gọi v và v+15 ( v >0)

Ta có pt
\(\frac{90}{v}=\frac{90}{v+15}+1\)

bạn tự giải nhá!

\(\Leftrightarrow v^2\)+ 15v - 1350 =0

\(\Leftrightarrow\orbr{\begin{cases}v=30\\v=-45\left(l\right)\end{cases}}\)

\(\Rightarrow v+15=45\)

Vậy....

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

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

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

\(A=\frac{a-1}{\sqrt{a}}.\frac{a}{\sqrt{a}+1}\)

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

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

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

=\(\left(\frac{a-1}{\sqrt{a}}\right)\). \(\frac{a}{\sqrt{a}+1}\)= \(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\)\(\frac{a}{\sqrt{a}+1}\)= \(\frac{\sqrt{a}-1}{\sqrt{a}}\)

Ta có \(P\left(1\right)=1^2+2m+m^2\)

\(Q\left(-1\right)=1-2m-1-1+m^2=m^2-2m-1\)

\(\Rightarrow m^2+2m+1=m^2-2m-1\Rightarrow4m=-2\Rightarrow m=-\frac{1}{2}\)

Vậy \(m=-\frac{1}{2}\)thì \(P\left(1\right)=Q\left(-1\right)\)

ai k mình k lại [ chỉ 3 người đầu tiên mà trên 10 điểm hỏi đáp ]

Đk:\(x\ge0\)

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

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

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

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

\(\Leftrightarrow\frac{x-1}{\sqrt{x+3}+2}+\frac{3\left(x-1\right)}{\sqrt{3x+1}-2}-\frac{4\left(x-1\right)}{2\sqrt{x}+2}-\frac{2\left(x-1\right)}{\sqrt{2x+2}+2}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{\sqrt{x+3}+2}+\frac{3}{\sqrt{3x+1}-2}-\frac{4}{2\sqrt{x}+2}-\frac{2}{\sqrt{2x+2}+2}\right)=0\)

Dễ thấy: \(\frac{1}{\sqrt{x+3}+2}+\frac{3}{\sqrt{3x+1}-2}-\frac{4}{2\sqrt{x}+2}-\frac{2}{\sqrt{2x+2}+2}>0\)

\(\Rightarrow x-1=0\Rightarrow x=1\)

Có \(\sqrt{8}\). 4 = \(\sqrt{\frac{128}{16}}\).4 > \(\sqrt{\frac{81}{16}}\).4 = 9/4 . 4 =9 = 3.3

<=> \(\frac{\sqrt{8}}{3}\)> 3/4 

ai k mình k lại [ chỉ 3 người đầu tiên mà trên 10 điểm hỏi đáp ]