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ta có
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{a+b+c}{abc}\right)=4\) (vì a+b=c=abc)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4\)
\(\Leftrightarrow M=2\)
A) P=\(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
=\(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b) Để P= -1 thì
\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=-1\)
=> \(\sqrt{x}-1=-\sqrt{x}-1\)
<=> \(\sqrt{x}+\sqrt{x}=-1+1\)
<=> \(2\sqrt{x}=0\Leftrightarrow x=0\) (m/t)
Vậy ..
ĐK x ≥ 0 , x ≠ 1; x ≠ -1
\(A=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}+\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\)
=\(\sqrt{x}+1+\sqrt{x}+1\)
=\(2\sqrt{x}+2\)
Để A có giá trị =6 thì
\(2\sqrt{x}+2=6\Leftrightarrow2\sqrt{x}=4\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)(t/m)
Vậy ....
| x+5| -6=9
<=> |x+5|= 15
TH1 x+5=15
=> x=10
TH2 x+5=-15
=> x= -20
Vậy ...
a) đkxđ x≥0 , x ≠1
\(K=\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{x-1-4\sqrt{x}+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
= \(\dfrac{x-3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)b)
\(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2-1}{\sqrt{x}-2}=1-\dfrac{1}{\sqrt{x}-2}\)
để K ∈ z thì \(\dfrac{-1}{\sqrt{x}-2}\) nguyên
=> √x -2 ∈ Ư(-1)={-1;1}
=> x ∈ {1; 9}
vậy ...
1) (x2+2xy+y2)+(x+y)
= (x+y)2 +(x+y)
= (x+y)(x+y+1)
\(Q=\dfrac{x^2-2}{x^2+1}=\dfrac{x^2+1-3}{x^2+1}=\dfrac{x^2+1}{x^2+1}-\dfrac{3}{x^2+1}=1-\dfrac{3}{x^2+1}\)
do x2 ≥ 0 ∀x
=> x2+1 ≥ 1 ∀x
=> \(\dfrac{-3}{x^2+1}\ge-3\)
=> \(1-\dfrac{3}{x^2+1}\ge1-3=-2\)
=> Q ≥ -2
min Q=-2 khi x2=0=> x=0