Gọi số học sinh lớp 9B là : \(x\) (học sinh) \(\left(x\inℕ^∗\right)\)

\(\Rightarrow\) Số học sinh lớp 9A là : \(x+10\) (học sinh)

Số cây học sinh lớp 9B trồng được : \(4x\) (cây)

Số cây học sinh lớp 9A trồng được : \(3\left(x+10\right)\) (cây)

Vì tổng số cây 2 lớp trồng được là : 275 cây

Nên ta có pt :

\(3\left(x+10\right)+4x=275\\ \Rightarrow3x+30+4x=275\\ \Rightarrow7x=245\\ \Rightarrow x=35\left(TMDK\right)\)

Vậy số HS lớp 9B là : 35 HS và lớp 9A là : 35+10=45(HS)

HS lớp 9B là 35hs. HS lớp 9A là 45hs.

Chào em, đó là nhân vật Lão Hạc chứ không phải Lão Hạt nhé!

bạn ghi sai lỗi chính tả rồi kìa!

tại vì đây là một cái tên do họ đặt

The new policy is expected to bring about positive changes in the healthcare system

I strongly agree with the idea :" wearing uniform is necessary for all students when they are at school "

Firstly , wearing uniform encourages students to be proud of being students of their school because they are wearing uniform with labels bearing their school's name. Secondly , wearing uniform can save money and time . Each morning or each day , students don't need to decide what to wear , they simply get dressed in their uniform. Furthermore , wearing uniform helps us feel equal in many ways , whether they are rich or poor . In addition, everyone wearing the same uniform allows students to be easily identified. If everyone was wearing different clothes , nobody would know where a students was from. And when you wearing uniform you look not scruffy . In conclusion wearing uniform is necessary for all students when they are at school.

Từ giả thiết \(\Rightarrow a+b=abc-c=c\left(ab-1\right)\Rightarrow c=\dfrac{a+b}{ab-1}\) (hiển nhiên \(ab-1>0\) do \(a+b>0\))

Đặt \(P=\dfrac{\sqrt{1+a^2}}{a}+\dfrac{\sqrt{1+b^2}}{b}-\sqrt{1+c^2}\)

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

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

\(\Rightarrow P< \dfrac{\sqrt{1+a^2}}{a}+\dfrac{\sqrt{1+b^2}}{b}-\dfrac{\sqrt{\left(a^2+1\right)\left(b^2+1\right)}}{ab}\)

Đặt \(\left\{{}\begin{matrix}\dfrac{\sqrt{1+a^2}}{a}=\sqrt{1+\dfrac{1}{a^2}}=x>1\\\dfrac{\sqrt{1+b^2}}{b}=\sqrt{1+\dfrac{1}{b^2}}=y>1\end{matrix}\right.\)

\(\Rightarrow P< x+y-xy=x+y-xy-1+1=\left(x-1\right)\left(1-y\right)+1\)

Do \(x>1;y>1\Rightarrow\left(x-1\right)\left(1-y\right)< 0\Rightarrow P< 1\)

\(B=\sqrt{\dfrac{8+\sqrt{15}}{2}}+\sqrt{\dfrac{8-\sqrt{15}}{2}}\) 

\(B=\dfrac{\sqrt{8+\sqrt{15}}}{\sqrt{2}}+\dfrac{\sqrt{8-\sqrt{15}}}{\sqrt{2}}\)

\(B=\dfrac{\sqrt{2}\cdot\sqrt{8+\sqrt{15}}}{\sqrt{2}\cdot\sqrt{2}}+\dfrac{\sqrt{2}\cdot\sqrt{8-\sqrt{15}}}{\sqrt{2}\cdot\sqrt{2}}\)

\(B=\dfrac{\sqrt{16+2\sqrt{15}}}{2}+\dfrac{\sqrt{16-2\sqrt{15}}}{2}\)

\(B=\dfrac{\sqrt{\left(\sqrt{15}\right)^2+2\cdot\sqrt{15}\cdot1+1^2}}{2}+\dfrac{\sqrt{\left(\sqrt{15}\right)^2-2\cdot\sqrt{15}\cdot1+1^2}}{2}\)

\(B=\dfrac{\sqrt{\left(\sqrt{15}+1\right)^2}}{2}+\dfrac{\sqrt{\left(\sqrt{15}-1\right)^2}}{2}\)

\(B=\dfrac{\sqrt{15}+1+\sqrt{15}-1}{2}\)

\(B=\dfrac{2\sqrt{15}}{2}\)

\(B=\sqrt{15}\)

ta có P(1)=1+a+b+c+d+e=3

        P(2)=32+16a+8b+4c+2d+e=9

       P(3)=243+81a+27b+9c+3d+e=19

       P(4)=1024+256a+64b+16c+4d+e=33

      P(5)=3125+625a+125b+25c+5d+e=51

<=> P(1)=a+b+c+d+e=2

      P(2)=16a+8b+4c+2d+e=-23

      P(3)=81a+27b+9c+3d+e=-224

      P(4)=256a+64b+16c+4d+e=-991

      P(5)=625a+125b+25c+5d+e=-3074

<=> 15a+7b+3c+d=-25

        65a+19b+5c+d=-201 

      175a+37b+7c+d=-767

       369a+61b+9c+d=-2083

 <=> a=-15

         b=85

         c=-223

         d=274

Nên e=-119

Vậy P(x)= x⁵-15x⁴+85x³-223x²+274x-119

=> P(6)=193

      P(7)=819

    P(8)=2649

    P(9)=6883

    P(10)=15321

   P(11)=30483