a)    \(VT=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)\(=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\) (vì  a+b+c = 1)

\(=3+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\)

\(=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)

C/m  BĐT phụ:   \(\frac{x}{y}+\frac{y}{x}\ge2\)   với  x,y dương

             \(\Leftrightarrow\)\(x^2+y^2\ge2xy\)

            \(\Leftrightarrow\) \(x^2-2xy+y^2\ge0\)

            \(\Leftrightarrow\) \(\left(x-y\right)^2\ge0\)  luôn đúng

Dấu "=" xảy ra   \(\Leftrightarrow\)\(x=y\)

Áp dụng BĐT trên ta có:   \(\frac{a}{b}+\frac{b}{a}\ge2;\) \(\frac{a}{c}+\frac{c}{a}\ge2;\) \(\frac{b}{c}+\frac{c}{b}\ge2\)

\(\Rightarrow\)\(VT=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge3+2+2+2=9\)

Dấu "=" xảy ra  \(\Leftrightarrow\)\(a=b=c\)

Vậy    \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)

Dấu "=" xảy ra  \(\Leftrightarrow\)\(a=b=c\)

Đk: \(x\ge1\)

\(\Leftrightarrow4\left(2\sqrt{x-1}-1\right)+\left(4x-5\right)\left(x+2\right)=0\)

\(\Leftrightarrow\dfrac{4\left(4x-5\right)}{2\sqrt{x-1}+1}+\left(4x-5\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(4x-5\right)\left(\dfrac{4}{2\sqrt{x-1}+1}+x+2\right)=0\)

\(\Leftrightarrow x=\dfrac{5}{4}\)(Dễ thấy ngoặc to lớn hơn 0 với \(x\ge1\))

Bạn làm chi tiết ra nữa đc khum? Như thế mình vẫn chưa hiểu lắm :((

x(3x-1)-(3x+2)(x-5)=0

<=> 3x^2-x-3x^2+15x-2x+10=0

<=>12x+10=0

<=>12x=-10

<=>x=-5/6

`(x+1)(x+3)=2x^2-2`

`<=>x^2+x+3x+3=2x^2-2`

`<=>x^2-4x-5=0`

`<=>x^2-5x+x-5=0`

`<=>x(x-5)+(x-5)=0`

`<=>(x-5)(x+1)=0`

`<=>` $\left[ \begin{array}{l}x=5\\x=-1\end{array} \right.$

Vậy `S={5,-1}`

Ta có: \(\left(x+1\right)\left(x+3\right)=2x^2-2\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-2x^2+2=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-2\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3\right)-2\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left[x+3-2\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+3-2x+2\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(5-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)

Vậy: S={-3;5}

bó tay

theo BĐT CÔ SI ta đc 

a+b+c \(\ge\)\(3\sqrt[3]{abc}\)

 1/ a + 1/ b + 1/c  \(\ge\)\(3\sqrt[3]{\frac{1}{abc}}\)

nhân vế vs vế ta đc ( a+ b+c) (  1/ a + 1/ b + 1/c ) \(\ge\)9

maf a +b+c = 1 nên ......bn tự lm nha

Theo đề bài thì ta có:

\(\hept{\begin{cases}3x_1^2+5x_1+4-m=0\\x_2^2-5x_2+4+m=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}9x_1^2+15x_1+12-3m=0\left(1\right)\\x_2^2-5x_2+4+m=0\left(2\right)\end{cases}}\)

Lấy (1) - (2) ta được

\(\left(9x_1^2-x_2^2\right)+\left(15x_1+5x_2\right)+8-4m=0\)

\(\Leftrightarrow\left(3x_1+x_2\right)\left(3x_1-x_2+5\right)+8-4m=0\)

\(\Leftrightarrow\left(3x_1+x_2\right)\left(3x_1+x_2-2x_2+5\right)+8-4m=0\)

\(\Leftrightarrow\left(6-2x_2\right)+8-4m=0\)

\(\Leftrightarrow x_2=7-2m\)

Thế lại vô (2) ta được

\(\left(7-2m\right)^2-5\left(7-2m\right)+4+m=0\)

\(\Leftrightarrow4m^2-17m+18=0\)

\(\Leftrightarrow\orbr{\begin{cases}m=2\\m=\frac{9}{4}\end{cases}}\)

Oh thanks you very muck!!!!

( 3x-1) ( x²+ 9) = (3x-1) (7x-10)

⇒( 3x-1) ( x²+ 9) - (3x-1) (7x-10) = 0

⇒( 3x-1) (( x²+ 9)-(7x-10)) = 0

⇒( 3x-1)(x²+9-7x+10)=0

⇒( 3x-1)(x²-7x+19)=0

⇒\(\left[{}\begin{matrix}3x-1=0\\x^2-7x+19=0\end{matrix}\right.\)

3x-1=0

⇒x=\(\dfrac{1}{3}\)

x²-7x+19=0

⇒ \(x^2-\dfrac{7}{2}x-\dfrac{7}{2}x+\left(\dfrac{7}{2}\right)^2+\dfrac{27}{4}=0\)

⇒ \(\left(x-\dfrac{7}{2}\right)^2+\dfrac{27}{4}=0\)

vì \(\left(x-\dfrac{7}{2}\right)^2\ge0\); \(\dfrac{27}{4}>0\)

⇒ \(\left(x-\dfrac{7}{2}\right)^2+\dfrac{27}{4}>0\)

⇒ x vô nghiệm

Vậy x= \(\dfrac{1}{3}\)

\(\left(3x-1\right)\left(x^2+9\right)=\left(3x-1\right)\left(7x-10\right)\\ \Leftrightarrow\left(3x-1\right)\left(x^2+9\right)-\left(3x-1\right)\left(7x-10\right)\\ \Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\\ \Leftrightarrow\left(3x-1\right)\left(x^2-4x-3x+12\right)=0\\ \Leftrightarrow\left(3x-1\right)\left[x\left(x-4\right)-3\left(x-4\right)\right]=0\\ \Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)