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1.: Áp dụng BĐT Cauchy-Schwarz cho 3 số dương
\(a+b+c\ge3\sqrt[3]{abc};\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\frac{1}{abc}}=9\)
5x-2>2(x+3)\(\Leftrightarrow\)5x-2>2x+6
\(\Leftrightarrow\) 5x-2x>6+2
\(\Leftrightarrow\)3x>8
\(\Leftrightarrow\)x>\(\dfrac{8}{3}\)
Chúc bn học tốt❤
\(ĐKXĐ:x\ne-1;x\ne2\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5x+5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow\frac{x-2-5x-5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow x-2-5x-5=15\)
\(\Leftrightarrow-4x=22\Leftrightarrow x=\frac{-11}{2}\)
Vậy \(S=\left\{\frac{-11}{2}\right\}\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(x-2\right)}\left(ĐKXĐ:x\ne-1;x\ne2\right)\)
\(\Leftrightarrow\frac{1\left(x-2\right)-5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\frac{x-2-5x-5}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\frac{-4x-7}{\left(x+1\right)\left(x-2\right)}=\frac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow-4x-7=15\)
\(\Leftrightarrow-4x=22\)
\(\Leftrightarrow x=22:\left(-4\right)\)
\(\Leftrightarrow x=\frac{-22}{4}=\frac{-11}{2}\)
Vậy tập nghiệm \(S=\left\{\frac{-11}{2}\right\}\)
a( ax + 1) = x( a + 2) + 2
⇔ a2x + a - ax- 2x = 2 - a
⇔ x( a2 - a - 2 ) = 2 - a
⇔ x( a2 + a - 2a - 2) = 2 - a
⇔ x[ a( a + 1) -2( a + 1) ] = 2 - a
⇔ x( a + 1)( a - 2) = 2 - a ( *)
+) Với : a # 2 ; a # - 1 , ta có :
( * ) ⇔ x = \(\dfrac{-1}{a+1}\)
+) Với : a = 2 , ta có :
( * ) ⇔ 0x= 0 ( Luôn đúng )
+) Với : a = - 1 , ta có :
( * ) ⇔0x = 3 ( Vô lý )
KL.....
<=>(a^2-a-2)x=2-a
[(a-1/2)^2-(3/2)^2]x=2-a
<=>(a+1)(a-2)x=2-a
a=2 ; =>moi x$R
a=-1 ; vo nghiem
a≠{-1;2}: x=-1/(a+1)
ĐK: a \(\ne\) 0
BPT tương đương
x +\(\frac{x}{a}\)- \(\frac{1}{a}\)- \(\frac{x}{a}\)- \(\frac{1}{a}\)+ (a - 2)x < 0
<=> x - \(\frac{2}{a}\)+ (a - 2) x < 0
<=> (a - 1)x < \(\frac{2}{a}\)
TH1: a = 1: BPT luôn đúng với mọi x
TH2: a > 1: BPT tương đương:
x < \(\frac{2}{a\left(a-1\right)}\)
TH3: a < 1 (a\(\ne\)0) BPT tương đương:
x > \(\frac{2}{a\left(a-1\right)}\)