\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)

A = x² - 8x + 1 = (x² - 8x + 16) - 15 = (x - 4)² - 15

Ta có: (x - 4)² \(\ge\)0 \(\forall\)x

=> (x - 4)² - 15 \(\ge\)-15 \(\forall\) x 

Dấu "=" xảy ra khi: x - 4 = 0 <=> x = 4

vậy Min của A = -15 tại x = 4

B = 9x² - 12x - 2 = 9(x² - 4/3x + 4/9) - 6 = 9(x - 2/3)² - 6

Ta có: (x - 2/3)² \(\ge\)0 \(\forall\)x ---> 9(x - 2/3)² \(\ge\)0 \(\forall\)x

=> 9(x - 2/3)² - 6 \(\ge\)-6 \(\forall\)x

Dấu "=" xảy ra khi: x - 2/3 = 0 <=> x = 2/3

vậy Min của B = -6 tại x = 2/3

a, 15x³ - 15x = 0    

15x(x²-1)=0

15x=0 hoặc x²-1=0  (tự tính nhoa)

b,3x²-6x+3=0

3(x²-2x+1)=0

x² -2x+1=0:3=3

x²-2x=3-1=2

x(x-2)=0

x=0 hoặc x-2=0 (tự tính nhoa)

Bài làm

a) 15x³-15x=0

<=> 15x( x² - 1 ) = 0

<=> \(\orbr{\begin{cases}15x=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}}\)

Vậy x = { 0; + 1 }

b) 3x² - 6x + 3 = 0

<=> 3( x² - 2x + 1 ) = 0

<=> x² - 2x + 1 = 0

<=> ( x - 1 )² = 0

<=> x - 1 = 0

<=> x = 1

Vậy x = 1

c) 5(x - 1) - 3x(1 - x) = 0

<=> 5(x - 1) + 3x(x - 1) = 0

<=> (5 + 3x)(x - 1) = 0

<=> \(\orbr{\begin{cases}5+3x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=1\end{cases}}}\)

Vậy x = { -5/3; 1 }

e) -7(x + 2) = 2x(x + 2) 

<=> -7(x + 2 ) - 2x( x + 2 ) = 0

<=> (x + 2)(-7 - 2x) = 0

<=> \(\orbr{\begin{cases}x+2=0\\-7-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{7}{2}\end{cases}}}\)

Vậy x = { -2; x = -7/2 }

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

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

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

<=> (3x + 5)(x - 2) = 0

<=> \(\orbr{\begin{cases}3x+5=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=2\end{cases}}}\)

Vậy x = { -5/3; 2 }

không ai trả lời thì mình giúp vậy 

\(c,5\left(x-1\right)-3x\left(1-x\right)=0\)

\(< =>5x-5-3x+3x^2=0\)

\(< =>3x^2+2x-5=0< =>3x^2-3x+5x-5=0\)

\(< =>3x\left(x-1\right)+5\left(x-1\right)=\left(2x+5\right)\left(x-1\right)=0\)

\(< =>\orbr{\begin{cases}2x+5=0\\x-1=0\end{cases}}< =>\orbr{\begin{cases}2x=-5\\x=1\end{cases}}< =>\orbr{\begin{cases}x=-\frac{5}{2}\\x=1\end{cases}}\)

\(d,6x-\left(x+2\right)^2-5=\left(1-3x\right)^3-12x\)

\(< =>6x-x^2-4x-9=1-27x^3+27x^2-21x\)

\(< =>-x^2+2x-9-x+27x^3-27x^2+21x=0\)

\(< =>27x^3-28x^2+22x-9=0\)(vô nghiệm)

\(e,-7\left(x+2\right)=2x\left(x+2\right)\)

\(< =>-7x-14-2x^2-4x=0\)

\(< =>-2x^2-11x-14=0\)

\(< =>-2\left(x^2+\frac{11}{2}x+\frac{121}{16}\right)+\frac{9}{8}=0\)

\(< =>\left(x+\frac{11}{4}\right)^2=-\frac{9}{8}=\frac{9}{16}\)

\(< =>\orbr{\begin{cases}x+\frac{11}{4}=\sqrt{\frac{9}{16}}\\x+\frac{11}{4}=-\sqrt{\frac{9}{16}}\end{cases}}< =>\orbr{\begin{cases}x=\frac{3}{4}-\frac{11}{4}=-2\\x=-\frac{3}{4}-\frac{11}{4}=-\frac{14}{4}=-\frac{7}{2}\end{cases}}\)

\(f,\left(2x-3\right)\left(3x+5\right)=\left(x-1\right)\left(3x+5\right)\)

\(< =>2x-3=x-1< =>2x-x-3+1=0< =>x=2\)

bài cuối mình chịu đấy 

a) 4x(3x-7)-6(2x²-5x+1)=12

 =>4x.3x-4x.7-6.2x²-6.(-5x)-6.1=12

 =>12x²-28x-12x²+30x-6=12

 =>2x-6                         =12

 =>2x                            =12+6

 =>2x                            =18

 =>x                              =18:2

 =>x                              =6

b)(5x+3)(4x-1)+(10x-7)(-2x+3)=27

=>5x.4x-5x.1+3.4x+3.(-1)+10x.(-2x)+10x.3-7.-(2x)-7.3=27

=>20x²-5x+12x-3-20x²+30x+14x-21=27

=>39x-36                                      =27

=>39x                                          =27+36

=>39x                                          =63

=>x                                              =63:39

=>x                                              =21/13

c) (8x-5)(3x+2)-(12x+7)(2x-1)=17

=>8x.3x+8x.2-5.3x-5.2-12x.2x-12x.(-1)+7.2x+7.(-1)=17

=>24x²+16x-15x-10-24x²+12x+14x-7=17

=>27x-17                                        =17

=>27x                                            =17+17

=>27x                                            =34

=>x                                                =34:27

=>x                                                =34/27

d) (5x+9)(6x-1)-(2x-3)(15x+1)=-190

=>30x²-5x+63x-9 - 30x²-2x-45x-3=-190

=>11x-12                                   =-190

=>11x                                        =-190+12

=>11x                                        =-178

=>x                                            = -178:11

=>x                                            =-178/11

\(a,A=x^2+15x-25\)

\(=x^2+2.x.\frac{15}{2}+\frac{125}{4}-\frac{125}{4}-25\)

\(=\left(x+\frac{15}{2}\right)^2-\frac{225}{4}\)

\(A_{min}=-\frac{225}{4}\Leftrightarrow\left(x+\frac{15}{2}\right)^2=0\)

\(\Leftrightarrow x=-\frac{15}{2}\)

\(A=x^2-6x+10=\left(x-3\right)^2+1\ge1\)

\(\Rightarrow A_{min}=1\Leftrightarrow x=3\)

\(B=4x^2-4x+25=\left(2x-1\right)^2+24\ge24\)

\(\Rightarrow B_{min}=24\Leftrightarrow x=\frac{1}{2}\)

\(C=3x^2+9x+12=3\left(x+\frac{3}{2}\right)^2+\frac{21}{4}\ge\frac{21}{4}\)

\(\Rightarrow C_{min}=\frac{21}{4}\Leftrightarrow x=\frac{-3}{2}\)

a) = 3(x²-2x+1) +1-3

GTNN = -2

B) tt