Cho a, b là 2 số dương thỏa mãn a + b \(\le\) \(\dfrac{4}{5}\).

Chứng minh a + b + \(\dfrac{a+b}{ab}\) \(\ge\) \(\dfrac{29}{5}\)

a) 3x+2(x-5)=-x+2

<=> 3x+2x+x=2+10

<=>6x=12

<=>x=2

b) 3x²-2x=0

<=>x(3x-2)=0

<=>\(\left[{}\begin{matrix}x=0\\3x-2=0\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

c) \(\dfrac{2x}{3}\)+\(\dfrac{x-4}{6}\)=2-\(\dfrac{x}{2}\)

<=>\(\dfrac{8x+2x-8}{12}\)=\(\dfrac{24-6x}{12}\)

<=> 8x+2x-8=24-6x

<=>8x+2x+6x=24+8

<=>16x=32

<=>x=2

d) \(\dfrac{x-2}{x+2}\)-\(\dfrac{3}{x-2}\)= -\(\dfrac{2\left(x-11\right)}{4-x^2}\) ( ĐKXĐ: x\(\ne\)\(\pm\)2)

<=> \(\dfrac{\left(x-2\right)^2-3\left(x+2\right)}{x^2-4}\)=\(\dfrac{2\left(x-11\right)}{x^2-4}\)

=> (x-2)²-3(x+2)=2(x-11)

<=> x²-4x+4-3x-6=2x-22

<=> x²-4x-3x-2x=-22-4+6

<=> x-9x+20=0

<=> (x-4)(x-5)=0

<=>\(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\) ( thỏa mãn diều kiện )

d) (x²+1)(x²-4x+4)=0

=> x²-4x+4=0 (x²+1\(\ge\)1 với mọi x)

=>(x-2)² =0

=>x=2

Cho a, b, c > 0 thỏa mãn a + b +c =3

Chứng minh : \(\dfrac{a}{b^2+1}\)+\(\dfrac{b}{c^2+1}\)+\(\dfrac{c}{a^2+1}\) \(\ge\) \(\dfrac{3}{2}\)

Tính giá trị của biểu thức A = \(\dfrac{yz}{x^2}\)+\(\dfrac{xz}{y^2}\)+\(\dfrac{xy}{z^2}\) , biết rằng xy+yz+xz=0 và xyz \(\ne\) 0

Do a, b, c >0

=> a+b+c>0 và \(\dfrac{a^2}{c}\)+\(\dfrac{b^2}{a}\)+\(\dfrac{c^2}{b}\) >0

Áp dụng bất đẳng thức Cô si ta có:

\(\dfrac{a^2}{c}\)+\(\dfrac{b^2}{a}\)+\(\dfrac{c^2}{b}\) \(\ge\) 3 \(\sqrt[3]{\dfrac{a^2b^2c^2}{abc}}\) = 3\(\sqrt[3]{abc}\)

a+b+c \(\ge\) 3 \(\sqrt[3]{abc}\)

=> \(\dfrac{a^2}{c}\)+\(\dfrac{b^2}{a}\)+\(\dfrac{c^2}{b}\) - (a+b+c) \(\ge\) 3\(\sqrt[3]{abc}\) - 3\(\sqrt[3]{abc}\)

=>\(\dfrac{a^2}{c}\)+\(\dfrac{b^2}{a}\)+\(\dfrac{c^2}{b}\)- (a+b+c) \(\ge\) 0

=> \(\dfrac{a^2}{c}\)+\(\dfrac{b^2}{a}\)+\(\dfrac{c^2}{b}\) \(\ge\) a+b+c (dpcm)

Ta có: \(\dfrac{\left(x-2\right)\left(x+3\right)}{x-1}\)>0

Th1:\(\left\{{}\begin{matrix}\left(x-2\right)\left(x+3\right)>0\\x-1>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-2>0\\x+3>0\\x-1>0\end{matrix}\right.\) Hoặc \(\left\{{}\begin{matrix}x-2< 0\\x+3< 0\\x-1>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>2\\x>-3\\x>1\end{matrix}\right.\) Hoặc \(\left\{{}\begin{matrix}x< 2\\x< -3\\x>1\end{matrix}\right.\)( Vô lý )

=>x>2

Th2:\(\left\{{}\begin{matrix}\left(x-2\right)\left(x+3\right)< 0\\x-1< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-2>0\\x+3< 0\\x-1< 0\end{matrix}\right.\) Hoặc \(\left\{{}\begin{matrix}x-2< 0\\x+3>0\\x-1< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>2\\x< -3\\x< 1\end{matrix}\right.\) ( Vô lý ) Hoặc \(\left\{{}\begin{matrix}x< 2\\x>-3\\x< 1\end{matrix}\right.\)

=> -3<x<1

Vậy nghiệm của phương trình là

x>2 hoặc -3<x<1