\(15-\left(3x-6\right)+\left(8+9x\right)=41\)

=\(15-3x+6+8+9x=41\)

=\(6x+29=41\)

=\(6x=12=>x=2\)

\(15-\left(3x-6\right)+\left(8+9x\right)=41\) 

      \(15-3x+6+8+9x=41\) 

                        \(-3x+9x=41-15-6-8\) 

                                   \(6x=12\) 

                                     \(x=12:6\) 

                                     \(x=2\)

\(A=\dfrac{3x^2+9x+17}{3x^2+9x+7}=1+\dfrac{10}{3x^2+9x+7}=1+\dfrac{10}{3\left(x^2+2.x.\dfrac{9}{2}+\dfrac{81}{4}\right)-\dfrac{215}{4}}\\ =1+\dfrac{10}{3\left(x+\dfrac{9}{2}\right)^2-\dfrac{215}{4}}\le\dfrac{35}{43}\)

Câu khác giải TT

a) x= - 15

b) x= - 2

c) x= -12 

d) x= -2 

1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)

\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)

\(=x^2+4x+3-\dfrac{12}{2x+3}\)

1: Sửa đề: 3x-5

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

2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

=5x^2+14x^2+12x+8

3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)

4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)

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

Bài 1:

a: \(\Leftrightarrow\left|3x-7\right|+\left|3x-15\right|=8\)

TH1: x<7/3

Pt sẽ là \(7-3x+15-3x=8\)

=>22-6x=8

=>6x=14

hay x=7/3(loại)

TH2: 7/3<=x<5

Pt sẽ là \(3x-7+15-3x=8\)

=>8=8(luôn đúng)

TH3: x>=5

Pt sẽ là 3x-7+3x-15=8

=>6x-22=8

hay x=5(nhận)

b: \(\Leftrightarrow\left|4x-98\right|+\left|4x-8\right|=90\)

TH1: x<2

Pt sẽ là 8-4x+98-4x=90

=>106-8x=90

=>x=2(loại)

TH2: 2<=x<49/2

Pt sẽ là 4x-8+98-4x=90

=>90=90(luôn đúng)

TH3: x>=49/2

Pt sẽ là 4x-8+4x-98=90

=>8x-106=90

=>8x=196

hay x=24,5(nhận)

Tìm x hả bn

a: \(\lim\limits_{x\rightarrow3}\dfrac{\sqrt{2x+10}-4}{3x-9}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{2x+10-16}{3x-9}\cdot\dfrac{1}{\sqrt{2x+10}+4}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{2\left(x-3\right)}{3\left(x-3\right)\cdot\left(\sqrt{2x+10}+4\right)}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{2}{3\left(\sqrt{2x+10}+4\right)}\)

\(=\dfrac{2}{3\cdot\sqrt{6+10}+3\cdot4}=\dfrac{2}{3\cdot4+3\cdot4}=\dfrac{2}{24}=\dfrac{1}{12}\)

b: \(\lim\limits_{x\rightarrow7}\dfrac{\sqrt{4x+8}-6}{x^2-9x+14}\)

\(=\lim\limits_{x\rightarrow7}\dfrac{4x+8-36}{\sqrt{4x+8}+6}\cdot\dfrac{1}{\left(x-2\right)\left(x-7\right)}\)

\(=\lim\limits_{x\rightarrow7}\dfrac{4x-28}{\left(\sqrt{4x+8}+6\right)\cdot\left(x-2\right)\left(x-7\right)}\)

\(=\lim\limits_{x\rightarrow7}\dfrac{4}{\left(\sqrt{4x+8}+6\right)\left(x-2\right)}\)

\(=\dfrac{4}{\left(\sqrt{4\cdot7+8}+6\right)\left(7-2\right)}\)

\(=\dfrac{4}{5\cdot12}=\dfrac{4}{60}=\dfrac{1}{15}\)

c: \(\lim\limits_{x\rightarrow5}\dfrac{x^2-8x+15}{2x^2-9x-5}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\left(x-3\right)\left(x-5\right)}{2x^2-10x+x-5}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{\left(x-3\right)\left(x-5\right)}{\left(x-5\right)\left(2x+1\right)}\)

\(=\lim\limits_{x\rightarrow5}\dfrac{x-3}{2x+1}=\dfrac{5-3}{2\cdot5+1}=\dfrac{2}{11}\)