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

\(=\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}\cdot1+1^2}+\left|\sqrt{2}-2\right|\)

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

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

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

\(=3\)

b) \(\dfrac{1}{5}\sqrt{50}-2\sqrt{96}-\dfrac{\sqrt{30}}{\sqrt{15}}+12\sqrt{\dfrac{1}{6}}\)

\(=\dfrac{1}{5}\cdot5\sqrt{2}-2\cdot4\sqrt{6}-\sqrt{\dfrac{30}{15}}+\sqrt{\dfrac{144}{6}}\)

\(=\sqrt{2}-8\sqrt{6}-\sqrt{2}+2\sqrt{6}\)

\(=-8\sqrt{6}+2\sqrt{6}\)

\(=-6\sqrt{6}\)

c) \(\left(\dfrac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\dfrac{4}{1+\sqrt{5}}+4\right)\)

\(=\left[\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}-2\right]\left[\dfrac{4\left(1-\sqrt{5}\right)}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}+4\right]\)

\(=\left(\sqrt{5}-1-2\right)\left(\dfrac{4\left(1-\sqrt{5}\right)}{1-5}+4\right)\)

\(=\left(\sqrt{5}-3\right)\left(\sqrt{5}-1+4\right)\)

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

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

\(=-4\)

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

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

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

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

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

\(A=\dfrac{\sqrt{a}+1}{\sqrt{a}}:\dfrac{\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

\(A=\dfrac{\sqrt{a}+1}{\sqrt{a}}\cdot\left(\sqrt{a}-1\right)\)

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

\(A=\dfrac{a-1}{\sqrt{a}}\)

b) Ta có:

\(a=4+2\sqrt{3}=\left(\sqrt{3}\right)^2+2\sqrt{3}\cdot1+1^2=\left(\sqrt{3}+1\right)^2\)

Thay vào A ta có:

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

c) \(A< 0\) khi:

\(\dfrac{a-1}{\sqrt{a}}< 0\)

Mà: \(\sqrt{a}\ge0\forall x\) (xác định) 

\(\Leftrightarrow a-1< 0\)

\(\Leftrightarrow a< 1\)

Kết hợp với đk:

\(0< a< 1\)

\(\left(x^2-\dfrac{y}{2}\right)^3\)

\(=\left(x^2\right)^3-3\cdot\left(x^2\right)^2\cdot\dfrac{y}{2}+3\cdot x^2\cdot\left(\dfrac{y}{2}\right)^2-\left(\dfrac{y}{2}\right)^3\)

\(=x^6-\dfrac{3x^4y}{2}+\dfrac{3x^2y^2}{4}-\dfrac{y^3}{8}\)

Em đã nhận giải thưởng phụ rồi ạ em cảm ơn cô Thương Hoài rất nhiều đã tổ chức ra một cuộc thi bổ ích và thú vị như vật ạ 

\(\dfrac{3}{2}+\left(\dfrac{4}{5}:x-\dfrac{1}{2}\right)=4\)

\(\dfrac{3}{2}+\dfrac{4}{5}:x-\dfrac{1}{2}=4\)

\(\left(\dfrac{3}{2}-\dfrac{1}{2}\right)+\dfrac{4}{5}:x=4\)

\(\dfrac{2}{2}+\dfrac{4}{5}:x=4\)

\(1+\dfrac{4}{5}:x=4\)

\(\dfrac{4}{5}:x=4-1\)

\(\dfrac{4}{5}:x=3\)

\(x=\dfrac{4}{5}:3\)

\(x=\dfrac{4}{15}\)

\(x+x\times\dfrac{1}{4}+x\times\dfrac{3}{4}=200\)

\(x\times\left(1+\dfrac{1}{4}+\dfrac{3}{4}\right)=200\)

\(x\times\left(1+1\right)=200\)

\(x\times2=200\)

\(x=200:2\)

\(x=100\)

\(A=x-x^2-1\)

\(A=-\left(x^2-x+1\right)\)

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

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

Mà: \(-\left(x-\dfrac{1}{2}\right)^2\le0\forall x\)

Và: \(-\dfrac{3}{4}< 0\)

\(\Rightarrow A=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}< 0\forall x\)

c) \(x^2-9=2\cdot\left(x+3\right)^2\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-2\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(x+3\right)\left[x-3-2\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-3-2x-6\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(-x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)

b) \(x^3-3x^2+3x-1=0\)

\(\Leftrightarrow x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

d) \(x^2-8x+3x-24=0\)

\(\Leftrightarrow\left(x^2-8x\right)+\left(3x-24\right)=0\)

\(\Leftrightarrow x\left(x-8\right)+3\left(x-8\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=8\end{matrix}\right.\)

Để:

1111111*33333333 chia hết cho 7 thì:

⇒ 11111111*3333333 + (3 x 5) phải chia hết cho 7

⇒ 11111111+* x 100000000 + 33333333 + 15 chia hết cho 7 

⇒ 44444459 + * x 100000000 chia hết cho 7 

⇒ * = 2 

Chiều dài thực tế của mảnh vườn:

\(6\times500=3000\left(cm\right)=30\left(m\right)\)

Chiều rộng thực tế của mảnh vườn:

\(4\times500=2000\left(cm\right)=20\left(m\right)\)

Diện tích của mảnh vườn:

\(30\times20=600\left(m^2\right)\)

Số lượng ngô thu hoạch được là:

\(600:5\times30=3600\left(kg\right)\)

Đổi: 3600(kg) = 360(tạ)

Đáp số: 360 tạ