Bài 1:

a) \(6x\left(3x+15\right)-2x\left(9x-2\right)=17\) (1)

\(\Leftrightarrow18x^2+90x-18x^2+4x=17\)

\(\Leftrightarrow94x=17\)

\(\Leftrightarrow x=\dfrac{17}{94}\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{17}{94}\right\}\)

b) \(\left(15x-2x\right)\left(4x+1\right)-\left(13x-4x\right)\left(2x-3\right)-\left(x-1\right)\left(x+2\right)+x+2=52\)

\(\Leftrightarrow\left(60x^2+15x-8x^2-2x\right)-\left(26x^2-39x-8x^2+12x\right)-\left(x^2+2x-x-2\right)+x+2=52\)

\(\Leftrightarrow60x^2+15x-8x^2-2x-26x^2+39x+8x^2-12x-x^2-2x+x+2+x+2=52\)

\(\Leftrightarrow33x^2+40x+4=52\)

\(\Leftrightarrow33x^2+40x=48\)

...

Bài 1 có ng làm rồi nên mình không làm nx nhé.

2) a) Rút gọn

P=\(3x\left(4x+1\right)+5x^2-4x\left(3x+9\right)+x\left(5x-5x^2\right)\)

P= \(12x^2+3x+5x^3-12x^3-36x+5x^2-5x^3\)

P= \(-33x\)

b) |x| = 2

\(\Rightarrow\left\{{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Với x = 2 \(\Rightarrow\) P = -33 . 2 = -66

Với x = -2 \(\Rightarrow\) P = -33 . (-2) = 66

c) Để P = 2017 \(\Rightarrow\) -33x = 2017 \(\Rightarrow\) x = \(-\dfrac{2017}{33}\)

Bài 3: Giải

f(x) = \(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

f(x) = \(\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

f(x) = \(\left(x^2+5x\right)^2-6^2\) ( Hằng đẳng thức số 3 )

f(x) = \(\left(x^2+5x\right)^2-36\ge-36\) với mọi x

Vậy \(Min_{f\left(x\right)}\) = -36 khi x = 0 hoặc x = -5

Nguyễn Huy Tú tl dùm đc k mk cần lắm

Thanks nhiều

\(A=x^2-4x+10=x^2-4x+4+6=\left(x-2\right)^2+6\ge6\)

Vậy GTNN A là 6 khi x - 2 = 0 <=> x = 2 

\(B=\left(1-x\right)\left(3x-4\right)=3x-4-3x^2+4x=-3x^2+7x-4\)

\(=-3\left(x^2-\frac{7}{3}x+\frac{4}{3}\right)=-3\left(x^2-2.\frac{7}{6}x+\frac{49}{36}-\frac{1}{36}\right)=-3\left(x-\frac{7}{6}\right)^2+\frac{1}{12}\ge\frac{1}{12}\)

\(=3\left(x-\frac{7}{6}\right)^2-\frac{1}{12}\le-\frac{1}{12}\)Vậy GTLN B là -1/12 khi x = 7/6 

\(C=3x^2-9x+5=3\left(x^2-3x+\frac{5}{3}\right)=3\left(x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{7}{12}\right)\)

\(=3\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\ge-\frac{7}{4}\)Vậy GTNN C là -7/4 khi x = 3/2 

\(D=-2x^2+5x+2=-2\left(x^2-\frac{5}{2}x-1\right)=-2\left(x^2-2.\frac{5}{4}x+\frac{25}{16}-\frac{41}{16}\right)\)

\(=-2\left(x-\frac{5}{4}\right)^2+\frac{21}{8}\le\frac{21}{8}\)Vậy GTLN D là 21/8 khi x = 5/4 

a.\(A=\frac{3x^2-x+3}{3x+2}=\frac{3x^2+2x-3x-2+5}{3x+2}=x-1+\frac{5}{3x+2}\)

là số nguyên khi 3x+2 là ước của 5 hay \(\orbr{\begin{cases}3x+2=\pm1\\3x+2=\pm5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

b.\(B=\frac{2x^3-9x^2+10x+4}{2x-1}=\frac{2x^3-x^2-8x^2+4x+6x-3+7}{2x-1}=x^2-4x+3+\frac{7}{2x-1}\)

là số nguyên khi 2x-1 là ước của 7 hay \(\orbr{\begin{cases}2x-1=\pm7\\2x-1=\pm1\end{cases}}\Leftrightarrow x\in\left\{-3,0,1,4\right\}\)

Mình nghĩ ra câu C rồi bạn nào giúp mình nghĩ nốt câu A,B hộ mình nhé mình cảm ơn!

a:6x-5-9x^2

=-(9x^2-6x+5)

=-(9x^2-6x+1+4)

=-(3x-1)^2-4<=-4

=>A>=2/-4=-1/2

Dấu = xảy ra khi x=1/3

b: \(B=\dfrac{4x^2-6x+4-1}{2x^2-3x+2}=2-\dfrac{1}{2x^2-3x+2}\)

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

=2(x^2-2*x*3/4+9/16+7/16)

=2(x-3/4)^2+7/8>=7/8

=>-1/2x^2-3x+2<=-1:7/8=-8/7

=>B<=-8/7+2=6/7

Dâu = xảy ra khi x=3/4

a)đkxđ: \(x+1\ne0\Leftrightarrow x\ne-1\)

 \(B=\frac{x^2-x+1}{x^2+2x+1}=\frac{x^2+2x+1-3x}{x^2+2x+1}=1-\frac{3x}{\left(x+1\right)^2}=1-\frac{3\left(x+1\right)-3}{\left(x+1\right)^2}\)

\(B=1-\frac{3}{x+1}+\frac{3}{\left(x+1\right)^2}\)

Đặt \(\frac{1}{x+1}=a\)\(\Rightarrow B=3a^2-3a+1=3\left(a^2-a+\frac{1}{3}\right)=3\left(a^2-2a.\frac{1}{2}+\frac{1}{4}+\frac{1}{12}\right)=3\left(a-\frac{1}{2}\right)^2+\frac{1}{4}\)

Vì \(\left(a-\frac{1}{2}\right)^2\ge0\Leftrightarrow B\ge\frac{1}{4}\)

Dấu "=" xảy ra khi \(a=\frac{1}{2}\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}\Leftrightarrow x+1=2\Leftrightarrow x=1\left(nhận\right)\)

Vậy GTNN của B là \(\frac{1}{4}\)khi \(x=1\)

b) đkxđ \(x-1\ne0\Leftrightarrow x\ne1\)\(E=\frac{3x^2-8x+6}{x^2-2x+1}=\frac{3\left(x^2-2x+1\right)-2x+3}{x^2-2x+1}=3-\frac{2x-3}{\left(x-1\right)^2}=3-\frac{2\left(x-1\right)-1}{\left(x-1\right)^2}\)

\(=3-\frac{2}{x-1}+\frac{1}{\left(x-1\right)^2}\)

Đặt \(\frac{1}{x-1}=b\)\(\Rightarrow E=b^2-2b+3=b^2-2b+1+2=\left(b-1\right)^2+2\)

Vì \(\left(b-1\right)^2\ge0\Leftrightarrow B\ge2\)

Dấu "=" xảy ra khi \(b-1=0\Leftrightarrow b=1\Leftrightarrow\frac{1}{x-1}=1\Leftrightarrow x-1=1\Leftrightarrow x=2\left(nhận\right)\)

Vậy GTNN của B là 2 khi x = 2

a) \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{9x^2-6x+1}\)

\(=-\dfrac{9x^2+3x+2x-6x^2}{\left(3x-1\right)\left(3x+1\right)}.\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)

\(=-\dfrac{x\left(3x+5\right)}{\left(3x-1\right)^2}.\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)

\(=\dfrac{-1}{2}\)

b) \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)

\(=\left(\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{3x-9-x^2}{3x\left(x+3\right)}\right)\)

\(=\dfrac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\dfrac{3x\left(x+3\right)}{-x^2+3x-9}\)

\(=\dfrac{x^2-3x+9}{x-3}.\dfrac{3}{-\left(x^2-3x+9\right)}\)

\(=-\dfrac{3}{x-3}\)