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Chi biet phan 5 thoi @
Vi 3a=5b=12suy ra a=4 ;b=2,4 ta co p=a.b suy ra p=4×2.4=9.6 suy ra p>[=9.6 gtln=9.6
\(a,P=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)
\(=\left(x^2+5x+5-1\right)\left(x^2+5x+5+1\right)+1\)
\(=\left(x^2+5x+5\right)^2-1+1\)
\(=\left(x^2+5x+5\right)^2\ge0\forall x\)
Vậy \(P\ge0\forall x\)
\(b,P=\left(x^2+5x+5\right)^2\left(cmt\right)\)
Thay \(x=\frac{\sqrt{7}-5}{2}\)vào P ta được
\(P=\left(\left(\frac{\sqrt{7}-5}{2}\right)^2+5.\frac{\sqrt{7}-5}{2}+5\right)^2\)
\(=\left(\frac{7-10\sqrt{7}+25}{4}+\frac{10\sqrt{7}-50}{4}+\frac{20}{4}\right)^2\)
\(=\left(\frac{32-10\sqrt{7}+10\sqrt{7}-50+20}{4}\right)^2\)
\(=\left(\frac{2}{4}\right)^2\)
\(=\frac{1}{4}\)
a,
P=(x+1)(x+2)(x+3)(x+4)+1
P=[(x+1).(x+4)].[(x+2).(x+3)]+1
P=(x^2+5x+4)(x^2+5x+6)+1
P=[(x^2+5x+5)-1].[(x^2+5x+5)+1]+1
P=(x^2+5x+5)^2-1+1
P=\(\left(x^2+5x+5\right)^2\) \(\ge\)0 với mọi x
Câu b thì thay x vào rồi bấm máy ra ra kết quả
\(A=\frac{x+\sqrt{x}}{x-2\sqrt{x}+1}\div\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{1}{1-\sqrt{x}}+\frac{2-x}{x-\sqrt{x}}\right)\)
ĐKXĐ : x > 1
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\left(\frac{\sqrt{x}+1}{\sqrt{x}}+\frac{1}{\sqrt{x}-1}+\frac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\left(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\left(\frac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\times\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(=\frac{x}{\sqrt{x}-1}\)
Để A = 9/2
=> \(\frac{x}{\sqrt{x}-1}=\frac{9}{2}\)( ĐK : x > 1 )
<=> 2x = 9( √x - 1 )
<=> 2x = 9√x - 9
<=> 2x + 9 = 9√x (1)
Bình phương hai vế
(1) <=> 4x2 + 36x + 81 = 81x
<=> 4x2 + 36x + 81 - 81x = 0
<=> 4x2 - 45x + 81 = 0
<=> 4x2 - 36x - 9x + 81 = 0
<=> 4x( x - 9 ) - 9( x - 9 ) = 0
<=> ( x - 9 )( 4x - 9 ) = 0
<=> \(\orbr{\begin{cases}x-9=0\\4x-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x=\frac{9}{4}\end{cases}}\)( tm )
Ta có: \(B=\frac{\sqrt{\frac{1}{9}}-3}{\sqrt{\frac{1}{9}}-1}\)
\(B=\frac{\frac{1}{3}-3}{\frac{1}{3}-1}\)
\(B=\frac{-\frac{8}{3}}{-\frac{2}{3}}=4\)
đkxđ: \(\hept{\begin{cases}x\ne1\\x\ne25\end{cases}}\)
Ta có:
\(A=\frac{x-21}{x-6\sqrt{x}+5}+\frac{1}{\sqrt{x}-1}+\frac{1}{5-\sqrt{x}}\)
\(A=\frac{x-21}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-5\right)}+\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}-5}\)
\(A=\frac{x-21+\sqrt{x}-5-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-5\right)}\)
\(A=\frac{x-25}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-5\right)}\)
\(A=\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-5\right)}\)
\(A=\frac{\sqrt{x}+5}{\sqrt{x}-1}\)
a, A\(=\left(\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+4\sqrt{x}\left(x-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\frac{x-1}{\sqrt{x}}\) ĐK x>0 ;\(x\ne1;x\ne-1\)
\(A=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1+4x\sqrt{x}-4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\sqrt{x}}{x-1}\)
\(A=\frac{4x\sqrt{x}}{x-1}.\frac{\sqrt{x}}{x-1}\)=\(\frac{4x^2}{\left(x-1\right)^2}\)
b, Để A =2 \(\Rightarrow\frac{4x^2}{\left(x-1\right)^2}=2\Rightarrow4x^2=2\left(x-1\right)^2\)
<=> \(4x^2=2x^2-4x+2\)
<=> \(2x^2+4x-2=0\)
<=> \(x^2+2x-1=0\)
\(\Delta=1^2-1.\left(-1\right)\) = 2
=> \(\orbr{\begin{cases}x_1=-1-\sqrt{2}\left(loại\right)\\x_2=-1+\sqrt{2}\left(nhận\right)\end{cases}}\)
Vậy x=\(-1+\sqrt{2}\)thì A =2
c, Thay x =\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)=2
=>A = \(\frac{4.2^2}{\left(2-1\right)^2}=16\)
Vậy A=16 thì x=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
A = (\(\dfrac{1}{x-\sqrt{x}}\) + \(\dfrac{1}{\sqrt{x}+1}\)) : \(\sqrt{x}\) + \(\dfrac{1}{x-2\sqrt{x}+1}\)
Có phải đề bài như này không em?