Vì \(180< 441\)\(\Rightarrow\)\(\sqrt{180}< \sqrt{441}\)

                               \(\Leftrightarrow\)\(14+6\sqrt{5}< 14+21\)

                               \(\Leftrightarrow\)\(9+6\sqrt{5}+5< 35\)

                               \(\Leftrightarrow\)\(\left(\sqrt{9}+\sqrt{5}\right)^2< 35\)

                               \(\Leftrightarrow\)\(\sqrt{9}+\sqrt{5}< \sqrt{35}\)

Vậy \(\sqrt{9}+\sqrt{5}< \sqrt{35}\)

 \(\tanh\)

Đề TST của KHTN lớp 10 :3 

Dễ có:\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Leftrightarrow ab+bc+ca\le3\)

\(P=\Sigma\frac{bc}{\sqrt[4]{a^2+3}}\ge\Sigma\frac{bc}{\sqrt[4]{a^2+ab+bc+ca}}=\Sigma\frac{bc}{\sqrt[4]{\left(a+b\right)\left(a+c\right)}}=\Sigma\frac{\sqrt{2}bc}{\sqrt[4]{\left(a+b\right)\left(a+c\right)2\cdot2}}\)

Đến đây khó quá huhu

Thật ra là hóa lớp 8 ạ, em gấp quá nên nhấn lộn TvT

a)

n Zn = \(\frac{m}{M}=\frac{13}{65}=0,2\left(mol\right)\) 

Zn + 2HCl \(\rightarrow\) ZnCl2 + H2 

0,2      0,4               0,2       0,2       ( mol ) 

b) 

m ct HCl = \(n\cdot M=0,4\cdot36,5=14,4\left(g\right)\) 

C% HCl = \(\frac{mct\cdot100\%}{mdd}=\frac{14,4\cdot100\%}{120}=12\%\) 

c) 

V H2 = \(n\cdot22,4=0,2\cdot22,4=4,48\left(l\right)\) 

d) 

Cu0 + H2 \(\rightarrow\) Cu + H20 

 0,2      0,2          0,2     0,2         

m Cu = \(n\cdot M=0,2\cdot64=12,8\left(g\right)\)           

\(P=\left(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}\right)\div\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-4}{1-x}\right)\)

ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

a) \(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\frac{x+2}{\sqrt{x}+1}\right)\div\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}-4}{x-1}\right)\)

\(P=\left(\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}\right)\div\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

\(P=\frac{\sqrt{x}-2}{\sqrt{x}+1}\div\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

\(P=\frac{\sqrt{x}-2}{\sqrt{x}+1}\div\frac{x-\sqrt{x}+\sqrt{x}-4}{x-1}\)

\(P=\frac{\sqrt{x}-2}{\sqrt{x}+1}\times\frac{x-1}{x-4}\)

\(P=\frac{\sqrt{x}-2}{\sqrt{x}+1}\times\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x-4}\)

\(P=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-4}\)

\(P=\frac{x-3\sqrt{x}+2}{x-4}\)

b) Để P < 0

=> \(\frac{x-3\sqrt{x}+2}{x-4}< 0\)

Xét hai trường hợp

I) \(\hept{\begin{cases}x-3\sqrt{x}+2>0\\x-4< 0\end{cases}}\)

+) \(x-3\sqrt{x}+2>0\)

<=> ( √x - 1 )( √x - 2 ) > 0

1. \(\hept{\begin{cases}\sqrt{x}-1>0\\\sqrt{x}-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}>1\\\sqrt{x}>2\end{cases}}\Leftrightarrow\hept{\begin{cases}x>1\\x>4\end{cases}}\Leftrightarrow x>4\)(1)

2. \(\hept{\begin{cases}\sqrt{x}-1< 0\\\sqrt{x}-2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}< 1\\\sqrt{x}< 2\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 1\\x< 4\end{cases}}\Leftrightarrow x< 1\)

Kết hợp ĐKXĐ : \(0\le x< 1\)(2)

+) x - 4 < 0 <=> x < 4 (3)

Từ (1), (2) và (3) => \(0\le x< 1\)

II) \(\hept{\begin{cases}x-3\sqrt{x}+2< 0\\x-4>0\end{cases}}\)

 +) \(x-3\sqrt{x}+2< 0\)

<=> ( √x - 1 )( √x - 2 ) < 0

1. \(\hept{\begin{cases}\sqrt{x}-1< 0\\\sqrt{x}-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}< 1\\\sqrt{x}>2\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 1\\x>4\end{cases}}\)( loại )

2. \(\hept{\begin{cases}\sqrt{x}-1>0\\\sqrt{x}-2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}>1\\\sqrt{x}< 2\end{cases}}\Leftrightarrow\hept{\begin{cases}x>1\\x< 4\end{cases}}\Leftrightarrow1< x< 4\)(1)

+) x - 4 > 0 <=> x > 4 (2)

Từ (1) và (2) => Không có giá trị của x thỏa mãn

Vậy với \(0\le x< 1\)thì P < 0 

\(\frac{a-b}{4b^2}\cdot\sqrt{\frac{4a^2b^4}{a^2-2ab+b^2}}\)

\(=\frac{a-b}{4b^2}\cdot\sqrt{\frac{\left(2ab^2\right)^2}{\left(a-b\right)^2}}\)

\(=\frac{a-b}{4b^2}\cdot\frac{2ab}{a-b}\)

\(=\frac{a}{2b}\)

a) \(B=\frac{x\sqrt{x}+1}{x-1}-\frac{x-1}{\sqrt{x}-1}\)

ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

\(B=\frac{x\sqrt{x}+\sqrt{x}-\sqrt{x}+x-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{x-1}{\sqrt{x}-1}\)

\(B=\frac{\left(x\sqrt{x}-x+\sqrt{x}\right)+\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{x-1}{\sqrt{x}-1}\)

\(B=\frac{\sqrt{x}\left(x-\sqrt{x}+1\right)+1\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x+1}\right)}-\frac{x-1}{\sqrt{x}-1}\)

\(B=\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{x-1}{\sqrt{x}-1}\)

\(B=\frac{x-\sqrt{x}+1-\left(x-1\right)}{\sqrt{x}-1}\)

\(B=\frac{x-\sqrt{x}+1-x+1}{\sqrt{x}-1}=\frac{2-\sqrt{x}}{\sqrt{x}-1}\)

b) Để A = B

=> \(\frac{2-\sqrt{x}}{\sqrt{x}-1}=-1\)

<=> \(2-\sqrt{x}=-1\left(\sqrt{x}-1\right)\)

<=> \(2-\sqrt{x}=1-\sqrt{x}\)

<=> \(-\sqrt{x}+\sqrt{x}=1-2\)

<=> \(0=-1\)( vô lí )

Vậy không có giá trị của x thỏa mãn

\(1,\sqrt{\left(2+\sqrt{7}\right)^2-\sqrt{\left(2-\sqrt{7}\right)^2}}\)    ( áp dụng hđt thứ 3 \(a^2-b^2=\left(a-b\right)\left(a+b\right)\))

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

\(=\sqrt{4\cdot\sqrt{7}}\)

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

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

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

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

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

\(=6\sqrt{5}\cdot\left(-10\sqrt{2}\right)\)

\(3,\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}\)

\(\Leftrightarrow\sqrt{10+2\sqrt{21}}=\sqrt{10-2\sqrt{21}}\)

\(\Leftrightarrow10+2\sqrt{21}=10-2\sqrt{21}\)

\(\Leftrightarrow4\sqrt{21}\)

cuối lười tính nên thôi nhá :>

tks :>