1)Tco ABCD là hình chữ nhật ( ADC=DCB=ABC=\(90^o\))

=> DC= AB=1,5(m)

=>AD=BC=4(m)

Xét tam giác ACE vuông tại A có đường cao AD

=>\(AD^2=DC.DE\)

\(\Leftrightarrow DE=\dfrac{AD^2}{DC}=\dfrac{16}{1,5}=10,7\)(m)

\(\Leftrightarrow CE=DE+DC=1,5+10,7=12,2\left(m\right)\)

Ta có \(y\left(x-1\right)=x^2+2\)

\(\Leftrightarrow y\left(x-1\right)-x^2=2\)

\(\Leftrightarrow y\left(x-1\right)-x^2+1=3\)

\(\Leftrightarrow y\left(x-1\right)-\left(x^2-1\right)=3\)

\(\Leftrightarrow y\left(x-1\right)-\left(x-1\right)\left(x+1\right)=3\)

\(\Leftrightarrow\left(x-1\right)\left(y-x-1\right)=3\)

Vì x,y nguyên nên ta có bảng

Vậy\(\left(x,y\right)=\left\{\left(4,6\right),\left(2,8\right),\left(0,2\right),\left(-2,4\right)\right\}\)thỏa mãn

Mình nghĩ là should go vf câu này là câu giả định S+ suggest+ ( that)+S+(Should)+V(bare inf)

Đk \(x\ge\dfrac{3}{2}\)

\(\Leftrightarrow3\sqrt{2x-3}-4\sqrt{2x-3}=1-2\sqrt{2x-3}\)

\(\Leftrightarrow\sqrt{2x-3}=1\)

\(\Leftrightarrow2x-3=1\)

\(\Leftrightarrow x=2\)

Vậy S=\(\left\{2\right\}\)

\(A=\dfrac{1}{6ab}\sqrt{36.a^4b^2}=\dfrac{1}{6ab}.6a^2b=a\)

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

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

\(=2\sqrt{x}-5\)

A=\(\dfrac{1}{6ab}.6a^2.b\)

=a

B=\(\sqrt{x}-2+\sqrt{x}-3\)

=\(2\sqrt{x}-5\)

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

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

\(=\dfrac{2}{x+\sqrt{x}+1}\)