\(\frac{x}{\frac{25}{16}}=\frac{y}{\frac{5}{6}}=\frac{z}{\frac{2}{5}}=\frac{2x-3y+4z}{\frac{2.25}{16}-\frac{3.5}{6}+\frac{4.2}{5}}=\frac{5,34}{\frac{89}{40}}=240\)

\(x=\frac{25}{16}.\frac{12}{5}=\frac{15}{4}\)

\(y=\frac{5}{6}.\frac{12}{5}=2\)

\(z=\frac{2}{5}.\frac{12}{5}=\frac{24}{25}\)

ta có : 3\(\sqrt{5}=\sqrt{45}\)

\(5\sqrt{3}=\sqrt{75}\)

ta thấy :

75>45 \(\Leftrightarrow\sqrt{75}>\sqrt{45}hay5\sqrt{3}>3\sqrt{5}\)

a) \(\sqrt{\left(\sqrt{7-2}\right)^2}=\sqrt{5}\)

b)\(\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2-3\sqrt{2}\right)^2}\)

=\(\sqrt{2}-1-2+3\sqrt{2}=4\sqrt{2}-3\)

c)\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)

=\(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}=2\sqrt{3}\)

d) hình như bn ghi sai

e)\(\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)

=\(\left(\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4-2\sqrt{3}}}+\dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{4+2\sqrt{3}}}\right):\sqrt{2}\)

=\(\left(\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{3}-1}+\dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{3}+1}\right):\sqrt{2}\)

=\(\dfrac{\sqrt{2+\sqrt{3}}\left(\sqrt{3}+1\right)+\sqrt{2-\sqrt{3}}\left(\sqrt{3}-1\right)}{2\sqrt{2}}\)

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

=\(\dfrac{3+\sqrt{2+\sqrt{3}}+\sqrt{3}-\sqrt{2+\sqrt{3}}}{2\sqrt{2}}\)

=\(\dfrac{3+\sqrt{3}}{2\sqrt{2}}\)

f) \(\sqrt{9a^2}+3a-7=-3a+3a-7=-7\)

g)\(\dfrac{\sqrt{4x^2-4x+1}}{4x-2}+3x+2\)

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

=\(\dfrac{1}{2}+3x+2=\dfrac{5}{2}+3x\)

h)\(\sqrt{\left(5a-1\right)^2}+2a-3\)

nếu a<0 :\(-5a+1+2a-3=-3a-2\)

nếu a>0 : \(5a-1+2a-3=7a-4\)

i)\(\sqrt{\dfrac{2a}{5}}.\sqrt{\dfrac{5a}{18}}+2\left(a-1\right)\)

=\(\sqrt{\dfrac{10a^2}{90}}+2a-2=\sqrt{\dfrac{a^2}{9}}+2a-2\)

=\(\dfrac{a}{3}+2a-2=\dfrac{7a}{3}-2\)

nguyên liệu chủ yếu là : quặng pirit sắt (FeS2 ) hoặc lưu huỳnh , oxi và nươc

1. sản xuất SO2

- đốt FeS₂ hoặc lưu huỳnh

4FeS₂ + 11O2 -t^o-> 2Fe2O3 +8SO2

S +O2 -t^o-> SO2

2. oxi hóa SO2

2SO2 +O2 -t^o-> 2SO3 (chất xúc tác là V2O5)

3. hấp thụ SO3 tạo thành H2SO4

- trong thực tế người ta ko dùng nước mà dùng dd H2SO4 đậm đặc để hấp thụ SO3 thu được oleum có công thức là H₂SO₄.nSO₃

H2SO4.nSO3 + nH2O --> (n+1)H2SO4

bạn ghi sai đề rồi , mik chữa lại là 3,2M

SO2 +2KOH-->K2SO3 +H2O (1)

SO2+KOH --> KHSO3(2)

n_SO2=\(\dfrac{25,6}{64}=0,4\left(mol\right)\)

n_KOH=\(3,2.250=0,8\left(mol\right)\)

theo (1) : \(\dfrac{nSO2}{nKOH}=\dfrac{1}{2}\), theo (2) :\(\dfrac{nSO2}{nKOH}=1\)

theo đề : \(\dfrac{nSO2}{nKOH}=\dfrac{0,4}{0,8}=\dfrac{1}{2}\)

=> muối thu được là K2SO3=>SO2 hết , KOH dư=>tính theo SO2

theo (1) : nK2SO3=nSO2=0,4(mol)

=>mK2SO3=0,4.158=63,2(g)

Đi 180 km hết số lít xăng là:

180:12x1=15 ( l xăng)

vậy số tiền phải mua xăng để ô tô đó đi được quãng được dài 180 km là:

15x7500= 112500 ( đồng)

mình nha bạn

1. a)

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

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

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

b) ta có : P>1/3

\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}>\dfrac{1}{3}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}-\dfrac{1}{3}>0\)

\(\Leftrightarrow\dfrac{6-\sqrt{x}-2}{\sqrt{x}+2}>0\Leftrightarrow\dfrac{4-\sqrt{x}}{\sqrt{x}+2}>0\)

\(\Leftrightarrow4-\sqrt{x}>0\Leftrightarrow-\sqrt{x}>-4\Leftrightarrow x< 16\)

kết hợp đk ta có :0<x<16 (trừ 4)

vậy 0<x<16 (trừ 4 ) khi P>1/3

c) ta có : Q=9/2P

\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}.\dfrac{9}{2}\Leftrightarrow\dfrac{9}{\sqrt{x}+2}\)

để Q nguyên thì \(\sqrt{x}+2\)phải là ước của 9

\(\Leftrightarrow\sqrt{x}+2\in\left\{\pm1,\pm3,\pm9\right\}\)

vậy \(x\in\left\{1,49\right\}\)

A=\(\dfrac{2a^2+4}{1-a^3}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}\)

=\(\dfrac{2a^2+4}{\left(1-a\right)\left(1+a+a^2\right)}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}\)=

\(\dfrac{2a^2+4}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}-\dfrac{\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}-\dfrac{\left(1+\sqrt{a}\right)\left(1+a+a^2\right)}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}\)

= \(\dfrac{2a^2+4-\left(1-\sqrt{a}+a-a\sqrt{a}+a^2-a^2\sqrt{a}\right)-\left(1+\sqrt{a}+a+a\sqrt{a}+a^2+a^2\sqrt{2}\right)}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}\)

=\(\dfrac{-2a-2}{\left(1-a\right)\left(1+a+a^2\right)}=\dfrac{-2}{1+a+a^2}\)

a) \(\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}\)

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

=\(\sqrt{2}-1+\sqrt{2}+1=2\sqrt{2}\)

b) \(\sqrt{9-4\sqrt{5}}+\sqrt{6+2\sqrt{5}}\)

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

= \(2-\sqrt{5}+\sqrt{5}+1=3\)

c) \(\sqrt{9-4\sqrt{2}}-\sqrt{11+6\sqrt{2}}\)

=\(\sqrt{\left(2\sqrt{2}+1\right)^2}-\sqrt{\left(3+\sqrt{2}\right)^2}\)

=\(2\sqrt{2}+1-3-\sqrt{2}=\sqrt{2}-2\)

d) \(\sqrt{12+8\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)

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

=\(2\sqrt{2}+2+2-\sqrt{2}=\sqrt{2}+4\)

1.

a) \(\dfrac{1}{1+\sqrt{2}+\sqrt{5}}=\dfrac{1+\sqrt{2}-\sqrt{5}}{\left(1+\sqrt{2}+\sqrt{5}\right)\left(1+\sqrt{2}-\sqrt{5}\right)}\)

=\(\dfrac{1+\sqrt{2}-\sqrt{5}}{\left(1+\sqrt{2}\right)^2-\left(\sqrt{5}\right)^2}=\dfrac{1+\sqrt{2}-\sqrt{5}}{1+2\sqrt{2}+2-5}\)

=\(\dfrac{1+\sqrt{2}-\sqrt{5}}{2\sqrt{2}-2}\)

b) \(\dfrac{1}{\sqrt{x}+\sqrt{x+1}}=\dfrac{\sqrt{x}-\sqrt{x+1}}{\left(\sqrt{x}+\sqrt{x+1}\right)\left(\sqrt{x}-\sqrt{x+1}\right)}\)

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

2.

a) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)

=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)

=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20}+3}}\)

=\(\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)=\(\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

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

=\(\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)

b)\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)

=\(\sqrt{6+2\sqrt{5-\sqrt{13+2\sqrt{12}}}}\)

=\(\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\)

=\(\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\)

=\(\sqrt{6+2\sqrt{4-\sqrt{12}}}\)

=\(\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)

=\(\sqrt{6+2\sqrt{3}-2}=\sqrt{4+2\sqrt{3}}\)

=\(\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

c) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

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

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

làm giống câu a

3. a=\(\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)

=\(\sqrt{3-\sqrt{5}}\left(3\sqrt{10}+5\sqrt{2}-3\sqrt{2}-\sqrt{10}\right)\)

=\(\sqrt{3-\sqrt{5}}\left(2\sqrt{10}+2\sqrt{2}\right)\)

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

=\(\sqrt{6-2\sqrt{5}}\left(2\sqrt{5}+2\right)=\left(\sqrt{5}-1\right)\left(2\sqrt{5}+2\right)\)

=\(10-2\sqrt{5}+2\sqrt{5}-2=8\)

vậy a là số tự nhiên