+b^4\)

<=> \(a^3b+ab^3\le a^4+b^4\)

<=> \(a^4-a^3b+b^4-ab^3\ge0\)

<=> \(a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)

<=> \(\left(a-b\right)\left(a^3-b^3\right)\ge0\)

<=> \(\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\) (Luôn đúng)

=> đpcm

MK hứng bài nào thì lm bài đấy nhé!

Bài 21:

Ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)

<=> \(\dfrac{ab+bc+ca}{abc}=0\)

<=> \(ab+bc+ac=0\)

<=> \(ab+bc+ac+c^2=c^2\)

<=> \(\sqrt{ab+bc+ac+c^2}=\sqrt{c^2}\)

<=> \(\sqrt{\left(a+c\right)\left(b+c\right)}=\left|c\right|\) (1)

Mặt khác: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) ; \(a,b>0;c\ne0\) => \(c< 0\) (2)

Từ (1); (2) => \(\sqrt{\left(a+c\right)\left(b+c\right)}=-c\)

<=> \(2\sqrt{\left(a+c\right)\left(b+c\right)}+2c=0\)

<=> \(\left(a+c\right)+2\sqrt{\left(a+c\right)\left(b+c\right)}+\left(b+c\right)=a+b\)

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

<=> \(\sqrt{a+c}+\sqrt{b+c}=\sqrt{a+b}\) => Đpcm

Vì đây là lần đầu tiên bn gửi câu hỏi nên mk đã kiên nhẫn dịch cái đề và hi vọng nó đúng!

Ta có: \(\left(\sqrt{8+2\sqrt{7}}+2.\sqrt{8-2\sqrt{7}}\right).\left(\sqrt{63}+1\right)\)

\(=\left(\sqrt{7+2\sqrt{7}+1}+2.\sqrt{7-2\sqrt{7}+1}\right).\left(\sqrt{63}+1\right)\)

\(=\left(\sqrt{\left(\sqrt{7}+1\right)^2}+2.\sqrt{\left(\sqrt{7}-1\right)^2}\right)\left(\sqrt{63}+1\right)\)

\(=\left(\left|\sqrt{7}+1\right|+2.\left|\sqrt{7}-1\right|\right).\left(\sqrt{63}+1\right)\)

\(=\left(\sqrt{7}+1+2\sqrt{7}-2\right)\left(\sqrt{63}+1\right)\)

\(=\left(3\sqrt{7}-1\right)\left(\sqrt{63}+1\right)\)

\(=\left(\sqrt{63}-1\right)\left(\sqrt{63}+1\right)=63-1=62\)

PTHH: 2SO₂ + O₂ --> 2SO₃ (1)

SO₃ + H₂O --> H₂SO₄ (2)

Theo PT(1): \(n_{SO_2}=n_{SO_3}=\dfrac{8}{22,4}=\dfrac{5}{14}mol\)

=> \(m_{SO_3}=\dfrac{5}{14}.80=\dfrac{200}{7}g\)

Theo PT(2): \(n_{H_2SO_4}=n_{SO_3}=\dfrac{5}{14}mol\)

=> \(m_{H_2SO_4}=\dfrac{5}{14}.98=35g\)

Mặt khác: \(m_{dd.H_2SO_4}\) ban đầu = 57,2.1,5 = 85,8g

=> \(m_{H_2SO_4}\) ( chất tan) = 85,8.60% = 51,48g

=> Tổng khối lượng H₂SO₄ ( chất tan) = 35 + 51,48 = 86,48g

=> m_{dd sau p/ứ} = \(85,8+\dfrac{200}{7}=\dfrac{4003}{35}g\)

=> C% của dd axit thu được = \(\dfrac{86,48}{\dfrac{4003}{35}}.100\%=75,61\%\)

Ôi chu choa mạ ơi! Cái đề kiểu chi ri???

39 + x = 64

x=64-39

x=25

a) ĐK: \(x\ge0;x\ne1\)

Ta có: \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right):\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

<=> \(A=\left[\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x+1}\right)}\right]:\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

\(=\left[\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right]:\dfrac{\sqrt{x}}{\sqrt{x}-1}\)

\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2}{\left(\sqrt{x}+1\right)^2}\)

b) Để A nguyên => \(\dfrac{2}{\left(\sqrt{x}+1\right)^2}\) nguyên

=> \(2⋮\left(\sqrt{x}+1\right)^2\) => \(\left(\sqrt{x}+1\right)^2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Mà \(\left(\sqrt{x}+1\right)^2\ge1\) => \(\left(\sqrt{x}+1\right)^2\in\left\{1;2\right\}\)

=> \(\left[{}\begin{matrix}\sqrt{x}+1=1\\\sqrt{x}+1=\sqrt{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=0\\x=3-2\sqrt{2}\end{matrix}\right.\) (TM)

Vậy ...................................

Ơ? T đủ hết ý r mà